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0 String: 34 /foo/bar_11,,, I want to insert "34" instead of the second comma. So my string should look like this: /foo/bar_11,34, How I can do it? sed share | improve this question edited Jan 28 at 14:14 Josef Klimuk asked Jan 28 at 14:08 Josef Klimuk Josef Klimuk 553 1 17

3 $begingroup$ Find the value of $$mathop{sumsum}_{0leq i<jleq n}binom{n}{i}cdot binom{n}{j} $$ I get the result: $$frac{1}{2}left(2^{2n}-binom{2n}{n}right)$$ via a numeric argument. My question is: Can we solve it using a combinational argument? My Numeric Argument: $$left(sum^{n}_{r=0}binom{n}{i}right)^2=sum^{n}_{r=0}binom{n}{i}^2+2mathop{sumsum}_{0leq i<jleq n}binom{n}{i}cdot binom{n}{j}$$ So here $$displaystyle sum^{n}_{r=0}binom{n}{i} = binom{n}{0}+binom{n}{1}+.....+binom{n}{n} = 2^n$$ and $$displaystyle sum^{n}_{r=0}binom{n}{i}^2=binom{n}{0}^2+binom{n}{1}^2+.....+binom{n}{n}^2 = binom{2n}{n}$$ above we have calculate Using $$(1+x)^n = binom{n}{0}+binom{n}{1}x+binom{n}{2}x^2+.....+binom{n}{n}x^n$$ and $$(x+1)^n = binom{n}{0}x^n+binom{n}{1}x^{n-1}+binom{n}{2}x^{n-2}+.....+binom{n}{n}x^0