JEE Main 2016 · 10 Apr · Shift 1 · Mathematics Q66

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MathematicsSingle correct+4 / -1Question 66 of 90

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

n^2+3n-108=0

n^2+5n-84=0

n^2+2n-80=0

n^2+n-110=0

Answer and worked solution

Official answer: A

$\frac{(n+2)!}{6!(n-4)!}\cdot\frac1{(n-2)(n-3)}=\frac{(n+2)(n+1)n(n-1)}{720}=11$ , so $(n+2)(n+1)n(n-1)=7920=11\cdot10\cdot9\cdot8$ and n = 9, which satisfies $n^2+3n-108=0$ .

Topics:Combinations Permutations

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JEE Main 2016 · 10 Apr · Shift 1, Mathematics question 66