Proving $\sum_{i=0}^{p}{\sum_{j=0}^{q+i}{\sum_{k=0}^{r+j}{\binom{p}{i}\binom{q+i}{j}\binom{r+j}{k}}}}=4^{p}3^{q}2^{r}$

Question

This problem came to me when I was solving another binomial coefficient summation problem in this site.

I want to prove that $\sum_{i=0}^{p}{\sum_{j=0}^{q+i}{\sum_{k=0}^{r+j}{\binom{p}{i}\binom{q+i}{j}\binom{r+j}{k}}}}=4^{p}3^{q}2^{r}$

Answer 1

$$S=\sum_{i=0}^{p} \sum_{j=0}^{q+i} \sum_{k=0}^{r+j} {p \choose i} {q+i \choose j} {r+j \choose k}$$

First do the $k$ sum, then $$S=\sum_{i=0}^{p} \sum_{j=0}^{q+i} {p \choose i} {q+i \choose j} 2^{r+j}$$ Next do the $j$ sum, then $$S=2^r 3^q\sum_{i=0}^{p} {p \choose i} 3^{i}=2^r 3^q 4^p $$ Lastly do the $i$ sum $$\implies S=2^r 3^q 4^p$$

Answer 2

$$ \begin{aligned} (3+x)^{p}(2+x)^{q}(1+x)^{r}&=(1+(2+x))^{p}(2+x)^{q}(1+x)^{r}\\ &=\sum_{i=0}^{p}{\binom{p}{i}(2+x)^{i}(2+x)^{q}(1+x)^{r}}\\ &=\sum_{i=0}^{p}{\binom{p}{i}(2+x)^{q+i}(1+x)^{r}}\\ &=\sum_{i=0}^{p}{\binom{p}{i}(1+(1+x))^{q+i}(1+x)^{r}}\\ &=\sum_{i=0}^{p}{\sum_{j=0}^{q+i}{\binom{p}{i}\binom{q+i}{j}(1+x)^{j}(1+x)^{r}}}\\ &=\sum_{i=0}^{p}{\sum_{j=0}^{q+i}{\binom{p}{i}\binom{q+i}{j}(1+x)^{r+j}}}\\ &=\sum_{i=0}^{p}{\sum_{j=0}^{q+i}{\sum_{k=0}^{r+j}{\binom{p}{i}\binom{q+i}{j}\binom{r+j}{k}x^{k}}}}\\ \end{aligned} $$

Now substitute $x=1$