Question

If ${}^{n}P_r=30240$ and ${}^{n}C_r=252$, then the ordered pair $\left(n,r\right)$ is equal to?

• A. $\left(12,6\right)$
• B. $\left(10,5\right)$
• C. $\left(9,4\right)$
• D. $\left(16,7\right)$

Answer 1

The correct option is B

$\left(10,5\right)$

Explanation for the correct option:

Step 1: Solve for the value of $r$

Given that ${}^{n}P_r=30240$ and ${}^{n}C_r=252$,

We know that ${}^{\mathbf{n}}\mathbf{P}_{\mathbf{r}}\mathbf{=}\frac{\mathbf{n}\mathbf{!}}{\left(n-r\right)\mathbf{!}}$ and ${}^{\mathbf{n}}\mathbf{C}_{\mathbf{r}}\mathbf{=}\frac{\mathbf{n}\mathbf{!}}{\mathbf{r}\mathbf{!}\left(n-r\right)\mathbf{!}}$

$\therefore \frac{n!}{\left(n-r\right)!}=30240$

Now ,

$\begin{array}{crcll}& \frac{n!}{r!\left(n-r\right)!}& =& 252& \\ \Rightarrow & \frac{30240}{r!}& =& 252& \left[\because \frac{n!}{\left(n-r\right)!}=30240\right]\\ \Rightarrow & r!& =& \frac{30240}{252}& \\ \Rightarrow & r!& =& 120& \\ \Rightarrow & r& =& 5& \mathbf{}\left[\because 5!=5\times 4\times 3\times 2\times 1=120\right]\end{array}$

Step 2: Solve for the value of $n$

We know that $r=5$

$\begin{array}{ccccc}\therefore & \frac{n!}{\left(n-5\right)!}& =& 30240& \mathbf{}\left[\because \frac{n!}{\left(n-r\right)!}=30240\right]\\ \Rightarrow & \frac{n\left(n-1\right)\left(n-2\right)\left(n-3\right)\left(n-4\right)\times \left(n-5\right)!}{\left(n-5\right)!}& =& 30240& \\ \Rightarrow & n\left(n-1\right)\left(n-2\right)\left(n-3\right)\left(n-4\right)& =& 30240& \end{array}$

Substitute $n=10$

$\begin{array}{cccc}\Rightarrow & 10\left(10-1\right)\left(10-2\right)\left(10-3\right)\left(10-4\right)& =& 30240\\ \Rightarrow & 10\times 9\times 8\times 7\times 6& =& 30240\\ \Rightarrow & 30240& =& 30240\end{array}$

$\Rightarrow$ L.H.S. $=$ R.H.S.

Thus the value of $n$is $10$

Therefore the ordered pair $\left(n,r\right)=\left(10,5\right)$

Hence the correct option is option(B).