prove-that-m-m-n-n-k-m-m-1-n-n-k-1-m-m-2-n-n-k-2-m-m-k-n-n-m-n-k-

Question Number 105625 by M±th+et+s last updated on 30/Jul/20

p r o v e t h a t

(\begin{matrix} m \\ m \end{matrix}) (\begin{matrix} n \\ n - k \end{matrix}) + (\begin{matrix} m \\ m - 1 \end{matrix}) (\begin{matrix} n \\ n - k + 1 \end{matrix})

+ (\begin{matrix} m \\ m - 2 \end{matrix}) (\begin{matrix} n \\ n - k + 2 \end{matrix}) + \dots \dots + (\begin{matrix} m \\ m - k \end{matrix}) (\begin{matrix} n \\ n \end{matrix})

= (\begin{matrix} m + n \\ k \end{matrix})

Answered by prakash jain last updated on 30/Jul/20

{(1 + x)}^{m + n} = {(1 + x)}^{m} {(1 + x)}^{n}

compare coefficient of x^{k} on LHS

and RHS .

LHS = C_{k}^{m + n}

RHS

(C_{0}^{m} x^{m} + C_{1}^{m} x^{m - 1} + . . + C_{m - k}^{m} x^{k} + . . + C_{m}^{m}) \times

(C_{0}^{n} x^{n} + C_{1}^{n} x^{n - 1} + . . + C_{n - k}^{n} x^{k} + . . + C_{n}^{n}) \times

c o e f f i c i e n t o f x^{k}

C_{m}^{m} C_{n - k}^{n} + . . + C_{n}^{n} C_{n - k}^{n}