Question

If \({ }^{n} P_{r}:^{n} C_{r}=120\) then find \(n\) and \(r\)

Answer 1

Given that

\[^{n}P_{r}=\dfrac{n!}{(n-r)!}\]

and

\[^{n}C_{r}=\dfrac{n!}{r!(n-r)!}\]

then

\[^{n} P_{r}:^{n} C_{r}= \dfrac{\dfrac{n!}{(n-r)!}}{\dfrac{n!}{r!(n-r)!}}\]

\[= \dfrac{n!}{(n-r)!} \times \dfrac{r!(n-r)!}{n!} = r! = 120\]

i.e.

\[r\cdot (r-1) \cdot (r-2) \cdot (r-3) \ldots (2) \cdot (1) = 120\]

\[\therefore r!=5!=120\]

Given the nature of the relation between combinations and permutations, \(n\) could be anything.