The sum `1/(1!9!) + 1/(3!7!) + ... + 1/(7!3!) + 1/(9!1!)` can be written in the form `2^a/(b!)` Find a and b.

Explanation

1/(1!9!) + 1/(3!7!) + ... + 1/(7!3!) + 1/(9!1!)`
`1/(10!)((10!)/(1!9!) + (10!)/(3!7!) + .... + (10!)/(9!1!))`
`1/(10!) ("^10C_1 + ^10C_3 + .... + ^10C_9)`
`1/(10!) 2^(10-1)`
`2^9/(10!)` = `2^a/(b!)` where a = 9, b = 10