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Question: If \({}^n{P_r} = 840\) , \({}^n{C_r} = 35\) then find the value of n- A.\(6\) B.\(7\) C.\(8\...

If nPr=840{}^n{P_r} = 840 , nCr=35{}^n{C_r} = 35 then find the value of n-
A.66
B.77
C.88
D.99

Explanation

Solution

Use the following formulae- nPr=n!n−r!{}^n{P_r} = \dfrac{{n!}}{{n - r!}} and nCr=n!r!n−r!{}^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.

Complete step-by-step answer:
Given, nPr=840{}^n{P_r} = 840- (i)
And nCr=35{}^n{C_r} = 35- (ii)
We have to find the value of n.
We know that- nPr=n!n−r!{}^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,
⇒n!n−r!=840\Rightarrow \dfrac{{n!}}{{n - r!}} = 840 - (iii)
Also nCr=n!r!n−r!{}^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,
⇒n!r!n−r!=35\Rightarrow \dfrac{{n!}}{{r!n - r!}} = 35 - (iv)
On substituting the value from eq. (iii) to eq. (iv), we get,
⇒840r!=35\Rightarrow \dfrac{{840}}{{r!}} = 35
On cross multiplication we get,
⇒\Rightarrow r!=84035r! = \dfrac{{840}}{{35}}
On dividing the numerator by denominator, we get
⇒r!=24\Rightarrow r! = 24
We can break 2424 into its factors then,
⇒r!=4×3×2×1=4!\Rightarrow r! = 4 \times 3 \times 2 \times 1 = 4!
This means that r=44
On substituting the value of r in eq. (i)
⇒n!n−4!=840\Rightarrow \dfrac{{n!}}{{n - 4!}} = 840
On opening factorial of numerator we get,
⇒n(n−1)(n−2)(n−3)(n−4)!n−4!=840\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,
⇒n(n−1)(n−2)(n−3)=840\Rightarrow n\left( {n - 1} \right)\left( {n - 2} \right)\left( {n - 3} \right) = 840
On breaking 840840 into factors, we get
⇒n(n−1)(n−2)(n−3)=42×20\Rightarrow n\left( {n - 1} \right)\left( {n - 2} \right)\left( {n - 3} \right) = 42 \times 20
On further breaking the factors we get,
⇒n(n−1)(n−2)(n−3)=7×6×5×4\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-
⇒n=7\Rightarrow n = 7
Hence, the correct option is ‘B’.

Note: We can also find the value of r in above question using the formula-
⇒nPr=nCr×r!\Rightarrow {}^n{P_r} = {}^n{C_r} \times r!
We can directly obtain value of r by putting the given values-
⇒r!=84035=24\Rightarrow r! = \dfrac{{840}}{{35}} = 24
We can then solve the question in the same manner as we solved in the above solution.