Question

If $P(n,r)=1680$ and $C(n,r)=70$ , then $69n+r!$ is equal to

• A. $128$
• B. $576$
• C. $256$
• D. $625$
• E. $1152$

Answer 1

The correct option is B

$576$

Explanation for correct option :

Step 1: Apply permutation and combination formula

$\Rightarrow {}^{n}p_r=1680$ $\to \frac{n!}{(n-r)!}=1680$ …..$\left(1\right)$ and ${}^{n}C_r=70$ $\to \frac{n!}{(n-r)!r!}=70$ ……$\left(2\right)$

On Solving equations $\left(1\right)$ and $\left(2\right)$, we get,

$$\begin{gathered} \Rightarrow r!=\frac{1680}{70}=24=4! \\ \Rightarrow r=4 \end{gathered}$$

Step 2:On putting value of $r$ in equation $\left(i\right)$, we get

$$\begin{gathered} \Rightarrow \frac{n!}{(n-4)!}=1680 \\ \Rightarrow n(n-1)(n-2)(n-3)=1680 \\ \Rightarrow n(n-1)(n-2)(n-3)=8(8-1)(8-2)(8-3) \\ \Rightarrow n=8 \end{gathered}$$

Therefore, putting the values in $69n+r!$ we get,

$69n+r!$$=69x\left(8\right)+4!$

$=552+24+576$

Hence option (B) is correct.