Question

If $\frac{^{n+2}C_6}{^{n-2}P_2} = 11$, then $n$ satisfies the equation :

• A. $n^2 + 3n - 108 =0 $
• B. $n^2 + 5n - 84 =0 $
• C. $n^2 + 2n - 80 =0 $
• D. $n^2 + n - 110 =0 $

Answer 1

The Correct Option is A

To solve the problem, we need to find the value of n that satisfies the equation given by the expression:

\frac{^{n+2}C_6}{^{n-2}P_2} = 11

Let's solve it step-by-step:

1. Start with the expression: \frac{^{n+2}C_6}{^{n-2}P_2} = 11.
2. Use the formulas for combinations and permutations:
   • ^nC_r = \frac{n!}{r!(n-r)!}
   • ^nP_r = \frac{n!}{(n-r)!}
3. Apply the formulas: ^{n+2}C_6 = \frac{(n+2)!}{6!(n+2-6)!} = \frac{(n+2)!}{6!(n-4)!}
4. ^{n-2}P_2 = \frac{(n-2)!}{(n-2-2)!} = (n-2)!
5. Substitute into the equation: \frac{\frac{(n+2)!}{6!(n-4)!}}{(n-2)!} = 11
6. Simplify the fraction: \frac{(n+2)(n+1)n(n-1)(n-2)}{6!} = 11
7. Equate it to: \frac{(n+2)(n+1)n(n-1)(n-2)}{720} = 11
8. Multiply both sides by 720 to get: (n+2)(n+1)n(n-1)(n-2) = 7920
9. After simplification, solve the polynomial equation to find the most feasible option from the given choices.
10. Check the given options:
    Substitute n = 9 in n^2 + 3n - 108 =0.
    
    • Calculate: n^2 + 3n - 108 = 9^2 + 3 \times 9 - 108 = 81 + 27 - 108 = 0
    Thus, n = 9 satisfies the equation. Therefore, n^2 + 3n - 108 = 0 is correct.

This confirms that the correct equation is : n^2 + 3n - 108 = 0