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# Question of the week - 30 (A group of 4 persons is to be selected ...)

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Question of the week - 30 (A group of 4 persons is to be selected ...)  [#permalink]

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Updated on: 03 Apr 2019, 04:00
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33% (03:04) correct 67% (02:54) wrong based on 111 sessions

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Question of the Week #30

A group of 4 persons is to be selected from 5 married couples, who are sitting in five different rows, A, B, C, D, and E. Each row can accommodate exactly one couple. If a person is selected from a row, which is marked by a vowel, then we need to select another person of opposite gender, from another row, which is also a vowel. In how many ways can the group be formed?

A. 15
B. 16
C. 46
D. 75
E. 76

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Originally posted by EgmatQuantExpert on 04 Jan 2019, 04:25.
Last edited by EgmatQuantExpert on 03 Apr 2019, 04:00, edited 3 times in total.
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Re: Question of the week - 30 (A group of 4 persons is to be selected ...)  [#permalink]

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04 Jan 2019, 08:07
1
3
A group of 4 persons is to be selected from 5 married couples, who are sitting in five different rows, A, B, C, D, and E. Each row can accommodate exactly one couple. If a person is selected from a row, which is marked by a vowel, then we need to select another person of opposite gender, from another row, which is also a vowel. In how many ways can the group be formed?

A,E are the two rows with vowels..So 4 people belong to Vowel Rows

Case(1) if non of the vowel row couples are selected
remaining 3 couples i.e., 6 persons can be selected in 6C4 = 15

Case(2) if one person from Vowel row is selected, he can be selected in 4C1 ways and the other person is obvious(1 way)
example
A - Male => other person is E - Female
Other two persons from non vowel rows can be selected in 6C2 ways
Total = 4C1*6C2 = 4*15 = 60

Case(3) If all the persons are selected from Vowel rows they can be selected in 4C4 ways
Selecting 4 vowels = 1 way

Total number of ways = 15+60+1 = 76 ways

Option E is correct
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Re: Question of the week - 30 (A group of 4 persons is to be selected ...)  [#permalink]

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Updated on: 06 Jan 2019, 09:55
1
IMO
we can have three cases to solve this question:

case 1: all 4 persons from non vowel rows i.e 6 people
6c4: 15
case 2:
Given If a person is selected from a row, which is marked by a vowel, then we need to select another person of opposite gender, from another row, which is also a vowel.

so we have 2 males and 2 females ( row A & E)
we can say
2c1*2c1*6c2 = 60

case 3:

all people from A & E : 2c2*2c2 = 1

total ways to form group 60+15+1= 76
IMO E

EgmatQuantExpert wrote:
A group of 4 persons is to be selected from 5 married couples, who are sitting in five different rows, A, B, C, D, and E. Each row can accommodate exactly one couple. If a person is selected from a row, which is marked by a vowel, then we need to select another person of opposite gender, from another row, which is also a vowel. In how many ways can the group be formed?

A. 15
B. 16
C. 60
D. 75
E. 76

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Originally posted by Archit3110 on 06 Jan 2019, 09:50.
Last edited by Archit3110 on 06 Jan 2019, 09:55, edited 1 time in total.
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Re: Question of the week - 30 (A group of 4 persons is to be selected ...)  [#permalink]

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06 Jan 2019, 09:54
Dare Devil wrote:
A group of 4 persons is to be selected from 5 married couples, who are sitting in five different rows, A, B, C, D, and E. Each row can accommodate exactly one couple. If a person is selected from a row, which is marked by a vowel, then we need to select another person of opposite gender, from another row, which is also a vowel. In how many ways can the group be formed?

