If sumlimits_{K = 1}^{12} {12K cdot {^{12}C_K | vedclass

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(D) We know that $K \cdot {^{12}C_K} = 12 \cdot {^{11}C_{K - 1}}$.Substituting this into the summation:$\sum\limits_{K = 1}^{12} {12(12 \cdot {^{11}C_

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$6$

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(D) We know that $K \cdot {^{12}C_K} = 12 \cdot {^{11}C_{K - 1}}$.Substituting this into the summation:$\sum\limits_{K = 1}^{12} {12(12 \cdot {^{11}C_{K - 1}}) \cdot {^{11}C_{K - 1}}} = 144 \sum\limits_{K = 1}^{12} {({^{11}C_{K - 1}})^2}$.Using the identity $\sum\limits_{r=0}^{n} ({^nC_r})^2 = {^{2n}C_n}$,we get:$144 \cdot {^{22}C_{11}} = 144 \cdot \frac{22!}{11!11!}$.Expanding $22! = 22 	imes 21 	imes 20 	imes 19 	imes \dots 	imes 1$ and simplifying:$144 \cdot \frac{22 	imes 21 	imes 20 	imes 19 	imes \dots 	imes 1}{11! \cdot 11!} = 144 \cdot \frac{22 	imes 20 	imes 18 	imes \dots 	imes 2}{11!} \cdot \frac{21 	imes 19 	imes \dots 	imes 1}{11!}$.This simplifies to $144 \cdot 2^{11} \cdot \frac{21 	imes 19 	imes \dots 	imes 3}{11!} = 12 \cdot 12 \cdot 2^{11} \cdot \frac{21 	imes 19 	imes \dots 	imes 3}{11!} = 6 \cdot 2^{12} \cdot 12 \cdot \frac{21 	imes 19 	imes \dots 	imes 3}{11!}$.Comparing with the given expression,we find $p = 6$.

If $\sum\limits_{K = 1}^{12} {12K \cdot {^{12}C_K} \cdot {^{11}C_{K - 1}}} $ is equal to $\frac{{12 \times 21 \times 19 \times 17 \times \dots \times 3}}{{11!}} \times {2^{12}} \times p$,then $p$ is:

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