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A dance delegation of 4 people must be chosen from 5 pairs of dance pa
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08 Apr 2016, 02:53

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A dance delegation of 4 people must be chosen from 5 pairs of dance partners. If 2 dance partners can never be together on the delegation, how many different ways are there to form the delegation?

Re: A dance delegation of 4 people must be chosen from 5 pairs of dance pa
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08 Apr 2016, 22:56

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FightToSurvive wrote:

Vyshak wrote:

FightToSurvive wrote:

A dance delegation of 4 people must be chosen from 5 pairs of dance partners. If 2 dance partners can never be together on the delegation, how many different ways are there to form the delegation?

A. 20 B. 30 C. 60 D. 70 E. 80

Bunnuel, from what I understood from the question stem is that 2 people who are dance partners cannot be in the delegation team of 4.

In total we have 10 people ( 5 pairs) out of which 2 dance partners cannot be on the team.

Hence to select a team of 4 people from 8. = 8C4 = 70. D IMO.

The method you have adopted is incorrect. You are just removing 2 members from the 10 members and then selecting a committee of 4 members. Out of 8 members, there is still a possibility of selecting members from the same pair.

Hope it helps.

I know that... I just thought that the question asked to remove any 2 people who are dance partners..

Anyways thanks for clarifying.

Could you please explain your approach to the solution?

Posted from my mobile device

Question: 2 dance partners can never be together on the delegation - It means we can select only one dancer from any given pair.

We have a total of 5 pairs of members. We have to select a delegation of 4 members. This can be done by choosing 4 members such that only one member will be picked up from one pair.

Out of 5 pairs, we will select 4 pairs. --> 5C4 ways

But each pair will have 2 dancers and hence there will be 2 ways of picking a member from each pair. So we have 4 pairs and we will get 2*2*2*2 = 16 ways of picking the members from the selected 4 pairs.

Hence, total number of ways = 5C4 * 2 * 2 * 2 * 2 = 80

Re: A dance delegation of 4 people must be chosen from 5 pairs of dance pa
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09 Apr 2016, 00:02

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i still have doubt 5 pair = 10 people so out of 10 we need to choose 4 hence total combinations are 10c4 which is 210 now out 5 pair choose any one pair at a time 5c1 * (out of 8 remaining peope that that is 4 pair choose any 2 people ) it will be 5c1 * 8c2 which is 140 now these many combinations when 1pair must be on the list, so subtract it 210 -140 = 70 please clarify Bunuel, i asked same question in one more post.

A dance delegation of 4 people must be chosen from 5 pairs of dance pa
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09 Apr 2016, 00:08

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sudhirmadaan wrote:

i still have doubt 5 pair = 10 people so out of 10 we need to choose 4 hence total combinations are 10c4 which is 210 now out 5 pair choose any one pair at a time 5c1 * (out of 8 remaining peope that that is 4 pair choose any 2 people ) it will be 5c1 * 8c2 which is 140 now these many combinations when 1pair must be on the list, so subtract it 210 -140 = 70 please clarify Bunuel, i asked same question in one more post.

5C1 * 8C2 will enable you to select only 1 pair and 2 members. Also, you are picking 1 complete pair from 5 pairs and the 2 members are picked from 8 members. The members can still be picked from the same pair.

Re: A dance delegation of 4 people must be chosen from 5 pairs of dance pa
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18 Apr 2016, 20:17

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onomyst wrote:

Can someone tell me what's wrong with my approach?

Let there be pairs from A-E and 1,2 be the different people in the couple: A1, A2, B1, B2, C1, C2, D1, D2, E1,E2

Now, there are 10 ways to select the first person (let's say A1).Then there are 8 people left to choose from (excluding A1 and A2). For the next person there are 6 options, and the next 4.

So 10*8*6*4

Hi, your method has repetitions in it ... say when you are choosing 1 out of 10, you chose A1 and then B1 and then C1 and then D1.. but then in next arrangement, you chose B1 initially and then A1, then B1 and then D1.. here in your method the two choosen are different, although they are the SAME.. _________________

Re: A dance delegation of 4 people must be chosen from 5 pairs
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12 Oct 2018, 11:31

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Top Contributor

Abhi077 wrote:

A dance delegation of 4 people must be chosen from 5 pairs of dance partners. If 2 dance partners can never be together on the delegation, how many different ways are there to form the delegation? A)20 B) 30 C) 60 D) 70 E) 80

Take the task of creating the dance delegation and break it into stages.

