1

There are 1 to 24 numbers (1,2,3...24). How many possible combinations of 12 combine numbers will result with a sum of 146 when you add those 12 numbers?

So meaning you have to combine 12 numbers from 1 to 24 and the total must be 146. You cannot use the same number twice.

Let's say (1+2+3+4+5........+4) this is not possible because you already used number 4.

Arthur
  • 199,419
abraham
  • 11

1 Answers1

2

This is the coefficient of $x^{146}y^{12}$ in $(1+xy)(1+x^2y)(1+x^3y)\dots(1+x^{24}y)$, though I do not see any easy way to compute that by hand.

Macavity
  • 46,381