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Question

Mathematics Question on Binomial theorem

A set contains (2n + 1) elements. If the number of subsets of this set which contain at most n elements is 4096, then the value of n is

A

6

B

15

C

21

D

None of these

Answer

6

Explanation

Solution

The number of subsets of the set which contain at most n elements is 2n+1C0+2n+1C1+2n+1C2+....+2n+1Cn=K(say)^{2n + 1}C_{0} + ^{2n + 1}C_{1} +^{ 2n + 1}C_{2} + .... + ^{2n + 1}C_{n} = K \left(say\right) We have 2K=2(2n+1C0+2n+1C1+2n+1C2+....+2n+1Cn)2K = 2 \left(^{2n + 1}C_{0 }+^{ 2n + 1}C_{1} +^{ 2n + 1}C_{2} + .... + ^{2n + 1}C_{n}\right) =(2n+1C0+2n+1C2n+1)+(2n+1C1+2n+1C2n)= \left(^{2n + 1}C_{0} + ^{2n + 1}C_{2n + 1}\right) + \left(^{2n + 1}C_{1} +^{ 2n + 1}C_{2n}\right) +...+(2n+1Cn+2n+1Cn+1)(∵nCr=nCn−r)+ ... + \left(^{2n + 1}C_{n} + ^{2n + 1}C_{n + 1}\right)\quad \left(\because ^{n}C_{r} = ^{n}C_{n -r}\right) =2n+1C0+2n+1C1+2n+1C2+....+2n+1C2n+1= ^{2n + 1}C_{0} + ^{2n + 1}C_{1} + ^{2n + 1}C_{2} + .... + ^{2n + 1}C_{2n + 1} =22n+1⇒K=22n= 2^{2n + 1} \Rightarrow K = 2^{2n}