Stage 1: Select 4 pairs of PARTNERS Since the order in which we select the PAIRS does not matter, we can use combinations. We can select 4 pairs from 5 pairs in 5C4 ways (= 5 ways) So, we can complete stage 1 in 5 ways

ASIDE: Now that we've selected 4 PAIRS, we'll select 1 person from each PAIR. This will ensure that we meet the condition that 2 dance partners can never be together on the delegation

Stage 2: From one chosen PAIR of dancers, select 1 person to be in the delegation There are 2 dancers in a PAIR, so we can complete stage 2 in 2 ways

Stage 3: From another PAIR of chosen dancers, select 1 person to be in the delegation There are 2 dancers in a PAIR, so we can complete stage 3 in 2 ways

Stage 4: From another PAIR of chosen dancers, select 1 person to be in the delegation There are 2 dancers in a PAIR, so we can complete stage 4 in 2 ways

Stage 5: From last remaining PAIR of chosen dancers, select 1 person to be in the delegation There are 2 dancers in a PAIR, so we can complete stage 5 in 2 ways

By the Fundamental Counting Principle (FCP), we can complete all 5 stages (and thus create the delegation) in (5)(2)(2)(2)(2) ways (= 80 ways)

Answer: E

Note: the FCP can be used to solve the MAJORITY of counting questions on the GMAT. So, be sure to learn it.

Re: A dance delegation of 4 people must be chosen from 5 pairs of dance pa
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08 Apr 2016, 19:51

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Since 2 dance partners can never be together, lets select 4 pairs from 5 pairs of dance partners to fill the 4 positions. Now, in each position, each pair can be selected in 2 different ways.

Therefore number of ways of forming a delegation such that the 2 dance partners are never together = 5C4*(2C1)^4 = 5 * 16 = 80

Re: A dance delegation of 4 people must be chosen from 5 pairs of dance pa
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08 Apr 2016, 21:38

A dance delegation of 4 people must be chosen from 5 pairs of dance partners. If 2 dance partners can never be together on the delegation, how many different ways are there to form the delegation?

A. 20 B. 30 C. 60 D. 70 E. 80

Bunnuel, from what I understood from the question stem is that 2 people who are dance partners cannot be in the delegation team of 4.

In total we have 10 people ( 5 pairs) out of which 2 dance partners cannot be on the team.

Hence to select a team of 4 people from 8. = 8C4 = 70. D IMO.

Re: A dance delegation of 4 people must be chosen from 5 pairs of dance pa
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08 Apr 2016, 21:46

FightToSurvive wrote:

A dance delegation of 4 people must be chosen from 5 pairs of dance partners. If 2 dance partners can never be together on the delegation, how many different ways are there to form the delegation?

A. 20 B. 30 C. 60 D. 70 E. 80

Bunnuel, from what I understood from the question stem is that 2 people who are dance partners cannot be in the delegation team of 4.

In total we have 10 people ( 5 pairs) out of which 2 dance partners cannot be on the team.

Hence to select a team of 4 people from 8. = 8C4 = 70. D IMO.

The method you have adopted is incorrect. You are just removing 2 members from the 10 members and then selecting a committee of 4 members. Out of 8 members, there is still a possibility of selecting members from the same pair.

Re: A dance delegation of 4 people must be chosen from 5 pairs of dance pa
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08 Apr 2016, 22:25

Vyshak wrote:

FightToSurvive wrote:

A dance delegation of 4 people must be chosen from 5 pairs of dance partners. If 2 dance partners can never be together on the delegation, how many different ways are there to form the delegation?

A. 20 B. 30 C. 60 D. 70 E. 80

Bunnuel, from what I understood from the question stem is that 2 people who are dance partners cannot be in the delegation team of 4.

In total we have 10 people ( 5 pairs) out of which 2 dance partners cannot be on the team.

Hence to select a team of 4 people from 8. = 8C4 = 70. D IMO.

The method you have adopted is incorrect. You are just removing 2 members from the 10 members and then selecting a committee of 4 members. Out of 8 members, there is still a possibility of selecting members from the same pair.

Hope it helps.

I know that... I just thought that the question asked to remove any 2 people who are dance partners..

Anyways thanks for clarifying.

Could you please explain your approach to the solution?

Re: A dance delegation of 4 people must be chosen from 5 pairs of dance pa
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09 Apr 2016, 00:22

Bunuel wrote:

A dance delegation of 4 people must be chosen from 5 pairs of dance partners. If 2 dance partners can never be together on the delegation, how many different ways are there to form the delegation?

A. 20 B. 30 C. 60 D. 70 E. 80

another way to look at the Q altough the concept is same.. there are 5 pairs and we have to choose 4 out of 10... let us choose 4 pairs out of these 5 pairs = 5C4.. each pair choosen can send either of the TWO = 2*2*2*2 = 16 total ways = 16*5 = 80..
_________________

Re: A dance delegation of 4 people must be chosen from 5 pairs of dance pa
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09 Apr 2016, 00:28

Vyshak wrote:

sudhirmadaan wrote:

i still have doubt 5 pair = 10 people so out of 10 we need to choose 4 hence total combinations are 10c4 which is 210 now out 5 pair choose any one pair at a time 5c1 * (out of 8 remaining peope that that is 4 pair choose any 2 people ) it will be 5c1 * 8c2 which is 140 now these many combinations when 1pair must be on the list, so subtract it 210 -140 = 70 please clarify Bunuel, i asked same question in one more post.

5C1 * 8C2 will enable you to select only 1 pair and 2 members. Also, you are picking 1 complete pair from 5 pairs and the 2 members are picked from 8 members. The members can still be picked from the same pair.

