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  • If nCr – 1 = 36, nCr = 84 and nCr + 1 = 126, then find rC2.

If  ^nC_{r - 1} = 36, ^nC_{r } = 84and ^nC_{r +1} = 126, then find ^rC_{2}.

[Hint: From equation \frac{{ }^{n} C_{r}}{{ }^{n} C_{r+1}} \text { and } \frac{{ }^{n} C_{r}}{{ }^{n} C_{r-1}} using  to find the value of r.]

Answers (1)

\\\frac{n_{C_{r}}}{n_{C_{r-1}}}=\frac{n-r+1}{r}=\frac{84}{36}=\frac{7}{3} \\\\ 3 n-3 r+3=7 r \\ 10 r-3 n=3 \ldots \ldots . .(i ) \\\\ \frac{n_{C_{r+1}}}{n_{C_{r}}}=\frac{n-(r+1)+1}{r+1}=\frac{126}{84} \\ \frac{n-r}{r+1}=\frac{3}{2} \\ 2 n-2 r=3 r+3 \\ 2 n-5 r=3 \ldots \ldots . .(i i)

Equation (i) + 2 Equation(ii)

10r - 3n +4n - 10r = 3 + 6

n = 9

r =3

^rC_2 = ^3C_2 = 3

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infoexpert21

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