Computing $\sum_{i=2}^n \sum_{j=2}^i {n \choose i}{i \choose i-j}{j \choose 2}$

Question

My current approach is to use a combinatorial argument. The equation can be seen as all possible ways of first choosing $i$ out of a total of $n$ items, and choosing $(i-j)$ items out of the $i$ chosen ones, then choosing $2$ items out of the remaining $j$ items not chosen in the previous step.

So we are choosing ${n \choose i-j+2}$. But we need to subtract the number of times where we are double counting if we first choose $2 = (i-j)$ items out of the $i$ chosen ones, then choosing $2$ items out of the remaining $j$ items not chosen in the previous step, which is a total of $\frac{{n \choose 2} {n-2 \choose 2}}{2}$. So the final answer is ${n \choose i-j+2} - \frac{{n \choose 2} {n-2 \choose 2}}{2}$

Is this argument valid? If not, could someone tell me how I could do it? Thanks. Also, I don't have the correct answer.

Answer 1

Using the subset-of-a-subset identity $\binom ab\binom bc=\binom ac\binom {a-c}{b-c}$, we have $$\begin{align} \binom ni \binom i{i-j} \binom j2 &=\binom ni\underbrace{\binom ij\binom j2}\\ &=\underbrace{\binom ni\binom i2}\binom {i-2}{j-2}\\ &=\binom n2\binom {n-2}{i-2}\binom {i-2}{j-2} \end{align}$$ Hence $$\begin{align} \sum_{i=2}^n \sum_{j=2}^i \binom ni \binom i{i-j} \binom j2 &=\sum_{i=2}^n \sum_{j=2}^i \binom n2 \binom {n-2}{i-2} \binom {i-2}{j-2}\\ &=\binom n2 \sum_{i=2}^n \binom {n-2}{i-2} \sum_{j=2}^i \binom {i-2}{j-2}\\ &=\binom n2\sum_{i=2}^n \binom {n-2}{i-2} 2^{i-2}\\ &=\binom n2 (1+2)^{n-2}\\ &=\color{red}{\binom n2 3^{n-2}} \end{align}$$

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Alternatively,

$$\begin{align} \binom ni\binom i{i-j}\binom i2 &=\binom n2\binom {n-2}{i-2}\binom {i-2}{j-2}\\ &=\binom n2\underbrace{\binom {n-2}{n-i}\binom {i-2}{i-j}\binom {j-2}{j-2}}\color{lightgrey}{1^{n-i}1^{i-j}1^{j-2}}\\ &=\binom n2 \binom {n-2}{n-i,i-j,j-2}\color{lightgrey}{1^{n-i}1^{i-j}1^{j-2}}\\ \end{align}$$ Using the multinomial summation identity $$\sum_{p+q+r=m}\binom m{p,q,r}x^py^qz^r=(x+y+z)^m$$ we have

$$\begin{align} \sum_{i=2}^n\sum_{j=2}^n\binom ni\binom i{i-j}\binom i2 &=\binom n2\sum_{(n-i)+(i-j)+(j-2)=n}\binom {n-2}{n-i,i-j,j-2}\color{lightgrey}{1^{n-i}1^{i-j}1^{j-2}}\\ &=\color{red}{\binom n2 3^{n-2}}\end{align}$$

Answer 2

Hint : note that

$${n\choose i}{i \choose i-j}{j \choose 2} = {n\choose i}{i \choose j}{j \choose 2} $$

and this is the number of ways to choose first, $i$ items among $n$ available, then $j$ items among those $i$, then $2$ among those $j$.

Answer 3

${n\choose i}{i \choose i-j}{j \choose 2} = {n\choose i}{i \choose j}{j \choose 2}$

=${n\choose i}\times$coefficient of $x^2$ in $\left({i \choose 2}{2 \choose 2}+{i \choose 3}{3 \choose 2}+{i \choose 4}{4 \choose 2}+....+{i \choose i}{i \choose 2}\right)$

=${n\choose i}\times$coefficient of $x^2$ in $\left({i \choose 2}(1+x)^2+{i \choose 3}(1+x)^3+{i \choose 4}(1+x)^4+....+{i \choose i}(1+x)^i\right)$

=${n\choose i}\times$coefficient of $x^2$ in $\left\{(1+(1+x))^i-{i\choose 0}-{i\choose 1}(1+x)\right\}$

=${n\choose i}\times$coefficient of $x^2$ in $(2+x)^i$

=${n\choose i}{i\choose 2}2^{i-2}$

Therefore,required sum

=$\sum_{i=2}^n\dfrac{i(i-1)}{2}{n\choose i}2^{i-2}$

=$\sum_{i=2}^n\dfrac{(i-1)}{2}i{n\choose i}2^{i-2}$

=$\sum_{i=2}^n\dfrac{(i-1)}{2}n{n-1\choose i-1}2^{i-2}$

=$\dfrac{n}{2}\sum_{i=2}^n(i-1){n-1\choose i-1}2^{i-2}$

=$\dfrac{n(n-1)}{2}\sum_{i=2}^n{n-2\choose i-2}2^{i-2}$

=$\dfrac{n(n-1)}{2}(1+2)^{n-2}$

=${n\choose 2}3^{n-2}$

Combinatorial Approach

We may think of the problem in upside down manner.

Select two objects from $n$ in ${n\choose 2}$ ways then the remaining $n-2$ objects have $3$ choices each i.e.

(i)An object may choose to be among $j$ objects.

(ii)An object may choose to be among $i$ objects but not among $j$ objects.

(iii)An object may choose to be among the $n$ objects but neither among $i$ objects nor among the $j$ objects.

Therefore number of choices for the remaining $n-2$ objects is $3^{n-2}$.

Therefore the value of the given sum is ${n\choose 2}3^{n-2}$.

I am not sure whether my answer is correct or not.

Answer 4

Your first paragraph is correct. Then you jump to saying you are picking $\binom{n}{i - j + 2}$. But this is not correct; you are picking $i$ things AND then $j$ things from those $i$ AND $2$ things out of those $j$, not just picking $i - j + 2$ objects. Besides, the answer should not contain $i$ or $j$ -- those are just dummy variables.