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

Question

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

GMATinsight · Accepted Answer

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

EMPOWERgmatRichC · Answer

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:  C

GMAT assassins aren't born, they're made, 
Rich

Tapesh03 · Answer

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.

soleimanian · Answer

xa!/x!*a^x
12!/6!*2^6=10395

Manvreddy · Answer

Can you/someone please explain this with more detail?

chetan2u · Answer

Formula- 
 If you divide 'xa' items in 'x' groups of 'a' items each, the number of ways=  \frac{(xa)!}{x!(a!)^x}

so if you are having 12 that is 6*2 items and you want to divide 6 groups of 2 each..
 \frac{(xa)!}{x!(a!)^x}=\frac{12!}{6!(2!)^6}

if you want to know HOW and WHY?..
 https://gmatclub.com/forum/topic215915.html

rsrighosh · Answer

I have found a formula which makes such question solve within few seconds

The number of arrangements where a number of people are to be divided into m teams with n people

\frac{(mn)!}{(n!)^m * m!}

Question says 12 people are to be divided in pairs

means no. of teams = m = 6; number of people in each team =n = 2

Going by the formula:-
 \frac{(6*2)!}{(2!)^6*6!}  = 10395

Sahil2208 · Answer

Bunuel 
Please help with the reasoning. Here's what my brain worked. Since we need to form groups of 2 people out of 12 people, a total of 6 pairs/groups will be formed. Now I will first choose 6 people out of 12 hence 12C6. After that, the rest 6 people will be paired with the chosen people in 6! ways. Hence total ways = (12C6)*6! =  *6! = 12!/6! = 12(11)(10)...(7).

chetan2u · Answer

There are repetitions in the above.

Say 12C6 gives you ABCDEF and you pair remaining 6 with these and you get AG as a pair. 
Again 12C6 will give you ABCDEH and you pair remaining 6 with these and you get AG again as a pair. 
Take smaller numbers say 6 people and do the above, you will realise where you are going wrong.

6C3 and pairing will give you 6*5*4 or 120

But the answer should be  \frac{6!}{(2!)^3*3!}  or 15.
Check 
ABCDEF
AB: Others can be (CD,EF), or (CE,DF) or (CF,DE)
Similarly AC, AD, AE and AF will give 3 groups each.

Thus total = 3*5=15

bellbell · Answer

Fast way that can reduce calculation:
(12C2 * 10C2 * 8C2 * 6C2 * 4C2 * 2C2)/6!
= (66*45*28*15*6)/(6*5*4*3*2)
= 11*9*7*5*3
From the above, we can conclude that:
*the correct answer has to be pretty big => eliminate A & B
*the correct answer has to be odd because odd*odd = odd & 11,9,7,5,3 are all odd numbers => eliminate D & E

mysterymanrog · Answer

Neat Question!

How I approached it: Realize that the groups here are not unique - i.e it doesn't matter if AB or BA.

Now pick 2 pairs from 12
12C2*10C2*...*2C2

Because groups are not unique, we can simply divide n!, where n is the total number of pairs. In this case, 6.

We get 12C2*10C2*...2C2/6!
Reduces to 
11*9*7*5*3
which is equal to C.

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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

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