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If ${ }^n P_r=30240$ and ${ }^n C_r=252$, then the ordered pair $(n, r)$ is equal to
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2942 Upvotes
Verified Answer
The correct answer is:
$(10,5)$
Given that, ${ }^n P_r=30240$ and ${ }^n C_r=252$
$$
\begin{aligned}
& \Rightarrow \frac{n !}{(n-r) !}=30240 \text { and } \frac{n !}{(n-r) ! r !}=252 \\
& \Rightarrow \quad r !=\frac{30240}{252}=120 \\
& \Rightarrow \quad r=5 \\
& \therefore \quad \frac{n !}{(n-5) !}=30240 \\
&
\end{aligned}
$$
$$
\begin{aligned}
& \Rightarrow \quad n(n-1)(n-2)(n-3)(n-4)=30240 \\
& \Rightarrow \quad n(n-1)(n-2)(n-3)(n-4) \\
& \quad=10(10-1)(10-2)(10-3)(10-4) \\
& \Rightarrow \quad n=10
\end{aligned}
$$
Hence, required ordered pair is $(10,5)$.
$$
\begin{aligned}
& \Rightarrow \frac{n !}{(n-r) !}=30240 \text { and } \frac{n !}{(n-r) ! r !}=252 \\
& \Rightarrow \quad r !=\frac{30240}{252}=120 \\
& \Rightarrow \quad r=5 \\
& \therefore \quad \frac{n !}{(n-5) !}=30240 \\
&
\end{aligned}
$$
$$
\begin{aligned}
& \Rightarrow \quad n(n-1)(n-2)(n-3)(n-4)=30240 \\
& \Rightarrow \quad n(n-1)(n-2)(n-3)(n-4) \\
& \quad=10(10-1)(10-2)(10-3)(10-4) \\
& \Rightarrow \quad n=10
\end{aligned}
$$
Hence, required ordered pair is $(10,5)$.
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