If (n+2C6/n-2P2) = 11, then n satisfies the equation :

Question

If (n+2C6/n-2P2) = 11, then n satisfies the equation : (A) n2 + 3n - 108 =0  (B) n2 + 5n - 84 =0  (C) n2 + 2n - 80 =0  (D) n2 + n - 110 =0

Tardigrade Admin · Accepted Answer

( n +2 C 6/ n -2 P 2) =11 ⇒ (( n +2) !/6 !( n -4) !)=11 .(( n -2) !/( n -4) !) ⇒ ( n +2) !=11 . 6 !( n -2) ! ⇒ ( n +2)( n +1) n ( n -1)=11.6 ! ⇒ ( n +2)( n +1) n ( n -1)=11.10 .9 .8 ⇒ n +2=11 ⇒ n =9 Which satifies the n2+3 n-108=0

Q. If $\frac{^{n + 2} C _{6}}{^{n - 2} P _{2}} = 11$ , then $n$ satisfies the equation :

$n^{2} + 3 n - 108 = 0$

$n^{2} + 5 n - 84 = 0$

$n^{2} + 2 n - 80 = 0$

$n^{2} + n - 110 = 0$

Q. If n−2P2​n+2C6​​=11, then n satisfies the equation :

n2+3n−108=0

n2+5n−84=0

n2+2n−80=0

n2+n−110=0

n−2P2​n+2C6​​=11 ⇒6!(n−4)!(n+2)!​=11.(n−4)!(n−2)!​ ⇒(n+2)!=11.6!(n−2)! ⇒(n+2)(n+1)n(n−1)=11.6! ⇒(n+2)(n+1)n(n−1)=11.10.9.8 ⇒n+2=11 ⇒n=9 Which satifies the n2+3n−108=0

Q. If $\frac{^{n + 2} C _{6}}{^{n - 2} P _{2}} = 11$ , then $n$ satisfies the equation :

$n^{2} + 3 n - 108 = 0$

$n^{2} + 5 n - 84 = 0$

$n^{2} + 2 n - 80 = 0$

$n^{2} + n - 110 = 0$