I don't get you, please correct me again My intention is to have ways in which any two partners from 1 pair included i choose 5c1 , so I get any one pair , rest two members can be from another pair or can be from different pairs, how does it matters , as i already one pair included i have to subtract so 5c1 * 8c2 will always have two members from any pair. Make sense ?

Re: A dance delegation of 4 people must be chosen from 5 pairs of dance pa
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09 Apr 2016, 00:29

sudhirmadaan wrote:

i still have doubt 5 pair = 10 people so out of 10 we need to choose 4 hence total combinations are 10c4 which is 210 now out 5 pair choose any one pair at a time 5c1 * (out of 8 remaining peope that that is 4 pair choose any 2 people ) it will be 5c1 * 8c2 which is 140 now these many combinations when 1pair must be on the list, so subtract it 210 -140 = 70 please clarify Bunuel, i asked same question in one more post.

Your approach is correct too, But there are repetitions in 5c1 * 8c2..

say the pairs are A,B ; C,D ; E,F ; G,H and I, J.. say 5C1 you choose A,B.. and 8C2 you have C,D.. in another combination you have C,D as 5C1 and A,B in 8C2.. so you are counting same combination as 2 different CASES... what do you do then? Subtract cases where there are two pairs = 5C2 = 5!/3!2! = 10..

Re: A dance delegation of 4 people must be chosen from 5 pairs of dance pa
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09 Apr 2016, 00:41

chetan2u wrote:

sudhirmadaan wrote:

i still have doubt 5 pair = 10 people so out of 10 we need to choose 4 hence total combinations are 10c4 which is 210 now out 5 pair choose any one pair at a time 5c1 * (out of 8 remaining peope that that is 4 pair choose any 2 people ) it will be 5c1 * 8c2 which is 140 now these many combinations when 1pair must be on the list, so subtract it 210 -140 = 70 please clarify Bunuel, i asked same question in one more post.

Your approach is correct too, But there are repetitions in 5c1 * 8c2..

say the pairs are A,B ; C,D ; E,F ; G,H and I, J.. say 5C1 you choose A,B.. and 8C2 you have C,D.. in another combination you have C,D as 5C1 and A,B in 8C2.. so you are counting same combination as 2 different CASES... what do you do then? Subtract cases where there are two pairs = 5C2 = 5!/3!2! = 10..

thus actual ways = 140 - 10 = 130..

ans = 210-130 = 80..

Hi Chetan2u I apologies, if my doubt seems silly to you but in your explanation you are considering 5c1 gives me A and B and 8c2 gives me c and d 5c1*8c2 can not have same result it will give me one pair either A and B or C and D

check similar question and my result below this question is exactly same and my answer is correct here please tell me th basic difference in two question if i m missing something

Re: A dance delegation of 4 people must be chosen from 5 pairs of dance pa
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09 Apr 2016, 00:45

sudhirmadaan wrote:

chetan2u wrote:

sudhirmadaan wrote:

i still have doubt 5 pair = 10 people so out of 10 we need to choose 4 hence total combinations are 10c4 which is 210 now out 5 pair choose any one pair at a time 5c1 * (out of 8 remaining peope that that is 4 pair choose any 2 people ) it will be 5c1 * 8c2 which is 140 now these many combinations when 1pair must be on the list, so subtract it 210 -140 = 70 please clarify Bunuel, i asked same question in one more post.

Your approach is correct too, But there are repetitions in 5c1 * 8c2..

say the pairs are A,B ; C,D ; E,F ; G,H and I, J.. say 5C1 you choose A,B.. and 8C2 you have C,D.. in another combination you have C,D as 5C1 and A,B in 8C2.. so you are counting same combination as 2 different CASES... what do you do then? Subtract cases where there are two pairs = 5C2 = 5!/3!2! = 10..

thus actual ways = 140 - 10 = 130..

ans = 210-130 = 80..

Hi Chetan2u I apologies, if my doubt seems silly to you but in your explanation you are considering 5c1 gives me A and B and 8c2 gives me c and d 5c1*8c2 can not have same result it will give me one pair either A and B or C and D

check similar question and my result below this question is exactly same and my answer is correct here please tell me th basic difference in two question if i m missing something

over here 4 married couple and we need to choose 3 people now i will do the same 8c3-(4c1 * 6c1) = 32

8c3 way to give number of any three people 4c1 ways always include 1 pair * 6c1 will one member from rest of the group

Hi, Why should your 8C2 not contain a pair.. who are these? C,D,E,F,G,H,I,J.. you are looking at choosing two from these people, There will ofcourse be combination where these 2 will be CD, EF, GH and IJ apart from CE, CF, DE, DF etc..

Do one thing -- Take a smaller number 3 pairs and work out your solution by writing down each combination.. You will realize your mistake..
_________________

Re: A dance delegation of 4 people must be chosen from 5 pairs of dance pa
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18 Apr 2016, 12:37

Can someone tell me what's wrong with my approach?

Let there be pairs from A-E and 1,2 be the different people in the couple: A1, A2, B1, B2, C1, C2, D1, D2, E1,E2

Now, there are 10 ways to select the first person (let's say A1).Then there are 8 people left to choose from (excluding A1 and A2). For the next person there are 6 options, and the next 4.