RACHVIK wrote:In how many ways can 8 people be divided into 4 teams of 2 people each??

Source: GMAT Prep

Kindly explain how to approach this problem??

Thanks

First person can be paired with 7 different people, giving us 7 possible pairs.
8-2 = 6 people left.
Next person can be paired with 5 different people, giving us 5 possible pairs.
6-2 = 4 people left.
Next person can be paired with 3 different people, giving us 3 possible pairs.
4-2 = 2 people left, giving us 1 more possible pair.

Multiplying the results above, we get 7*5*3*1 = 105 possible pairs.

Thanks for your reply.

I tried approaching the problem using Combination method and found self stuck after finding 28 ways for selecting 2 out of 8, 15 ways for selecting 2 out of 6, and 6 ways for selecting 2 out of 4. Why do we need to divide by 4! to get 105 pairs of 4 teams??

Whats the catch in this method?? Multiplying above gives 2520 ways of forming 2 member groups, but question asks for forming 4 teams of 2 each??

Can you please explain.

Thanks

RACHVIK wrote:Thanks for your reply.

I tried approaching the problem using Combination method and found self stuck after finding 28 ways for selecting 2 out of 8, 15 ways for selecting 2 out of 6, and 6 ways for selecting 2 out of 4. Why do we need to divide by 4! to get 105 pairs of 4 teams??

Whats the catch in this method?? Multiplying above gives 2520 ways of forming 2 member groups, but question asks for forming 4 teams of 2 each??

Can you please explain.

Thanks

8C2 * 6C2 * 4C2 * 2C2 = 28*15*6*1 = 2520 ways to arrange all the possible pairs that can be made from 8 people. This would be the correct answer if we cared about the order of the pairs: AB - CD - EF - GH is a different arrangement from CD - AB - GH - EF.

But in this problem the order of the pairs doesn't matter: AB - CD - EF - GH gives us the same 4 pairs as CD - AB - GH - EF. When order doesn't matter, we divide by (number of choices being made)!. Since we're choosing 4 pairs, we need to divide by 4!, giving us (28*15*6*1)/(4*3*2*1) = 105 possible pairs.

Many test-takers find this distinction confusing. That's why in my earlier post I offered a different approach that I think is easier to follow.

Thanks a ton. What is puzzling is that how order comes into picture when we are using combination formula?? we have formed 2520 pairs of 2 people and each pair is different from the other. Why not just divide by 4??

My apology for bothering you with my query.

Thanks

RACHVIK wrote:Thanks a ton. What is puzzling is that how order comes into picture when we are using combination formula?? we have formed 2520 pairs of 2 people and each pair is different from the other. Why not just divide by 4??

My apology for bothering you with my query.

Thanks

There is a distinction between the following 2 questions:

Q: How many pairs can be made from 4 people?
A: 4C2 = 6.
Given ABCD, we can form the following pairs: AB, AC, AD, BC, BD, CD.

Q: How many ways can 4 people be divided into pairs?
A: (4C2*2C2)/2! = 3.
Given ABCD, we can divide them into pairs as follows: AB and CD, AC and BD, AD and BC.
The result is smaller because of the 4C2 = 6 pairs that can be formed from the 4 people, each pair has only 1 possible complement: AB must be combined with CD, AC must be combined with BD, and AD must be combined with BC. So there are only 3 ways to divide the 4 people into pairs.

Is the distinction clear?

Thanks a lot.

GMATGuruNY wrote:

RACHVIK wrote:Thanks a ton. What is puzzling is that how order comes into picture when we are using combination formula?? we have formed 2520 pairs of 2 people and each pair is different from the other. Why not just divide by 4??

My apology for bothering you with my query.

Thanks

There is a distinction between the following 2 questions:

Q: How many pairs can be made from 4 people?
A: 4C2 = 6.
Given ABCD, we can form the following pairs: AB, AC, AD, BC, BD, CD.

Q: How many ways can 4 people be divided into pairs?
A: (4C2*2C2)/2! = 3.
Given ABCD, we can divide them into pairs as follows: AB and CD, AC and BD, AD and BC.
The result is smaller because of the 4C2 = 6 pairs that can be formed from the 4 people, each pair has only 1 possible complement: AB must be combined with CD, AC must be combined with BD, and AD must be combined with BC. So there are only 3 ways to divide the 4 people into pairs.

Is the distinction clear?

Excellent explanation GMATGuruNY!

Hi mitch i know this is a very simple calculation but could you please explain to me how to calculate 8C2 * 6C2 * 4C2 * 2C2.I mean how do we calculate 8C2?
Please help me on this one

Thnx
Gaurav

golu23 wrote:Hi mitch i know this is a very simple calculation but could you please explain to me how to calculate 8C2 * 6C2 * 4C2 * 2C2.I mean how do we calculate 8C2?
Please help me on this one

Thnx
Gaurav

Perhaps my posts here will help:

https://www.beatthegmat.com/counting-problem-t72690.html

Pay special attention to the second post, which offers a step-by-step description of the slot method.

Hi GMATGuru,

I beleive we have a formula of dividing (m + n ) people into two groups with m and n people respectively = (m + n)!/(m! * n!)
Now,if each of these groups are identical we divide by 2!.

So if i apply this formula to the above problem,
No of ways = 8!/(2!*2!* 2!*2!*4!) =105.
Divided by 4! because all 4 groups are identical.

Is my approach correct??

RACHVIK wrote:In how many ways can 8 people be divided into 4 teams of 2 people each??

Source: GMAT Prep

Kindly explain how to approach this problem??

Thanks

1st team=8C2
2nd team=6C2
3rd team=4C2
4th team=2C2

multiply everything to get the required, 8C2*6C2*4C2*2C2 = 8!/6!2! *6!/4!2! *4!/2!2! *2!/2!0! here numerators cancel out denominators, 8!/(2!)^4 is the answer when order DOES matter, when the order is NOT important, the answer is 8!/4!*(2!)^4