First slide

Combinations

Question

If $^{15} C_{r + 1} :^{15} C_{3 r} = 3 : 11$ then value of $r$ is

Moderate

A

2
B

3
C

4
D

5

Solution

Note that $0 \leq 3 r \leq 15$
$\Rightarrow 0 \leq r \leq 5 . Also, \frac{15!}{(14 - r)! (r + 1)!} \times \frac{(3 r)! (15 - 3 r)!}{15!} = \frac{3}{11}$
$\Rightarrow \frac{(3 r)! (15 - 3 r)!}{(14 - r)! (r + 1)!} = \frac{3}{11} 1$
For the denominator, to be divisible by $11, 11 ∣ (14 - r)!$
Setting $14 - r = 11 .$
$\Rightarrow r = 3$
Putting $r = 3,$ in LHS of (1) we get
LHS of $(1) \frac{9! 6!}{11! 4!} = \frac{6 \times 5}{11 \times 10} = \frac{3}{11} =$ = RHS of (1)
$∴ r = 3$

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