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In how many different ways can a group of twelve people be split into

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In how many different ways can a group of twelve people be split into [#permalink]

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In how many different ways can a group of twelve people be split into pairs ?

A 105
B 132
C 10,395
D 11,880
E 665,280
[Reveal] Spoiler: OA

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In how many different ways can a group of twelve people be split into [#permalink]

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Alex75PAris wrote:
In how many different ways can a group of twelve people be split into pairs ?

A 105
B 132
C 10,395
D 11,880
E 665,280


The FIRST pair can be chosen in 12C2 ways

The SECOND pair can be chosen in 10C2 ways as 10 members are left after choosing 1st pair

The THIRD pair can be chosen in 8C2 ways as 8 members are left after choosing 1st two pair and so on...

Therefore total ways of splitting 12 members in 6 pairs = (12C2 * 10C2 * 8C2 * 6C2 * 4C2 * 2C2) / 6!

The entire expression is divided by 6! because the arrangements of pair have been accounted for, which need to be excluded. Arrangement for example Same pair which came at first place can be chosen at third place or forth places also while we are choosing different pairs at different places

Required answer = (66 * 45 * 28 * 15 * 6 * 1)/720 = 10395

Answer: option C
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Re: In how many different ways can a group of twelve people be split into [#permalink]

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New post 18 Nov 2017, 15:53
Hi Alex75PAris,

We're asked for the number of ways to divide a group of 12 people into groups of 2. This is an exceptionally 'step-heavy' question and it's unlikely that you would see something like this on the Official GMAT. That having been said, here's how you can solve it.

The primary issue is really about all of the 'duplicate' entries that would occur - duplicates that we have to remove by using the Combination Formula REPEATEDLY.

The first group of 2 would be 12c2 = 12!/10!2! = 66 options
The second group of 2 would be 10c2 = 10!/8!2! = 45 options
The third group of 2 would be 8c2 = 8!/6!2! = 28 options
The fourth group of 2 would be 6c2 = 6!/4!2! = 15 options
The fifth group of 2 would be 4c2 = 4!/2!2! = 6 options
The sixth group of 2 would be 2c2 = 2!/0!2! = 1 options

Now, the next issue is that the 6 groups formed (above) could appear in any order - for example, if we called the unique pairings A, B, C, D, E and F....
ABCDEF and FEDCBA are the SAME set of 6 pairings, but we're not allowed to count this one option twice. With 6 pairs, there are 6! repeats, so we have to divide those out. The end calculation would be....

(66)(45)(28)(15)(6)(1)/6!

Once you cancel out terms and do the necessary multiplication, you end up with the final answer...

Final Answer:
[Reveal] Spoiler:
C


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Re: In how many different ways can a group of twelve people be split into [#permalink]

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New post 26 Nov 2017, 02:23
Hi, Please help me identify what's wrong with the reasoning I applied below

Since there are 12 persons,
1st person can pair in 11 ways by pairing with all different 11 persons
2nd person can pair in 10 ways
3rd person can pair in 9 ways and so.
.
.
so finally, I got 11! as my answer, which as per above explanations, I got that it is wrong, but don't understand why it is wrong.

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Re: In how many different ways can a group of twelve people be split into   [#permalink] 26 Nov 2017, 02:23
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