Heres a maths ques plz help me out..no of terms in (1+x+x^2+....+x^10)^5?

Question

Heres a maths ques plz help me out..no of terms in (1+x+x^2+....+x^10)^5?

Answer 1

here, n=10;r=5

so, the number of terms is 14C9

Answer 2

Note that this expression is equal to

(1 + x + x^2 + ... + x^10)(1 + x + x^2 + ... + x^10)(1 + x + x^2 + ... + x^10)(1 + x + x^2 + ... + x^10)(1 + x + x^2 + ... + x^10)

and that a term x^n appears in the result whenever we can pick x^a from the first set of brackets, x^b from the second, x^c from the third, x^d from the fourth, and x^e from the fifth, such that a + b + c + d + e = n. That is, if we can write some number n as the sum of five numbers chosen from [0,10], then a x^n term appears in the product. (Note that since all signs are positive, no cancellation can occur). It is not hard to see that, therefore, there are terms of all degrees from 0 to 50, hence, there are 51 terms in the product.