The coefficient of x^{11} in the expansion of | vedclass

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(C) We need to find the coefficient of $x^{11}$ in the expansion of $(1+x^2)^4(1+x^3)^7(1+x^4)^{12}$.This is equivalent to finding the number of non-n

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$1113$

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(C) We need to find the coefficient of $x^{11}$ in the expansion of $(1+x^2)^4(1+x^3)^7(1+x^4)^{12}$.This is equivalent to finding the number of non-negative integer solutions to $2a + 3b + 4c = 11$,where $0 \le a \le 4$,$0 \le b \le 7$,and $0 \le c \le 12$.We list the possible combinations of $(a, b, c)$:$1$. If $b=1$,then $2a + 4c = 11 - 3 = 8 \implies a + 2c = 4$. Possible $(a, c)$ are $(4, 0), (2, 1), (0, 2)$.$2$. If $b=3$,then $2a + 4c = 11 - 9 = 2 \implies a + 2c = 1$. Possible $(a, c)$ is $(1, 0)$.Now calculate the coefficients using the binomial theorem $\binom{n}{r}$:- For $(a, b, c) = (4, 1, 0)$: $\binom{4}{4} 	imes \binom{7}{1} 	imes \binom{12}{0} = 1 	imes 7 	imes 1 = 7$.- For $(a, b, c) = (2, 1, 1)$: $\binom{4}{2} 	imes \binom{7}{1} 	imes \binom{12}{1} = 6 	imes 7 	imes 12 = 504$.- For $(a, b, c) = (0, 1, 2)$: $\binom{4}{0} 	imes \binom{7}{1} 	imes \binom{12}{2} = 1 	imes 7 	imes 66 = 462$.- For $(a, b, c) = (1, 3, 0)$: $\binom{4}{1} 	imes \binom{7}{3} 	imes \binom{12}{0} = 4 	imes 35 	imes 1 = 140$.Summing these values: $7 + 504 + 462 + 140 = 1113$.

The coefficient of $x^{11}$ in the expansion of $(1+x^2)^4(1+x^3)^7(1+x^4)^{12}$ is:

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