Question

If ${}^n{P}_r=840$ , ${}^n{C}_r=35$ then find the value of n-
A.$6$
B.$7$
C.$8$
D.$9$

Answer 1

Hint: Use the following formulae- ${}^n{P}_r={\displaystyle \frac{n!}{n-r!}}$ and ${}^n{C}_r={\displaystyle \frac{n!}{r!n-r!}}$ where n=total number of things
And r=number of things to be selected. Put the given values and solve for n.

Complete step-by-step answer:
Given, ${}^n{P}_r=840$- (i)
And ${}^n{C}_r=35$- (ii)
We have to find the value of n.
We know that- ${}^n{P}_r={\displaystyle \frac{n!}{n-r!}}$ where n=total number of things and r=number of things to be selected.
On putting the given value we get in the formula we get,
$\Rightarrow {\displaystyle \frac{n!}{n-r!}}=840$ - (iii)
Also ${}^n{C}_r={\displaystyle \frac{n!}{r!n-r!}}$ where n=total number of things and r=number of things to be selected.
On putting the value in this formula we get,
$\Rightarrow {\displaystyle \frac{n!}{r!n-r!}}=35$ - (iv)
On substituting the value from eq. (iii) to eq. (iv), we get,
$\Rightarrow {\displaystyle \frac{840}{r!}}=35$
On cross multiplication we get,
$\Rightarrow$ $r!={\displaystyle \frac{840}{35}}$
On dividing the numerator by denominator, we get
$\Rightarrow r!=24$
We can break $24$ into its factors then,
$\Rightarrow r!=4\times 3\times 2\times 1=4!$
This means that r=$4$
On substituting the value of r in eq. (i)
$\Rightarrow {\displaystyle \frac{n!}{n-4!}}=840$
On opening factorial of numerator we get,
$\Rightarrow {\displaystyle \frac{n\left(n-1\right)\left(n-2\right)\left(n-3\right)\left(n-4\right)!}{n-4!}}=840$
On solving we get,
$\Rightarrow n\left(n-1\right)\left(n-2\right)\left(n-3\right)=840$
On breaking $840$ into factors, we get
$\Rightarrow n\left(n-1\right)\left(n-2\right)\left(n-3\right)=42\times 20$
On further breaking the factors we get,
 $\Rightarrow n\left(n-1\right)\left(n-2\right)\left(n-3\right)=7\times 6\times 5\times 4$
By observing the above equation we can see that the left hand side become equal to right hand side only when-
$\Rightarrow n=7$
Hence, the correct option is ‘B’.

Note: We can also find the value of r in above question using the formula-
$\Rightarrow {}^n{P}_r={}^n{C}_r\times r!$
We can directly obtain value of r by putting the given values-
$\Rightarrow r!={\displaystyle \frac{840}{35}}=24$
We can then solve the question in the same manner as we solved in the above solution.