A,E are the two rows with vowels..So 4 people belong to Vowel Rows

Case(1) if non of the vowel row couples are selected
remaining 3 couples i.e., 6 persons can be selected in 6C4 = 15

Case(2) if one person from Vowel row is selected, he can be selected in 4C1 ways and the other person is obvious(1 way)
example
A - Male => other person is E - Female
Other two persons from non vowel rows can be selected in 6C2 ways
Total = 4C1*6C2 = 4*15 = 60

Case(3) If all the persons are selected from Vowel rows they can be selected in 4C4 ways
Selecting 4 vowels = 1 way

Total number of ways = 15+60+1 = 76 ways

Option E is correct

Dare Devil
daredevil
Other two persons from non vowel rows can be selected in 6C2 ways
Total = 4C1*6C2 = 4*15 = 60

total are 4 people to be selected so the solution should be 2c1*2c1*6c2 = 60 ... 2 males & 2 females in A & E can only 1 from each can be chosen..

also question specifically says Given If a person is selected from a row, which is marked by a vowel, then we need to select another person of opposite gender, from another row, which is also a vowel.

where as in your case 3
If all the persons are selected from Vowel rows they can be selected in 4C4 ways
Selecting 4 vowels = 1 way

it would be 2c2*2c2= 1

total then would be 76..
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Re: Question of the week - 30 (A group of 4 persons is to be selected ...)  [#permalink]

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Updated on: 10 Jan 2019, 02:39
Archit3110 wrote:
Dare Devil wrote:
A group of 4 persons is to be selected from 5 married couples, who are sitting in five different rows, A, B, C, D, and E. Each row can accommodate exactly one couple. If a person is selected from a row, which is marked by a vowel, then we need to select another person of opposite gender, from another row, which is also a vowel. In how many ways can the group be formed?

A,E are the two rows with vowels..So 4 people belong to Vowel Rows

Case(1) if non of the vowel row couples are selected
remaining 3 couples i.e., 6 persons can be selected in 6C4 = 15

Case(2) if one person from Vowel row is selected, he can be selected in 4C1 ways and the other person is obvious(1 way)
example
A - Male => other person is E - Female
Other two persons from non vowel rows can be selected in 6C2 ways
Total = 4C1*6C2 = 4*15 = 60

Case(3) If all the persons are selected from Vowel rows they can be selected in 4C4 ways
Selecting 4 vowels = 1 way

Total number of ways = 15+60+1 = 76 ways

Option E is correct

Dare Devil
daredevil
Other two persons from non vowel rows can be selected in 6C2 ways
Total = 4C1*6C2 = 4*15 = 60

total are 4 people to be selected so the solution should be 2c1*2c1*6c2 = 60 ... 2 males & 2 females in A & E can only 1 from each can be chosen..

also question specifically says Given If a person is selected from a row, which is marked by a vowel, then we need to select another person of opposite gender, from another row, which is also a vowel.

where as in your case 3
If all the persons are selected from Vowel rows they can be selected in 4C4 ways
Selecting 4 vowels = 1 way

it would be 2c2*2c2= 1

total then would be 76..

In vowel rows there are 4 people, I may randomly select any person, so the first person can be selected in 4C1 ways
Next person you can't choose because, that person should be from another vowel row and of opposite gender

I am approaching it differently from you that is, you have first selected the two rows(A/E) in two ways, then (Male/Female in two ways). But I am considering each of them as an individual A-Male, A-Female, E-Male and E-Female...both of Us are right, but the approach is different

Posted from my mobile device

Originally posted by Dare Devil on 06 Jan 2019, 10:27.
Last edited by Dare Devil on 10 Jan 2019, 02:39, edited 1 time in total.
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Question of the week - 30 (A group of 4 persons is to be selected ...)  [#permalink]

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Updated on: 03 Apr 2019, 04:03

Solution

Given:
• There are 5 married couples, and each couple is sitting in a different row among A, B, C, D and E
• A group of 4 is to be selected from these 5 married couples
• If a person is selected from a row, which is a vowel, then we need to select another person of opposite gender from another row, which is also a vowel

To find:
• The number of ways, in which a group of 4 can be selected

Approach and Working:
Case 1: When no person is selected from A or E
• In this case, all four persons are selected from B, C and D
o Total number of ways = $$^6C_4 = 15$$ ways

Case 2: When one person is selected from either A or E
• The selected person of one gender can be either from A or E, so 2 ways
o Then the second person of the opposite gender can be selected from the other row in only one way.

• The remaining 2 must be selected from rows, B, C and D
o This can be done in $$^6C_2$$ ways.

• Thus, the total number of ways = $$2 * 1 * ^6C_2 = 60$$ ways[/list]

Case 3: When two persons are selected from A and E
• In this case all the four must be chosen from A and E, according to the given condition.
• So, there is only one way in which we can select four persons from A and E.

