Originally Posted by

**Archie Meade** If you observe your work more closely,

you'll see a pattern from which you can derive a formula.

**k=2**

(1,2) one pair

**k=4**

(1,2) and the other pair

(1,3) and the other pair

(1,4) and the other pair

That's 3 pairs

**k=6**

(1,2) + (3,4) and the other pair

(1,2) + (3,5) and the other pair

(1,2) + (3,6) and the other pair

Repeat this for (1,3), (1,4), (1,5) and (1,6)

That's 5(3) pairs

**k=8**

(1,2) + (3,4) + (5,6) and the other pair

(1,2) + (3,4) + (5,7) and the other pair

(1,2) + (3,4) + (5,8) and the other pair

In this sequence, beginning with (1,2).....5 can be paired with the three numbers 6, 7, 8

..... 3 can be paired with five others, 4, 5, 6, 7, 8.

1 can be paired with 7 others, 2, 3, 4, 5, 6, 7.

Therefore the number of possible sets of pairs is 7(5)3.

For k=8, 10, 12 etc, the number of sets of pairs is (k-1)(k-3)(k-5)...5(3)1

To express this as a formula...

$\displaystyle \displaystyle\huge\frac{k(k-1)(k-2)(k-3)(k-4)(k-5)...5(4)3(2)1}{k(k-2)(k-4)..4(2)1}$

$\displaystyle =\displaystyle\huge\frac{k!}{2\left(\frac{k}{2}\ri ght)2\left(\frac{k-2}{2}\right)...2(3)2(2)2(1)}$

$\displaystyle =\displaystyle\huge\frac{k!}{2^{\frac{k}{2}}\left( \frac{k}{2}\right)!}$

which can be written neater if you let k represent the number of terms instead of pairs.