If n C r -1-36- n C r -84 and n C r -1-126 then r isA- 2B- 3C- none of theseD- 1

The correct option is B 3^nC_r 1=36, ^nC_r=84 and ^nC_r+1=126 We know that ^nC_r 1/^nC_r = r/n r+1 ∴36/84 = r/n r+1 or r/n r+1=3/7 or 3n 10r + 3 = 0 Also, ...

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Byju's Answer

Standard VII

Mathematics

nCr Definitions and Properties

If n C r -1=3...

Question

If $^{n} C_{r - 1} = 36,^{n} C_{r} = 84 a n d^{n} C_{r + 1} = 126$ then r is

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none of these

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Solution

The correct option is C 3
$^{n} C_{r - 1} = 36,^{n} C_{r} = 84 a n d^{n} C_{r + 1} = 126$
We know that
$\frac{^{n} C_{r - 1}}{^{n} C_{r}} = \frac{r}{n - r + 1} ∴ \frac{36}{84} = \frac{r}{n - r + 1} o r \frac{r}{n - r + 1} = \frac{3}{7} o r 3 n - 10 r + 3 = 0 A l s o, \frac{^{n} C_{r}}{^{n} C_{r + 1}} = \frac{r + 1}{n - r} = \frac{84}{126} = \frac{2}{3} o r 2 n - 5 r - 3 = 0 S o l v i n g (1) a n d (2), w e g e t n = 9 a n d r = 3$

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If nCr−1=36, nCr=84 and nCr+1=126 then r is

The correct option is C 3nCr−1=36, nCr=84 and nCr+1=126 We know that nCr−1nCr=rn−r+1∴3684=rn−r+1or rn−r+1=37or 3n−10r+3=0Also,nCrnCr+1=r+1n−r=84126=23or 2n−5r−3=0Solving (1) and (2), we get n=9 and r=3

If $^{n} C_{r - 1} = 36,^{n} C_{r} = 84 a n d^{n} C_{r + 1} = 126$ then r is