Originally Posted by

**HallsofIvy** By the binomial theorem (a+ b)^n has n terms. (x+ (y+z))^20, counting "y+ z" as a single term, has 20 terms. But then, expanding (y+ z)^n, with n going from 1 to 20, we have 1+ 2+ 3+ 4+ ...+ 20= 20(21)/2= 210 terms.

Yes, this is the same as asking "how many non-negative sets of (a, b, c) are there that add to 20".

By using that formula what? Are you thinking of the "a", "b", "c" as the three toy and the "20th power" as the 20 children? $\displaystyle C(22, 2)= \frac{22!}{2! 20!}= \frac{21*20}{2}= 21*10= 210$ as I said before.

Solve what? Getting out of the jungle? I doubt there **is** any "formula" for "getting out of the jungle"! I would probably scream really, really loud!