Question

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

Answer 1

Hint: By using the formula of permutation ${}^n{P_r} = \dfrac{{n!}}{{(n - r)!}}$and combination ${}^n{C_r} = \dfrac{{n!}}{{r!(n - r)!}}$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 $\dfrac{{{}^{n + 2}{C_6}}}{{{}^{n - 2}{P_2}}} = 11$
We know the formula of ${}^n{C_r} = \dfrac{{n!}}{{r!(n - r)!}}$
Here n = n+2 and r = 6
We get ,
$
   \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 ${}^n{P_r} = \dfrac{{n!}}{{(n - r)!}}$
Here n = n – 2 and r = 2
$
   \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 $\dfrac{{{}^{n + 2}{C_6}}}{{{}^{n - 2}{P_2}}} = 11$
$
   \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
 ${n^2} + n - 110 = 0$
$ \Rightarrow {9^2} + 9 - 110 = 81 + 9 - 110 = 90 - 110 = - 20$
Therefore the value of n does not satisfy this equation
${n^2} + 2n - 80 = 0$
$ \Rightarrow {9^2} + 2(9) - 80 = 81 + 18 - 80 = 99 - 80 = 19$
Therefore the value of n does not satisfy this equation
${n^2} + 3n - 108 = 0$
$ \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.