Question

$\sum _{r=0}^{n-p}(-1{)}^{p+r}\cdot {}^n{C}_r\cdot {}^{n-r}{C}_p\cdot 5^{n-p-r}\cdot 6^r$ is equal to

• A. $(-1{)}^p\cdot 5^{n-p}$
• B. ${}^n{C}_p(-1{)}^n$
• C. $(-1{)}^n\cdot 5^{n-p}$
• D. ${}^n{C}_p(-1{)}^p$

Answer 1

Answer: (B)

${}^n{C}_r\cdot {}^{n-r}{C}_p=\frac{n!}{r!(n-r)!}\times \frac{(n-r)!}{p!(n-r-p)!}\times \frac{(n-p)!}{(n-p)!}=\frac{n!}{p!(n-p)!}\cdot \frac{(n-p)!}{(n-p-r)!r!}={}^n{C}_p\cdot {}^{n-p}{C}_r$
$\Rightarrow \sum _{r=0}^{n-p}(-1{)}^{p+r}\cdot {}^n{C}_r\cdot {}^{n-r}{C}_p\cdot 5^{n-p-r}\cdot 6^r$
$\Rightarrow (-1{)}^{p\cdot {}^n{C}_p}\cdot \sum _{r=0}^{n-p}{}^{n-p}{C}_p\cdot 5^{n-p-r}\cdot (-6{)}^r$
$\Rightarrow (-1{)}^p\cdot {}^n{C}_p\cdot (5-6{)}^{n-p}={}^n{C}_p(-1{)}^{p+n-p}$
$\Rightarrow {}^n{C}_p(-1{)}^n$