If n is a positive integer and the coefficien | vedclass

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(C) In the expansion of $(1+x)^{15}$,the coefficient of $x^{10}$ is given by ${}^{15}C_{10}$.The expansion of $(1-x)^{-n}$ is given by $1 + nx + \frac

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$11$

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(C) In the expansion of $(1+x)^{15}$,the coefficient of $x^{10}$ is given by ${}^{15}C_{10}$.The expansion of $(1-x)^{-n}$ is given by $1 + nx + \frac{n(n+1)}{2!}x^2 + \dots + \frac{n(n+1)\dots(n+r-1)}{r!}x^r + \dots$.Thus,the coefficient of $x^5$ in $(1-x)^{-n}$ is $\frac{n(n+1)(n+2)(n+3)(n+4)}{5!}$.According to the problem,$\frac{n(n+1)(n+2)(n+3)(n+4)}{5!} = {}^{15}C_{10}$.We know that ${}^{15}C_{10} = {}^{15}C_{5} = \frac{15 	imes 14 	imes 13 	imes 12 	imes 11}{5 	imes 4 	imes 3 	imes 2 	imes 1} = 3003$.So,$n(n+1)(n+2)(n+3)(n+4) = 5! 	imes {}^{15}C_{5} = 120 	imes 3003 = 360360$.Alternatively,$n(n+1)(n+2)(n+3)(n+4) = 11 	imes 12 	imes 13 	imes 14 	imes 15$.Comparing both sides,we get $n = 11$.

If $n$ is a positive integer and the coefficient of $x^{10}$ in the expansion of $(1+x)^{15}$ is equal to the coefficient of $x^5$ in the expansion of $(1-x)^{-n}$,then $n=$

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