Therefore, the total number of ways = 15 + 30 + 1= 46

Hence, the correct answer is option C.

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Originally posted by EgmatQuantExpert on 09 Jan 2019, 02:35.
Last edited by EgmatQuantExpert on 03 Apr 2019, 04:03, edited 1 time in total.
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Re: Question of the week - 30 (A group of 4 persons is to be selected ...)  [#permalink]

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09 Jan 2019, 19:12
Dare Devil wrote:
A group of 4 persons is to be selected from 5 married couples, who are sitting in five different rows, A, B, C, D, and E. Each row can accommodate exactly one couple. If a person is selected from a row, which is marked by a vowel, then we need to select another person of opposite gender, from another row, which is also a vowel. In how many ways can the group be formed?

A,E are the two rows with vowels..So 4 people belong to Vowel Rows

Case(1) if non of the vowel row couples are selected
remaining 3 couples i.e., 6 persons can be selected in 6C4 = 15

Case(2) if one person from Vowel row is selected, he can be selected in 4C1 ways and the other person is obvious(1 way)
example
A - Male => other person is E - Female
Other two persons from non vowel rows can be selected in 6C2 ways
Total = 4C1*6C2 = 4*15 = 60

Case(3) If all the persons are selected from Vowel rows they can be selected in 4C4 ways
Selecting 4 vowels = 1 way

Total number of ways = 15+60+1 = 76 ways

Option E is correct

Let me take take case 2 :Since we take one person from one vowel row, it is said we have 4C1 ways to do so.Lets consider the 4 cases :
(1)AM-EF
(2)AF-EM
(3)EM-AF
(4)EF-AM
but isn't (1) and (4) the same?since it is selection?Similarly (2) and(3) the same?Can u please clarify?
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Re: Question of the week - 30 (A group of 4 persons is to be selected ...)  [#permalink]

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10 Jan 2019, 02:33
1
vishnu90 wrote:
Dare Devil wrote:
A group of 4 persons is to be selected from 5 married couples, who are sitting in five different rows, A, B, C, D, and E. Each row can accommodate exactly one couple. If a person is selected from a row, which is marked by a vowel, then we need to select another person of opposite gender, from another row, which is also a vowel. In how many ways can the group be formed?

A,E are the two rows with vowels..So 4 people belong to Vowel Rows

Case(1) if non of the vowel row couples are selected
remaining 3 couples i.e., 6 persons can be selected in 6C4 = 15

Case(2) if one person from Vowel row is selected, he can be selected in 4C1 ways and the other person is obvious(1 way)
example
A - Male => other person is E - Female
Other two persons from non vowel rows can be selected in 6C2 ways
Total = 4C1*6C2 = 4*15 = 60

Case(3) If all the persons are selected from Vowel rows they can be selected in 4C4 ways
Selecting 4 vowels = 1 way

Total number of ways = 15+60+1 = 76 ways

Option E is correct

Let me take take case 2 :Since we take one person from one vowel row, it is said we have 4C1 ways to do so.Lets consider the 4 cases :
(1)AM-EF
(2)AF-EM
(3)EM-AF
(4)EF-AM
but isn't (1) and (4) the same?since it is selection?Similarly (2) and(3) the same?Can u please clarify?

In Case (2) we have to select 2 persons among 4 vowel row persons, and 4 among 6 non-vowels
First person from 4 vowel row persons can be any of AM,AF,EM,EF
So I can Select him in 4 ways

Do you have a choice to select the second person no, since it is given "select another person of opposite gender, from another row, which is also a vowel"
Suppose we choose AM, second will be EF

I am not considering the selection of a row, I am considering each of them as an individual

(1)AM-EF
(2)AF-EM
(3)EM-AF
(4)EF-AM

But in your case you have to choose the row in two ways (A OR E)
(1)AM-EF
(4)EF-AM
----------Which makes them two different selections
Then select M/F again in two ways
Then, (two ways row selection)*(two ways male/Female selection) = 4
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Re: Question of the week - 30 (A group of 4 persons is to be selected ...)  [#permalink]

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02 Mar 2019, 18:04
EgmatQuantExpert wrote:

