Solveeit Logo
Login

Question

Question: If \(\dfrac{{{}^{n + 2}{C_6}}}{{{}^{n - 2}{P_2}}} = 11\) then n satisfies the equation A.\({n^2} +...

If n+2C6n−2P2=11\dfrac{{{}^{n + 2}{C_6}}}{{{}^{n - 2}{P_2}}} = 11 then n satisfies the equation
A.n2+n−110=0{n^2} + n - 110 = 0
B.n2+2n−80=0{n^2} + 2n - 80 = 0
C.n2+3n−108=0{n^2} + 3n - 108 = 0
D.n2+5n−84=0{n^2} + 5n - 84 = 0

Explanation

Solution

By using the formula of permutation nPr=n!(n−r)!{}^n{P_r} = \dfrac{{n!}}{{(n - r)!}}and combination nCr=n!r!(n−r)!{}^n{C_r} = \dfrac{{n!}}{{r!(n - r)!}}in the given expression we can find the value of n. By substituting that value in the given options we can find the equation satisfied by that value.

Complete step-by-step answer:
We are given that n+2C6n−2P2=11\dfrac{{{}^{n + 2}{C_6}}}{{{}^{n - 2}{P_2}}} = 11
We know the formula of nCr=n!r!(n−r)!{}^n{C_r} = \dfrac{{n!}}{{r!(n - r)!}}
Here n = n+2 and r = 6
We get ,
⇒n+2C6=(n+2)!6!(n+2−6)! ⇒n+2C6=(n+2)(n+1)n(n−1)(n−2)(n−3)(n−4)!6∗5∗4∗3∗2∗1∗(n−4)! ⇒n+2C6=(n+2)(n+1)n(n−1)(n−2)(n−3)6∗5∗4∗3∗2∗1  \Rightarrow {}^{n + 2}{C_6} = \dfrac{{(n + 2)!}}{{6!(n + 2 - 6)!}} \\\ \Rightarrow {}^{n + 2}{C_6} = \dfrac{{(n + 2)(n + 1)n(n - 1)(n - 2)(n - 3)(n - 4)!}}{{6*5*4*3*2*1*(n - 4)!}} \\\ \Rightarrow {}^{n + 2}{C_6} = \dfrac{{(n + 2)(n + 1)n(n - 1)(n - 2)(n - 3)}}{{6*5*4*3*2*1}} \\\
We know the formula of nPr=n!(n−r)!{}^n{P_r} = \dfrac{{n!}}{{(n - r)!}}
Here n = n – 2 and r = 2
⇒n−2P2=(n−2)!(n−2−2)! ⇒n−2P2=(n−2)(n−3)(n−4)!(n−4)! ⇒n+2C6=(n−2)(n−3)  \Rightarrow {}^{n - 2}{P_2} = \dfrac{{(n - 2)!}}{{(n - 2 - 2)!}} \\\ \Rightarrow {}^{n - 2}{P_2} = \dfrac{{(n - 2)(n - 3)(n - 4)!}}{{(n - 4)!}} \\\ \Rightarrow {}^{n + 2}{C_6} = (n - 2)(n - 3) \\\
Substituting the values in n+2C6n−2P2=11\dfrac{{{}^{n + 2}{C_6}}}{{{}^{n - 2}{P_2}}} = 11
⇒(n+2)(n+1)n(n−1)(n−2)(n−3)6∗5∗4∗3∗2∗1(n−2)(n−3)=11 ⇒(n+2)(n+1)n(n−1)6∗5∗4∗3∗2∗1=11 ⇒(n+2)(n+1)n(n−1)=11∗2∗3∗5∗4∗3∗2 ⇒(n+2)(n+1)n(n−1)=11∗10∗9∗8  \Rightarrow \dfrac{{\dfrac{{(n + 2)(n + 1)n(n - 1)(n - 2)(n - 3)}}{{6*5*4*3*2*1}}}}{{(n - 2)(n - 3)}} = 11 \\\ \Rightarrow \dfrac{{(n + 2)(n + 1)n(n - 1)}}{{6*5*4*3*2*1}} = 11 \\\ \Rightarrow (n + 2)(n + 1)n(n - 1) = 11*2*3*5*4*3*2 \\\ \Rightarrow (n + 2)(n + 1)n(n - 1) = 11*10*9*8 \\\
From this we get that n = 9
Now lets substitute this value of n in the options given
n2+n−110=0{n^2} + n - 110 = 0
⇒92+9−110=81+9−110=90−110=−20\Rightarrow {9^2} + 9 - 110 = 81 + 9 - 110 = 90 - 110 = - 20
Therefore the value of n does not satisfy this equation
n2+2n−80=0{n^2} + 2n - 80 = 0
⇒92+2(9)−80=81+18−80=99−80=19\Rightarrow {9^2} + 2(9) - 80 = 81 + 18 - 80 = 99 - 80 = 19
Therefore the value of n does not satisfy this equation
n2+3n−108=0{n^2} + 3n - 108 = 0
⇒92+3(9)−108=81+27−108=108−108=0\Rightarrow {9^2} + 3(9) - 108 = 81 + 27 - 108 = 108 - 108 = 0
From this we get that the value of n satisfies this equation
Therefore the correct option is C

Note: Permutations and combinations, the various ways in which objects from a set may be selected, generally without replacement, to form subsets. This selection of subsets is called a permutation when the order of selection is a factor, a combination when order is not a factor.