Problem Solving

BTGmoderatorDC 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

Since there are 12 people, the first person selected can be paired with any of the 11 other people, yielding 11 POSSIBLE PAIRS.
Since 2 of the 12 people have been selected thus far, the number of people remaining = 12-2 = 10.

Since 10 people remain, the next person selected can be paired with any of the 9 other people, yielding 9 POSSIBLE PAIRS.
Since 2 of the 10 remaining people were just selected, the number of people remaining = 10-2 = 8.

Since 8 people remain, the next person selected can be paired with any of the 7 other people, yielding 7 POSSIBLE PAIRS.
Since 2 of the 8 remaining people were just selected, the number of people remaining = 8-2 = 6.

Since 6 people remain, the next person selected can be paired with any of the 5 other people, yielding 5 POSSIBLE PAIRS.
Since 2 of the 6 remaining people were just selected, the number of people remaining = 6-2 = 4.

Since 4 people remain, the next person selected can be paired with any of the 3 other people, yielding 3 POSSIBLE PAIRS.
Since 2 of the 4 remaining people were just selected, the number of people remaining = 4-2 = 2.

The last 2 people yield 1 additional pair.

To combine the blue options above, we multiply:
11*9*7*5*3*1 = 11*63*15 = a very large integer with a units digit of 5.

The correct answer is C.

BTGmoderatorDC 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

First pair can be created 12 x 11 ways. Second pair, 10X9, etc. So total ways to create pairs is 12!.

However, this gives credit to the ordering of people within a pair, which is irrelevant, so need to divide by 2^6

Similarly, 12! also gives credit to the ordering of the 6 pairs, which is also irrelevant, so need to divide by 6! also.

So, 12!/(2^6x6!) = 12x11x10x9x8x7/2^6 = 3X11x5x9x7 = c, 10395