Solution

Given:
• There are 5 married couples, and each couple is sitting in a different row among A, B, C, D and E
• A group of 4 is to be selected from these 5 married couples
• If a person is selected from a row, which is a vowel, then we need to select another person of opposite gender from another row, which is also a vowel

To find:
• The number of ways, in which a group of 4 can be selected

Approach and Working:
Case 1: When no person is selected from A or E
• In this case, all four persons are selected from B, C and D
o Total number of ways = $$^6C_4 = 15$$ ways

Case 2: When one person is selected from either A or E
• The selected person can be either from A or E, so 2 ways
• And, that person can be either male or female = 2 ways
o Then the second person can be selected from the other row in one way

• The remaining 2 must be selected from rows, B, C and D
o This can be done in $$^6C_2$$ ways.

• Thus, the total number of ways = $$2 * 2 * 1 * ^6C_2 = 60$$ ways

Case 3: When two persons are selected from A and E
• In this case all the four must be chosen from A and E, according to the given condition.
• So, there is only one way in which we can select four persons from A and E.

Therefore, the total number of ways = 15 + 60 + 1= 76

Hence, the correct answer is option E.

Hi Payal,
How are we ruling out possibility of selecting first 3 from non-vowels & last one from vowels.
Let N be non vowel, V be vowel

N N N V

No. of ways to select first three - 6C3 = 20
No. of ways to select last one - = 4
no. of ways to select as per NNNV = 20 x 4 = 80

Total no. of ways should be 80 + 76 = 156

Pls correct me if wrong.
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Re: Question of the week - 30 (A group of 4 persons is to be selected ...)  [#permalink]

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19 Mar 2019, 13:39
EgmatQuantExpert wrote:

Solution

Given:
• There are 5 married couples, and each couple is sitting in a different row among A, B, C, D and E
• A group of 4 is to be selected from these 5 married couples
• If a person is selected from a row, which is a vowel, then we need to select another person of opposite gender from another row, which is also a vowel

To find:
• The number of ways, in which a group of 4 can be selected

Approach and Working:
Case 1: When no person is selected from A or E
• In this case, all four persons are selected from B, C and D
o Total number of ways = $$^6C_4 = 15$$ ways

Case 2: When one person is selected from either A or E
• The selected person can be either from A or E, so 2 ways
• And, that person can be either male or female = 2 ways
o Then the second person can be selected from the other row in one way

• The remaining 2 must be selected from rows, B, C and D
o This can be done in $$^6C_2$$ ways.

• Thus, the total number of ways = $$2 * 2 * 1 * ^6C_2 = 60$$ ways

Case 3: When two persons are selected from A and E
• In this case all the four must be chosen from A and E, according to the given condition.
• So, there is only one way in which we can select four persons from A and E.

Therefore, the total number of ways = 15 + 60 + 1= 76

Hence, the correct answer is option E.

Ma'am,

in case II,

since this is a selection problem, how does the order matters?? in can be in one of 2 ways (A Male-E female)and (A female-E male).....so 2*6C2=30 ways.... can you clarify?

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Re: Question of the week - 30 (A group of 4 persons is to be selected ...)  [#permalink]

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19 Mar 2019, 21:40
1
AKY13 wrote:
EgmatQuantExpert wrote:

Solution

Given:
• There are 5 married couples, and each couple is sitting in a different row among A, B, C, D and E
• A group of 4 is to be selected from these 5 married couples
• If a person is selected from a row, which is a vowel, then we need to select another person of opposite gender from another row, which is also a vowel

To find:
• The number of ways, in which a group of 4 can be selected

Approach and Working:
Case 1: When no person is selected from A or E
• In this case, all four persons are selected from B, C and D
o Total number of ways = $$^6C_4 = 15$$ ways

Case 2: When one person is selected from either A or E
• The selected person can be either from A or E, so 2 ways
• And, that person can be either male or female = 2 ways
o Then the second person can be selected from the other row in one way

• The remaining 2 must be selected from rows, B, C and D
o This can be done in $$^6C_2$$ ways.

