If C_0, C_1, C_2, ldots, C_{10} represent the | vedclass

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(B) We know that the binomial coefficients satisfy the property $C_r = C_{n-r}$.Thus,$C_6 = C_{10-6} = C_4$,$C_7 = C_3$,$C_8 = C_2$,$C_9 = C_1$,and $C

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$4845$

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(B) We know that the binomial coefficients satisfy the property $C_r = C_{n-r}$.Thus,$C_6 = C_{10-6} = C_4$,$C_7 = C_3$,$C_8 = C_2$,$C_9 = C_1$,and $C_{10} = C_0$.The given expression is $S = C_0 C_6 + C_1 C_7 + C_2 C_8 + C_3 C_9 + C_4 C_{10}$.Substituting the values,we get $S = C_0 C_4 + C_1 C_3 + C_2 C_2 + C_3 C_1 + C_4 C_0 = 2(C_0 C_4 + C_1 C_3) + C_2^2$.Using the identity $\sum_{k=0}^{r} C_k C_{n-k} = C_{2n, n}$,we look for the coefficient of $x^{10}$ in $(1+x)^{10}(1+x)^{10} = (1+x)^{20}$.The coefficient of $x^{10}$ in $(1+x)^{20}$ is $^{20}C_{10} = 184756$.However,the given sum is $C_0 C_6 + C_1 C_7 + C_2 C_8 + C_3 C_9 + C_4 C_{10} = \sum_{k=0}^{4} C_k C_{10-(6-k)} = \sum_{k=0}^{4} C_k C_{4-k}$.This is the coefficient of $x^4$ in the expansion of $(1+x)^{10}(1+x)^{10} = (1+x)^{20}$.The coefficient of $x^4$ in $(1+x)^{20}$ is $^{20}C_4 = \frac{20 	imes 19 	imes 18 	imes 17}{4 	imes 3 	imes 2 	imes 1} = 4845$.

If $C_0, C_1, C_2, \ldots, C_{10}$ represent the binomial coefficients in the expansion of $(1+x)^{10}$,then $C_0 C_6+C_1 C_7+C_2 C_8+C_3 C_9+C_4 C_{10}=$

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