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:

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

Rich

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