Question

If $S{=}^{404}{C}_4{-}^4{C}_1{.}^{303}{C}_4{+}^4{C}_2{.}^{202}{C}_5{-}^4{C}_3{.}^{101}{C}_4=(101{)}^k$ then k equal to

• A. $1$
• B. $2$
• C. $4$
• D. $6$

Answer 1

The correct option is A $1$

$S{=}^{404}{C}_4{-}^4{C}_1{.}^{303}{C}_4{-}^4{C}_2{.}^{202}{C}_5{-}^4{C}_3{.}^{101}{C}_4={\left(101\right)}^k$

$\Rightarrow \left(101\right)={\left(101\right)}^k\left[{\because }^n{C}_1=\frac{n!}{r!\left(n-1\right)!}\right]$

$\Rightarrow K=1$

Option $\left(A\right)$ is correc t answer.