Question

If ${}^{10}{P_r}$= 604800 and ${}^{10}{C_r}$= 120. Then the value of r ?

Answer 1

Hint: Start by applying the formula for ${}^n{P_r}$ and ${}^n{C_r}$ , when ‘n’ distinct objects are taken ‘r’ at a time. Substitute the values and mark them as equation 1 and 2 respectively . Divide these two equations in order to find the value of \[r!\], and the value of r can be found by hit and trial method.

Complete step-by-step answer:
Given,
${}^{10}{P_r}$= 604800 and ${}^{10}{C_r}$= 120
We know , That the value of ${}^n{P_r}$ can be found by using the formula
${}^n{P_r} = \dfrac{{n!}}{{\left( {n - r} \right)!}}$
On comparison we get n = 10
Substituting this value in above formula , we get
${}^{10}{P_r} = \dfrac{{10!}}{{\left( {10 - r} \right)!}} = 604800 \to eqn.1$
We know , That the value of ${}^n{C_r}$ can be found by using the formula
${}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$
On comparison we get n = 10
Substituting this value in above formula , we get
${}^{10}{C_r} = \dfrac{{10!}}{{r!\left( {10 - r} \right)!}} = 120 \to eqn.2$
Now, dividing the eqn.1 by eqn. 2 , we get
$
  \dfrac{{{}^{10}{P_r}}}{{{}^{10}{C_r}}} = \dfrac{{\dfrac{{10!}}{{\left( {10 - r} \right)!}}}}{{\dfrac{{10!}}{{r!\left( {10 - r} \right)!}}}} = \dfrac{{604800}}{{120}} \\
   \Rightarrow \dfrac{{{}^{10}{P_r}}}{{{}^{10}{C_r}}} = \dfrac{1}{{\left( {\dfrac{1}{{r!}}} \right)}} = 5040 \\
   \Rightarrow r! = 5040 \\
$
Now , we need to find the value of r for which $r! = 5040$.
So by hit and trial method , we find that the value of r = 7 , satisfies the condition.
Therefore , The value of r is 7.

Note: Similar questions can be solved by following the same procedure as above. Sometimes one might get quadratic equations as well in order to find the solution , in that case use the discriminant rule or splitting the middle terms. Attention must be given while substituting the terms and also while cancelling the factorial terms. Also the value of r and n can never be negative .