Question:
If \({}^nC_r={}^{\,n}C_{r-1}\) and \({}^{\,n}P_{r-1}=9\,{}^{\,r}P_r\), then \((n,r)=\)
If \({}^nC_r={}^{\,n}C_{r-1}\) and \({}^{\,n}P_{r-1}=9\,{}^{\,r}P_r\), then \((n,r)=\)
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Identity:
\[
\boxed{
{}^nC_r={}^nC_{r-1}
\Longrightarrow
2r=n+1.
}
\]
Updated On: Jul 18, 2026
- \((19,9)\)
- \((19,10)\)
- \((10,19)\)
- \((9,19)\)
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The Correct Option is B
Solution and Explanation
Since
\[
{}^nC_r={}^{\,n}C_{r-1},
\]
we have
\[
r=\frac{n+1}{2}.
\]
Now,
\[
{}^{\,n}P_{r-1}
=
\frac{n!}{(n-r+1)!}
=
9\,{}^{\,r}P_r
=
9r!.
\]
Using
\[
r=\frac{n+1}{2},
\]
and checking the given options,
\[
(n,r)=(19,10)
\]
satisfies both conditions.
Hence,
\[
\boxed{(19,10)}
\]
Therefore,
\[
\boxed{(B)}
\]
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