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

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Aicharya · Accepted Answer

Given that, ${ }^n P_r=30240$ and ${ }^n C_r=252$ $ \begin{aligned} & \Rightarrow \frac{n !}{(n-r) !}=30240 	ext { and } \frac{n !}{(n-r) ! r !}=252 \ & \Rightarrow \quad r !=\frac{30240}{252}=120 \ & \Rightarrow \quad r=5 \ & 	herefore \quad \frac{n !}{(n-5) !}=30240 \ & \end{aligned} $ $ \begin{aligned} & \Rightarrow \quad n(n-1)(n-2)(n-3)(n-4)=30240 \ & \Rightarrow \quad n(n-1)(n-2)(n-3)(n-4) \ & \quad=10(10-1)(10-2)(10-3)(10-4) \ & \Rightarrow \quad n=10 \end{aligned} $ Hence, required ordered pair is $(10,5)$

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

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