Double Sum and Double Product

$\binom{n}{m}$ counts the number of ways to pick $m$ out of $n$ balls to have non-red color (the rest are red), and $\binom{m}{p}$ counts the number of ways to pick $p$ out of $m$ non-red balls to have blue color (the rest are green). So $\binom{n}{m} \binom{m}{p}$ counts the number of ways to color $n$ balls into $n-m$ red, $m-p$ green, and $p$ blue.

Summing over all $m$ means we count the number of ways to color $n$ balls in red, green, and blue so that $p$ are blue. (The rest are red and/or green; the value of $m$ follows what coloring we get.) Summing over all $p$ , except for $p = 0$ , means we color $n$ balls in red, green, and/or blue, such that there is at least one blue ball. The number of ways is simply $3^n - 2^n$ : the number of ways to color $n$ balls in 3 colors, minus the number of ways to color $n$ in 2 colors (without blue). So $a + b = 3 + 2 = 5$ .

We have $\sum_{p=1}^n \sum_{m=p}^n \binom{n}{m}\binom{m}{p} \; = \; \sum_{1 \le p \le m \le n} \frac{n!}{p! (m-p)! (n-m)!} \; = \; \sum_{{a+b+c=n} \atop {a \ge 1, b,c \ge 0}} \frac{n!}{a! b! c!} \; = \; \sum_{{a+b+c=n} \atop {a,b,c \ge 0}} \frac{n!}{a! b! c!} - \sum_{{b+c=n} \atop {b,c \ge 0}} \frac{n!}{b! c!} \; = \; 3^n - 2^n$ making the answer $\boxed{5}$ .

$\displaystyle \sum_{p=1}^n\displaystyle \sum_{m=p}^n{n \choose m}{m \choose p}=\displaystyle \sum_{p=1}^n\displaystyle \sum_{m=p}^n\frac{n!}{p!(m-p)!(n-m)!}$

now this can be rewritten as $\displaystyle \sum_{p=1}^n{n \choose p}{p \choose 0}+{n \choose p+1}{p+1 \choose 1}+{n \choose p+2}{p+2 \choose 2}...........+{n \choose n}{n \choose p}$

This is the coefficient of $x^p$ in the expansion of $(1+(1+x))^n$ .

Thus we are required to to find the sum of coefficents of all powers of $x$ except $x=0$

Total sum of coefficients is obtained by putting the value of $x=1$ ,in $(1+(1+x))^n$ which equals $3^n$ $(1)$

The sum of coefficients of terms independent of $x$ is given by putting $x=0$ ,in $(1+(1+x))^n$ which equals $2^n$ $(2)$

Thus the required sum is $(1)-(2)=3^n-2^n$