• Thus, the total number of ways = $$2 * 2 * 1 * ^6C_2 = 60$$ ways

Case 3: When two persons are selected from A and E
• In this case all the four must be chosen from A and E, according to the given condition.
• So, there is only one way in which we can select four persons from A and E.

Therefore, the total number of ways = 15 + 60 + 1= 76

Hence, the correct answer is option E.

Hi Payal,
How are we ruling out possibility of selecting first 3 from non-vowels & last one from vowels.
Let N be non vowel, V be vowel

N N N V

No. of ways to select first three - 6C3 = 20
No. of ways to select last one - = 4
no. of ways to select as per NNNV = 20 x 4 = 80

Total no. of ways should be 80 + 76 = 156

Pls correct me if wrong.

The notion to take care of ALL the possible scenarios is paramount as well as correct.
However, in this case, when you have selected 3 !V and 1 V, you are going against the said-rule:
If a person is selected from a row, which is marked by a vowel, then we need to select another person of opposite gender, from another row, which is also a vowel.

Your mind has decided to select 1V at the end in the hope to cover all the possible cases, resulting in the extra count and going against the rule.
Because whether you select 1V at the beginning or at the end, the crux of the matter is 1V is already included.
Hence, the other V (by default) should be included.
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Re: Question of the week - 30 (A group of 4 persons is to be selected ...)  [#permalink]

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20 Mar 2019, 03:40
Even if you can select the vowel members of the group in 4 different ways in Case 2, essentially these 4 different ways would form only 2 different groups. So the answer should be 46 not 76.

If indeed 4 should be considered instead of 2 in case 2 as the total possible different ways to select from vowel members, then it should be 30 different ways to select from nonvowel members not 15. This would though turn the question into a totally different question, that is a question of permutation not combinations. The correct answer would be 484 in this case.
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Re: Question of the week - 30 (A group of 4 persons is to be selected ...)  [#permalink]

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20 Mar 2019, 04:21
avikroy wrote:
EgmatQuantExpert wrote:

Solution

Given:
• There are 5 married couples, and each couple is sitting in a different row among A, B, C, D and E
• A group of 4 is to be selected from these 5 married couples
• If a person is selected from a row, which is a vowel, then we need to select another person of opposite gender from another row, which is also a vowel

To find:
• The number of ways, in which a group of 4 can be selected

Approach and Working:
Case 1: When no person is selected from A or E
• In this case, all four persons are selected from B, C and D
o Total number of ways = $$^6C_4 = 15$$ ways

Case 2: When one person is selected from either A or E
• The selected person can be either from A or E, so 2 ways
• And, that person can be either male or female = 2 ways
o Then the second person can be selected from the other row in one way

• The remaining 2 must be selected from rows, B, C and D
o This can be done in $$^6C_2$$ ways.

• Thus, the total number of ways = $$2 * 2 * 1 * ^6C_2 = 60$$ ways

Case 3: When two persons are selected from A and E
• In this case all the four must be chosen from A and E, according to the given condition.
• So, there is only one way in which we can select four persons from A and E.

Therefore, the total number of ways = 15 + 60 + 1= 76

Hence, the correct answer is option E.

Ma'am,

in case II,

since this is a selection problem, how does the order matters?? in can be in one of 2 ways (A Male-E female)and (A female-E male).....so 2*6C2=30 ways.... can you clarify?

I think you'r right , the total should be 46
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Re: Question of the week - 30 (A group of 4 persons is to be selected ...)  [#permalink]

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20 Mar 2019, 04:28
Dare Devil wrote:
vishnu90 wrote:
Dare Devil wrote:
A group of 4 persons is to be selected from 5 married couples, who are sitting in five different rows, A, B, C, D, and E. Each row can accommodate exactly one couple. If a person is selected from a row, which is marked by a vowel, then we need to select another person of opposite gender, from another row, which is also a vowel. In how many ways can the group be formed?

