Question

If ${}^{n}P_r=840$ and ${}^{n}C_r=35$, then $n$ is equal to?

• A. $1$
• B. $3$
• C. $5$
• D. $7$

Answer 1

The correct option is D

$7$

Explanation for the correct option:

Step 1: Solve for the value of $r$

Given ${}^{n}P_r=840$ and ${}^{n}C_r=35$,

We know that ${}^{n}P_r=\frac{n!}{\left(n-r\right)!}$ and ${}^{n}C_r=\frac{n!}{r!\left(n-r\right)!}$

$\therefore \frac{n!}{\left(n-r\right)!}=840$

Now,

$\begin{array}{crclc}& \frac{n!}{r!\left(n-r\right)!}& =& 35& \\ \Rightarrow & \frac{840}{r!}& =& 35& \left[\because \frac{n!}{\left(n-r\right)!}=840\right]\\ \Rightarrow & r!& =& \frac{840}{35}& \\ \Rightarrow & r!& =& 24& \\ \Rightarrow & r& =& 4& \end{array}$

Step 2: Solve for the value of $n$

We know that $r=4$

$\begin{array}{cccc}\therefore & \frac{n!}{\left(n-4\right)!}& =& 840\\ \Rightarrow & \frac{n\left(n-1\right)\left(n-2\right)\left(n-3\right)\times \left(n-4\right)!}{\left(n-4\right)!}& =& 840\\ \Rightarrow & n\left(n-1\right)\left(n-2\right)\left(n-3\right)& =& 840\end{array}$

Putting the values of $n$ from options, let us substitute $n=7$

$\begin{array}{cccc}\Rightarrow & 7\left(7-1\right)\left(7-2\right)\left(7-3\right)& =& 840\\ \Rightarrow & 7\times 6\times 5\times 4& =& 840\\ \Rightarrow & 840& =& 840\end{array}$

Therefore the value of $n$is $7$

Hence, the correct option is option(D)