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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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04 Jan 2019, 03:25
00:00

Difficulty:

95% (hard)

Question Stats:

24% (01:49) correct 76% (02:04) wrong based on 54 sessions

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

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04 Jan 2019, 07:07
1
1
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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Question of the week - 30 (A group of 4 persons is to be selected ...)  [#permalink]

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Updated on: 06 Jan 2019, 08:55
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, 08:50.
Last edited by Archit3110 on 06 Jan 2019, 08:55, 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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06 Jan 2019, 08:54
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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Question of the week - 30 (A group of 4 persons is to be selected ...)  [#permalink]

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Updated on: 10 Jan 2019, 01:39
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, 09:27.
Last edited by Dare Devil on 10 Jan 2019, 01:39, 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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09 Jan 2019, 01:35

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.

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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, 18:12
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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Question of the week - 30 (A group of 4 persons is to be selected ...)  [#permalink]

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10 Jan 2019, 01:33
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
Question of the week - 30 (A group of 4 persons is to be selected ...) &nbs [#permalink] 10 Jan 2019, 01:33
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