The value of $$\displaystyle\mathop{\sum\sum}_{0\leq i< j\leq n}(-1)^{i-j+1}\binom{n}{i}\binom{n}{j} = $$
$\bf{My\; Try::}$ Let $$S=\mathop{\sum\sum}_{0\leq i<j\leq n}(-1)^{i-j+1}\binom{n}{i}\binom{n}{j} $$
So $$S=(-1)^{0-1+1}\binom{n}{0}\binom{n}{1}+(-1)^{0-2+1}\binom{n}{0}\binom{n}{1}+(-1)^{1-2+1}\binom{n}{1}\binom{n}{2}+.....$$
after that i did not understand how can i calculate above sum
Help me
Thanks
You’ve written the original expression incorrectly: it should be
$$\sum_{0\le i<j\le n}(-1)^{i-j+1}\binom{n}i\binom{n}j\;.$$
You could work directly on this, first converting it to a double summation, but there’s an easier approach. For $0\le i,j\le n$ let
$$a(i,j)=(-1)^{i-j+1}\binom{n}i\binom{n}m\;,$$
so that
$$\sum_{0\le i<j\le n}(-1)^{i-j+1}\binom{n}i\binom{n}j=\sum_{0\le i<j\le n}a(i,j)\;.$$
If you check, you’ll find that $a(i,j)=a(j,i)$, so
$$\begin{align*} \sum_{0\le i,j\le n}a(i,j)&=\sum_{0\le i<j\le n}a(i,j)+\sum_{0\le i=j\le n}a(i,j)+\sum_{0\le j<i\le n}a(i,j)\\ &=2\sum_{0\le i<j\le n}a(i,j)+\sum_{i=0}^na(i,i)\;\tag{1} \end{align*}$$
and therefore
$$\begin{align*} \sum_{0\le i<j\le n}(-1)^{i-j+1}\binom{n}i\binom{n}j&=\sum_{0\le i<j\le n}a(i,j)\\ &=\frac12\left(\sum_{0\le i,j\le n}a(i,j)-\sum_{i=0}^na(i,i)\right)\\ &=\frac12\left(\sum_{i=0}^n\sum_{j=0}^na(i,j)-\sum_{i=0}^na(i,i)\right)\\ &=\frac12\left(\sum_{i=0}^n(-1)^i\binom{n}i\sum_{j=0}^n(-1)^{1-j}\binom{n}j-\sum_{i=0}^n(-1)\binom{n}i^2\right)\;. \end{align*}$$
Can you finish the evaluation from there?
Added: I meant to include this diagram to help explain the idea. Your original sum is over the black entries in the diagram, the ones above the diagonal. Each of the red entries is equal to its mirror-image black entry across the diagonal, so the red sum is equal to the black sum. Thus, the sum over the whole square is twice the black sum plus the blue diagonal sum, as in $(1)$.
$$\begin{array}{c|cc} &0&1&2&3&\ldots&n\\ \hline 0&\color{blue}{a(0,0)}&a(0,1)&a(0,2)&a(0,3)&\ldots&a(0,n)\\ 1&\color{red}{a(1,0)}&\color{blue}{a(1,1)}&a(1,2)&a(1,3)&\ldots&a(1,n)\\ 2&\color{red}{a(2,0)}&\color{red}{a(2,1)}&\color{blue}{a(2,2)}&a(2,3)&\ldots&a(2,n)\\ 3&\color{red}{a(3,0)}&\color{red}{a(3,1)}&\color{red}{a(3,2)}&\color{blue}{a(3,3)}&\ldots&a(3,n)\\ \vdots&\color{red}\vdots&\color{red}\vdots&\color{red}\vdots&\color{red}\vdots&\color{blue}{\ddots}&\vdots\\ n&\color{red}{a(n,0)}&\color{red}{a(n,1)}&\color{red}{a(n,2)}&\color{red}{a(n,3)}&\color{red}{\ldots}&\color{blue}{a(n,n)}\\ \end{array}$$
