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

Last updated date: 19th Jun 2024
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Hint: Use the following formulae- ${}^n{P_r} = \dfrac{{n!}}{{n - r!}}$ and ${}^n{C_r} = \dfrac{{n!}}{{r!n - r!}}$ where n=total number of things
And r=number of things to be selected. Put the given values and solve for n.

Given, ${}^n{P_r} = 840$- (i)
And ${}^n{C_r} = 35$- (ii)
We have to find the value of n.
We know that- ${}^n{P_r} = \dfrac{{n!}}{{n - r!}}$ 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,
$\Rightarrow \dfrac{{n!}}{{n - r!}} = 840$ - (iii)
Also ${}^n{C_r} = \dfrac{{n!}}{{r!n - r!}}$ where n=total number of things and r=number of things to be selected.
On putting the value in this formula we get,
$\Rightarrow \dfrac{{n!}}{{r!n - r!}} = 35$ - (iv)
On substituting the value from eq. (iii) to eq. (iv), we get,
$\Rightarrow \dfrac{{840}}{{r!}} = 35$
On cross multiplication we get,
$\Rightarrow$ $r! = \dfrac{{840}}{{35}}$
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)
$\Rightarrow \dfrac{{n!}}{{n - 4!}} = 840$
On opening factorial of numerator we get,
$\Rightarrow \dfrac{{n\left( {n - 1} \right)\left( {n - 2} \right)\left( {n - 3} \right)\left( {n - 4} \right)!}}{{n - 4!}} = 840$
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-
$\Rightarrow r! = \dfrac{{840}}{{35}} = 24$
We can then solve the question in the same manner as we solved in the above solution.