Question

Find $n$ and $r$, if ${}^{n}{{P}_{r}}=720$ and ${}^{n}{{C}_{r}}=120$.

Answer 1

Hint: The given question is based on permutations and combinations. In order to solve this question we need to apply the formulas of permutation and combinations, which are given by
\[{}^{n}{{P}_{r}}=\dfrac{n!}{\left( n-r \right)!}\]
${}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$
Where, $n=$ total number of objects and
$r=$ number of objects to be selected.
Then, substituting the values given in the question and solving further we get the desired answer.

Complete step-by-step answer:
We have been given that ${}^{n}{{P}_{r}}=720$ and ${}^{n}{{C}_{r}}=120$.
We have to find the values of $n$ and $r$.
Now, we know that the formula of permutation is given by \[{}^{n}{{P}_{r}}=\dfrac{n!}{\left( n-r \right)!}\]
Substituting the value, we have \[720=\dfrac{n!}{\left( n-r \right)!}..........(i)\]
Also, we know that the formula of combination is given by ${}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$
Substituting the value, we have $120=\dfrac{n!}{r!\left( n-r \right)!}$
Or we can write that $120\times r!=\dfrac{n!}{\left( n-r \right)!}............(ii)$
Now, when we compare equation (i) and equation (ii), we analyze that R.H.S. of both the equations is equal so the L.H.S. of both the equations will also be equal,
So, we have $120\times r!=720$
Or we can write that
$\begin{align}
  & r!=\dfrac{720}{120} \\
 & r!=6 \\
\end{align}$
Now, we can write $6$ as $3\times 2\times 1$ which is equal to $3!$,
So, we have $r!=3!$
Or $r=3$ .
Now, substituting the value of $r$ into equation (i), we get
\[720=\dfrac{n!}{\left( n-3 \right)!}\]
Now, we know that we can write $n!=n\left( n-1 \right)\left( n-2 \right)\left( n-3 \right)!$
Now, we have \[720=\dfrac{n\left( n-1 \right)\left( n-2 \right)\left( n-3 \right)!}{\left( n-3 \right)!}\]
On solving we get
\[720=n\left( n-1 \right)\left( n-2 \right)\]
Now, we can factorize $720$ as $10\times 9\times 8$
So, we have \[n\left( n-1 \right)\left( n-2 \right)=10\times 9\times 8\]
On comparing we get $n=10$.
The value of $r=3$ and $n=10$.

Note: Alternatively one can directly calculate the value of $r$ by using the relation ${}^{n}{{P}_{r}}={}^{n}{{C}_{r}}\times r!$.
Substituting the values and solving further to get the value of $r$ and solving the remaining part as we solved above to get the value of $n$.