# Help me figure this out?

Prob simple but i forget and i used a formula to determine the 7th term for a binomial and I get 10!/6!4!x^4(-2)^6

can some explain how they go from that to this answer.....? 10!/6!4!*64x^6 like I get the first coefficient part or w/e but i dont understand how they got the powers and why -2^6 is positive shouldn't it be negative? can you help me?

### 1 Answer

- simplicitusLv 71 decade agoFavorite Answer
1. X^6 = (X^2)^3. Since X^2 is positive, so is X^6

2. (a + b)^n = sum over k = 0, n of C(n, k) a^k b^(n-k)

where C(n,k) = n! / (k! (n-k)!)

http://en.wikipedia.org/wiki/Binomial_coefficient

But since C(n, k) = C(n, n-k), this is also equal to:

(a + b)^n = sum over k = 0, n of C(n, k) a^(n-k) b^k

so the 10!/6!4! (actually 10! / (6!4!)) says that you are dealing with a situation n = 10 and k = either 6 or 4

10!/6!4!x^4(-2)^6 looks like C(10, 6) x^4 (-2)^6 which does look like the seventh term of the expansion of (x - 2)^10

10!/6!4!*64x^6 looks wrong. 10!/6!4!*16x^6 would be

C(10, 4) x^6 (-2)^4 which would make sense but (-2)^6 x^6 does not.

- Login to reply the answers