A,E are the two rows with vowels..So 4 people belong to Vowel Rows

Case(1) if non of the vowel row couples are selected
remaining 3 couples i.e., 6 persons can be selected in 6C4 = 15

Case(2) if one person from Vowel row is selected, he can be selected in 4C1 ways and the other person is obvious(1 way)
example
A - Male => other person is E - Female
Other two persons from non vowel rows can be selected in 6C2 ways
Total = 4C1*6C2 = 4*15 = 60

Case(3) If all the persons are selected from Vowel rows they can be selected in 4C4 ways
Selecting 4 vowels = 1 way

Total number of ways = 15+60+1 = 76 ways

Option E is correct

Let me take take case 2 :Since we take one person from one vowel row, it is said we have 4C1 ways to do so.Lets consider the 4 cases :
(1)AM-EF
(2)AF-EM
(3)EM-AF
(4)EF-AM
but isn't (1) and (4) the same?since it is selection?Similarly (2) and(3) the same?Can u please clarify?

In Case (2) we have to select 2 persons among 4 vowel row persons, and 4 among 6 non-vowels
First person from 4 vowel row persons can be any of AM,AF,EM,EF
So I can Select him in 4 ways

Do you have a choice to select the second person no, since it is given "select another person of opposite gender, from another row, which is also a vowel"
Suppose we choose AM, second will be EF

I am not considering the selection of a row, I am considering each of them as an individual

(1)AM-EF
(2)AF-EM
(3)EM-AF
(4)EF-AM

But in your case you have to choose the row in two ways (A OR E)
(1)AM-EF
(4)EF-AM
----------Which makes them two different selections
Then select M/F again in two ways
Then, (two ways row selection)*(two ways male/Female selection) = 4

, why you have to choose the row again since the result is the same ? we are counting the number of ways the group can be made ,so order doesn't matter ,
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Re: Question of the week - 30 (A group of 4 persons is to be selected ...)  [#permalink]

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23 Mar 2019, 23:48
Dare Devil wrote:
vishnu90 wrote:
Dare Devil wrote:
A group of 4 persons is to be selected from 5 married couples, who are sitting in five different rows, A, B, C, D, and E. Each row can accommodate exactly one couple. If a person is selected from a row, which is marked by a vowel, then we need to select another person of opposite gender, from another row, which is also a vowel. In how many ways can the group be formed?

A,E are the two rows with vowels..So 4 people belong to Vowel Rows

Case(1) if non of the vowel row couples are selected
remaining 3 couples i.e., 6 persons can be selected in 6C4 = 15

Case(2) if one person from Vowel row is selected, he can be selected in 4C1 ways and the other person is obvious(1 way)
example
A - Male => other person is E - Female
Other two persons from non vowel rows can be selected in 6C2 ways
Total = 4C1*6C2 = 4*15 = 60

Case(3) If all the persons are selected from Vowel rows they can be selected in 4C4 ways
Selecting 4 vowels = 1 way

Total number of ways = 15+60+1 = 76 ways

Option E is correct

Let me take take case 2 :Since we take one person from one vowel row, it is said we have 4C1 ways to do so.Lets consider the 4 cases :
(1)AM-EF
(2)AF-EM
(3)EM-AF
(4)EF-AM
but isn't (1) and (4) the same?since it is selection?Similarly (2) and(3) the same?Can u please clarify?

In Case (2) we have to select 2 persons among 4 vowel row persons, and 4 among 6 non-vowels
First person from 4 vowel row persons can be any of AM,AF,EM,EF
So I can Select him in 4 ways

Do you have a choice to select the second person no, since it is given "select another person of opposite gender, from another row, which is also a vowel"
Suppose we choose AM, second will be EF

I am not considering the selection of a row, I am considering each of them as an individual

(1)AM-EF
(2)AF-EM
(3)EM-AF
(4)EF-AM

But in your case you have to choose the row in two ways (A OR E)
(1)AM-EF
(4)EF-AM
----------Which makes them two different selections
Then select M/F again in two ways
Then, (two ways row selection)*(two ways male/Female selection) = 4

Hi
I am not clear why is AM-EF different from EF-AM? No matter which one you select, the team will be represented by same people. Had arrangement been important, we would have had 15*4! ways of picking 4 people from BCD,(considering that arrangement is important). Also can someone please explain how there are 4 ways of selecting from rows with vowels?

If I select a male from A, I get female from E ( AM, EF)
If I select female from A, I get male from E ( AF,EM)

having say ( BM,CF,AM,EF) will not be different from having (BM,CF,EF,AM)

If arrangement is not important then it should be: 6C4+6C2*2+ 1 = 46

If arrangement is important then then answer will be completely different.

