Question 422053
It comes from the Binomial theorem. The expansion of {{{(x+y)^n}}} is given by


{{{sum((nCi)(x^i)(y^(n-i)), i=0, n)}}} (here, {{{nCi = n!/((n-i)!i!)}}}, number of ways to choose i objects out of n total)


In this case,


{{{(x+y)^5 = (5C0)x^5 + (5C1)(x^4)y + (5C2)(x^3)(y^2) + (5C3)(x^2)(y^3) + (5C4)xy^4 + (5C5)y^5}}}


{{{(x+y)^5 = x^5 + 5(x^4)y + 10(x^3)(y^2) + 10(x^2)(y^3) + 5xy^4 + y^5}}}