First slide

Combinations

Question

The number of ordered pairs (r,k) for which $6.^{35} C_{r} = (k^{2} - 3) .^{36} C_{r + 1}$ , where k is an integer is:

Moderate

A

6
B

2
C

3
D

4

Solution

$W e h a v e \frac{36}{r + 1} \times {}^{35}C_{r} (k^{2} - 3) = 6 {}^{35}C_{r} (∵ {}^{n + 1}C_{r + 1} = \frac{n + 1}{r + 1} {}^{n}C_{r})$
$\Rightarrow k^{2} - 3 = \frac{r + 1}{6} \Rightarrow k^{2} = 3 + \frac{r + 1}{6}$
$\Rightarrow r = 5, 35$
For r = 5, $k = \pm 2$
and for $r = 35, k = \pm 3$
$∴ T h e n u m b e r o f o r d e r e d p a i r s (r, k) = 4$

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