Problem Solving

Hi AAPL,

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

AAPL 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 OA is C.

Please, can any expert assist me with this PS question? I don't have it clear and I appreciate if any explain it for me. Thanks.

Twelve people can form 6 pairs.

The first pair can be formed in 12 x 11 ways, the second pair can be formed in 10 x 9 ways, and so on. Therefore, the 6 pairs can be formed 12! ways if order matters. However, since the order of the pairs does not matter, we have to divide 12! by 6!.

Furthermore, the order within each pair also does not matter, so we have to also divide by 2!. Since there are 6 pairs, we have to divide by (2!)^6 = 2^6.

Therefore, the number of ways we can split 12 people into 6 pairs is:

12!/(6! x 2^6) = (12 x 11 x 10 x 9 x 8 x 7)/2^6 = 3 x 11 x 5 x 9 x 7

We see that the number is pretty large and its units digit must be 5, so choice C is the correct answer.

Answer: C