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Re: Question of the week - 30 (A group of 4 persons is to be selected ...)  [#permalink]

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24 Mar 2019, 05:44
Xylan wrote:
AKY13 wrote:
EgmatQuantExpert wrote:

Solution

Given:
• There are 5 married couples, and each couple is sitting in a different row among A, B, C, D and E
• A group of 4 is to be selected from these 5 married couples
• If a person is selected from a row, which is a vowel, then we need to select another person of opposite gender from another row, which is also a vowel

To find:
• The number of ways, in which a group of 4 can be selected

Approach and Working:
Case 1: When no person is selected from A or E
• In this case, all four persons are selected from B, C and D
o Total number of ways = $$^6C_4 = 15$$ ways

Case 2: When one person is selected from either A or E
• The selected person can be either from A or E, so 2 ways
• And, that person can be either male or female = 2 ways
o Then the second person can be selected from the other row in one way

• The remaining 2 must be selected from rows, B, C and D
o This can be done in $$^6C_2$$ ways.

• Thus, the total number of ways = $$2 * 2 * 1 * ^6C_2 = 60$$ ways

Case 3: When two persons are selected from A and E
• In this case all the four must be chosen from A and E, according to the given condition.
• So, there is only one way in which we can select four persons from A and E.

Therefore, the total number of ways = 15 + 60 + 1= 76

Hence, the correct answer is option E.

Hi Payal,
How are we ruling out possibility of selecting first 3 from non-vowels & last one from vowels.
Let N be non vowel, V be vowel

N N N V

No. of ways to select first three - 6C3 = 20
No. of ways to select last one - = 4
no. of ways to select as per NNNV = 20 x 4 = 80

Total no. of ways should be 80 + 76 = 156

Pls correct me if wrong.

The notion to take care of ALL the possible scenarios is paramount as well as correct.
However, in this case, when you have selected 3 !V and 1 V, you are going against the said-rule:
If a person is selected from a row, which is marked by a vowel, then we need to select another person of opposite gender, from another row, which is also a vowel.

Your mind has decided to select 1V at the end in the hope to cover all the possible cases, resulting in the extra count and going against the rule.
Because whether you select 1V at the beginning or at the end, the crux of the matter is 1V is already included.
Hence, the other V (by default) should be included.

hi Xylan
The rule that opposite gender has to be selected is not applicable for non-vowels.
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Re: Question of the week - 30 (A group of 4 persons is to be selected ...)  [#permalink]

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24 Mar 2019, 07:16
AKY13 wrote:
Xylan wrote:
EgmatQuantExpert wrote:

Solution

Given:
• There are 5 married couples, and each couple is sitting in a different row among A, B, C, D and E
• A group of 4 is to be selected from these 5 married couples
• If a person is selected from a row, which is a vowel, then we need to select another person of opposite gender from another row, which is also a vowel

To find:
• The number of ways, in which a group of 4 can be selected

Approach and Working:
Case 1: When no person is selected from A or E
• In this case, all four persons are selected from B, C and D
o Total number of ways = $$^6C_4 = 15$$ ways

Case 2: When one person is selected from either A or E
• The selected person can be either from A or E, so 2 ways
• And, that person can be either male or female = 2 ways
o Then the second person can be selected from the other row in one way

• The remaining 2 must be selected from rows, B, C and D
o This can be done in $$^6C_2$$ ways.

• Thus, the total number of ways = $$2 * 2 * 1 * ^6C_2 = 60$$ ways

Case 3: When two persons are selected from A and E
• In this case all the four must be chosen from A and E, according to the given condition.
• So, there is only one way in which we can select four persons from A and E.

Therefore, the total number of ways = 15 + 60 + 1= 76

Hence, the correct answer is option E.

The notion to take care of ALL the possible scenarios is paramount as well as correct.
However, in this case, when you have selected 3 !V and 1 V, you are going against the said-rule:
If a person is selected from a row, which is marked by a vowel, then we need to select another person of opposite gender, from another row, which is also a vowel.

