BUT!!! what if the question asks for 2 only, not 3, 4 or 5 of the same suit, then there would be a probability less than one, but the question doesn't specify, so im just being difficult
expand then integrate \binom{n-1}{0}x+\binom{n-1}{1}x^{2}/2+\binom{n-1}{2}x^{3}/3+...+\binom{n-1}{n-1}x^{n}/n
let x=0 to evaluate C, C=-1
let x=7 LHS= as u have shown
RHS = ((8)^n)/n -1 =((2^3)^n)/n -1 as required
yeh i gave up on the syntax
well if u consider (1+x)^n(1+1/x)^n, that gives you square coefficients for x^n, but i dont know where the (n+1) comes from, because that would hint that there is a differential, but then it should be n not (n+1) gahh