The value of sum_{r=0}^{6} left({}^{6}C_{r} c | vedclass

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Question

(D) The given expression is $\sum_{r=0}^{6} {}^{6}C_{r} \cdot {}^{6}C_{6-r}$.Using the property of binomial coefficients,${}^{n}C_{r} = {}^{n}C_{n-r}$

Vedclass · Accepted Answer

$924$

Vedclass · Answer

(D) The given expression is $\sum_{r=0}^{6} {}^{6}C_{r} \cdot {}^{6}C_{6-r}$.Using the property of binomial coefficients,${}^{n}C_{r} = {}^{n}C_{n-r}$,we can write ${}^{6}C_{6-r} = {}^{6}C_{r}$.Thus,the sum becomes $\sum_{r=0}^{6} {}^{6}C_{r} \cdot {}^{6}C_{r} = \sum_{r=0}^{6} ({}^{6}C_{r})^2$.Alternatively,using Vandermonde's Identity,$\sum_{k=0}^{r} {}^{m}C_{k} \cdot {}^{n}C_{r-k} = {}^{m+n}C_{r}$.Here,$m=6, n=6$,and $r=6$.Therefore,$\sum_{r=0}^{6} {}^{6}C_{r} \cdot {}^{6}C_{6-r} = {}^{6+6}C_{6} = {}^{12}C_{6}$.Calculating the value: ${}^{12}C_{6} = \frac{12 	imes 11 	imes 10 	imes 9 	imes 8 	imes 7}{6 	imes 5 	imes 4 	imes 3 	imes 2 	imes 1} = 924$.

The value of $\sum_{r=0}^{6} \left({}^{6}C_{r} \cdot {}^{6}C_{6-r}\right)$ is equal to:

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