Your mind has decided to select 1V at the end in the hope to cover all the possible cases, resulting in the extra count and going against the rule.
Because whether you select 1V at the beginning or at the end, the crux of the matter is 1V is already included.
Hence, the other V (by default) should be included.

hi Xylan
The rule that opposite gender has to be selected is not applicable for non-vowels.

I hope to do justice to your confusion with my explanation.

NO where in the resolution we are denying the fact that:
1) the rule is applicable ONLY for vowels.
2) the rule need not be applicable for non-vowels.

The Conditional-statement:
If a person is selected from a row, which is a vowel, then we need to select another person of the opposite gender from another row, which is also a vowel
The gist of the above conditional statement: IF X --------> Y
If V is selected --------------> Another V MUST be selected

Let's break down all the possible cases:
Case 1: When no person is selected from A or E
• In this case, all four persons are selected from B, C and D
By your convention: All 4 selections from non-vowels(N):
NNNN

Case 2: When one person is selected from either A or E
• The selected person can be either from A or E
• The remaining 2 must be selected from rows, B, C and D
Implication of the above condition on this case:
If V is selected --------------> Another V MUST be selected
NNV ----corrsponds to----> NNVV
NNVV

Case 3: When all 4 are selected from A and E (vowels)
• In this case all the four must be chosen from A and E, according to the given condition.
VVVV

The case of NNNV is NOT at all possible because once you have included V in your selection-scenario,
you are activating the if-condition of the conditional statement, resulting in the then-statement of the conditional statement to bound to be TRUE.

Implication of the above condition on this case:
If V is selected --------------> Another V MUST be selected
NNNV ----corrsponds to----> NNNVV

Since the question statement asks us to select 4 persons, the above case is INVALID.
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Re: Question of the week - 30 (A group of 4 persons is to be selected ...)  [#permalink]

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28 Mar 2019, 18:32
The notion to take care of ALL the possible scenarios is paramount as well as correct.
However, in this case, when you have selected 3 !V and 1 V, you are going against the said-rule:
If a person is selected from a row, which is marked by a vowel, then we need to select another person of opposite gender, from another row, which is also a vowel.

Your mind has decided to select 1V at the end in the hope to cover all the possible cases, resulting in the extra count and going against the rule.
Because whether you select 1V at the beginning or at the end, the crux of the matter is 1V is already included.
Hence, the other V (by default) should be included.[/quote]

hi Xylan
The rule that opposite gender has to be selected is not applicable for non-vowels.[/quote]

I hope to do justice to your confusion with my explanation.

NO where in the resolution we are denying the fact that:
1) the rule is applicable ONLY for vowels.
2) the rule need not be applicable for non-vowels.

The Conditional-statement:
If a person is selected from a row, which is a vowel, then we need to select another person of the opposite gender from another row, which is also a vowel
The gist of the above conditional statement: IF X --------> Y
If V is selected --------------> Another V MUST be selected

Let's break down all the possible cases:
Case 1: When no person is selected from A or E
• In this case, all four persons are selected from B, C and D
By your convention: All 4 selections from non-vowels(N):
NNNN

Case 2: When one person is selected from either A or E
• The selected person can be either from A or E
• The remaining 2 must be selected from rows, B, C and D
Implication of the above condition on this case:
If V is selected --------------> Another V MUST be selected
NNV ----corrsponds to----> NNVV
NNVV

Case 3: When all 4 are selected from A and E (vowels)
• In this case all the four must be chosen from A and E, according to the given condition.
VVVV

The case of NNNV is NOT at all possible because once you have included V in your selection-scenario,
you are activating the if-condition of the conditional statement, resulting in the then-statement of the conditional statement to bound to be TRUE.

Implication of the above condition on this case:
If V is selected --------------> Another V MUST be selected
NNNV ----corrsponds to----> NNNVV

Since the question statement asks us to select 4 persons, the above case is INVALID.[/quote]

Hi Xylan
Thanks for pointing out a perspective overlooked by me.
However, I am not sure whether it can be assumed that we can't have last selection of vowel.
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Re: Question of the week - 30 (A group of 4 persons is to be selected ...)  [#permalink]

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03 Apr 2019, 04:07
Re: Question of the week - 30 (A group of 4 persons is to be selected ...)   [#permalink] 03 Apr 2019, 04:07
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