If n-2 C6-n-2-11- then n satisfies the equation-A- n2-2 n-80-0B- n2-5 n-84-0C- n2-3 n-108-0D- n 2- n -110-0

The correct option is C n^2+3n 108=0 nbsp; nbsp;^n+2C_6/^n 2P_2=11 ⇒ ^n+2C_6=11.^n 2P_2 ⇒(n+2)!/6!(n 4)!=11.(n 2)!/(n 4)! ⇒ (n+2)(n+1)(n)(n 1)=11 × 6! ⇒ ...

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Byju's Answer

Standard XII

Mathematics

nCr Definitions and Properties

If n+2 C6/n-2...

Question

$If \frac{^{n + 2} C_{6}}{^{n - 2} P_{2}} = 11$ , then $n$ satisfies the equation:

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

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

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

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n^{2} + n - 110 = 0

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Solution

The correct option is C $n^{2} + 3 n - 108 = 0$
$\frac{^{n + 2} C_{6}}{^{n - 2} P_{2}} = 11 \Rightarrow^{n + 2} C_{6} = {11.}^{n - 2} P_{2} \Rightarrow \frac{(n + 2)!}{6! (n - 4)!} = 11. \frac{(n - 2)!}{(n - 4)!} \Rightarrow (n + 2) (n + 1) (n) (n - 1) = 11 \times 6! \Rightarrow (n + 2) (n + 1) (n) (n - 1) = 11 \times 10 \times 9 \times 8 \Rightarrow n = 9 \to which satisfies n^{2} + 3 n - 108 = 0$

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Similar questions

Q. If

^{n^{2} - n} C_{2} =^{n^{2} - n} C_{10}

, then

$n^{t h}$ term of the series 2 + 4 + 7 + 11 + ...... will be

lim n \to \infty (\frac{n}{n^{2} + 1^{2}} + \frac{n}{n^{2} + 2^{2}} + \frac{n}{n^{2} + 3^{2}} . . . . . . \frac{1}{5 n})

Q. If

S = {tan}^{- 1} (\frac{1}{n^{2} + n + 1}) + {tan}^{- 1} (\frac{1}{n^{2} + 3 n + 3}) + \dots + {tan}^{- 1} (\frac{1}{1 + (n + 19) (n + 20)})

then

tan S

is equal to

{lim}_{n \to \infty} (\frac{n}{n^{2}} + \frac{n}{n^{2} + 1} + \frac{n}{n^{2} + 2^{2}} + . . . . . . + \frac{n}{2 n^{2} - 2 n + 1})

is equal to

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If n+2C6n−2P2=11, then n satisfies the equation:

The correct option is C n2+3n−108=0 n+2C6n−2P2=11⇒n+2C6=11.n−2P2⇒(n+2)!6!(n−4)!=11.(n−2)!(n−4)!⇒(n+2)(n+1)(n)(n−1)=11×6!⇒(n+2)(n+1)(n)(n−1)=11×10×9×8⇒n=9→which satisfies n2+3n−108=0

$If \frac{^{n + 2} C_{6}}{^{n - 2} P_{2}} = 11$ , then $n$ satisfies the equation: