A) \[npq\]
B) \[pq\]
C) \[{{2}^{n}}pq\]
D) 0
Correct Answer: D
Solution :
We have, \[p{{q}^{n}}{{C}_{0}}-(p-1)\,(q-1){{\,}^{n}}{{C}_{1}}+\,(p-2)\,\,(q-2){{\,}^{n}}{{C}_{2}}\] \[-(p-3)\,(q-3){{\,}^{n}}{{C}_{3}}+...\] \[+{{(-1)}^{n}}\,(p-n)\,(q-n){{\,}^{n}}{{C}_{n}}\] \[=\sum\limits_{k=0}^{n}{{{(-1)}^{k}}\,(p-k)\,(q-k){{\,}^{n}}{{C}_{k}}}\] \[=\sum\limits_{k=0}^{n}{{{(-1)}^{k}}\,\{pq-\,(p+q)\,k+{{k}^{2}})\cdot {{\,}^{n}}{{C}_{k}}}\] \[=pq\sum\limits_{k=0}^{n}{{{(-1)}^{k}}{{\,}^{n}}{{C}_{k}}-(p+q)\,\sum\limits_{k=0}^{n}{{{(-1)}^{k}}k\cdot {{\,}^{n}}{{C}_{k}}}}\] \[=\,\sum\limits_{k=0}^{n}{{{(-1)}^{k}}\,{{k}^{2}}\cdot {{\,}^{n}}{{C}_{k}}}\] = 0 \[[\because \,\sum\limits_{k=0}^{n}{{{(-1)}^{k}}{{\,}^{n}}{{C}_{k}}=0=\,\sum\limits_{k=0}^{n}{{{(-1)}^{k}}\,k{{\,}^{n}}{{C}_{k}}}}\] \[=\,\sum\limits_{k=0}^{n}{{{(-1)}^{k}}\,{{k}^{2}}{{\,}^{n}}{{C}_{k}}]}\]
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