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Question: If \(12P_{r} = 11880\), then \(\sum_{r = 1}^{r}{rC_{i}}\) equals...

If 12Pr=1188012P_{r} = 11880, then ∑r=1rrCi\sum_{r = 1}^{r}{rC_{i}} equals

A

16

B

15

C

32

D

31

Answer

16

Explanation

Solution

12!(12−r)!\frac{12!}{(12 - r)!} = 11880 = 12 x 11 x 10 x 9

⇒ 12!(12−r) 6mu!= 6mu12!8!\frac{12!}{(12 - r)\mspace{6mu}!} = \mspace{6mu}\frac{12!}{8!}

⇒ r = 4

∴ Required sum = ∑i=1rrCi=4C1+4C2+4C3+4C4\sum_{i = 1}^{r}{rC_{i} =^{4}C_{1} +^{4}C_{2} +^{4}C_{3} +^{4}C_{4}}

= 24 – 1 = 15