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Question: A set contains \[\left( {2n + 1} \right)\] elements. The number of subsets of the set which contains...

A set contains (2n+1)\left( {2n + 1} \right) elements. The number of subsets of the set which contains at most nn elements is
A) 2n{2^n}
B) 2n+1{2^{n + 1}}
C) 22n−1{2^{2n - 1}}
D) 22n{2^{2n}}

Explanation

Solution

Here, we have to find the number of subsets. We will use the formula for finding the number of subsets of sets. Then by using the Binomial Theorem we will find the number of subsets for sets which contain (2n+1)\left( {2n + 1} \right) elements. We will then solve the equation further to find the number of subsets of the set.

Formula Used:
We will use the following formulas:

  1. The number of subsets of the set which contain at most nn elements is 2n+1C0+2n+1C1+2n+1C2+2n+1C3+.....+2n+1Cn=K{}^{2n + 1}{C_0} + {}^{2n + 1}{C_1} + {}^{2n + 1}{C_2} + {}^{2n + 1}{C_3} + ..... + {}^{2n + 1}{C_n} = K
  2. Property of Binomial Theorem is given by nC0+nC1+nC2+........+nCn=2n{}^n{C_0} + {}^n{C_1} + {}^n{C_2} + ........ + {}^n{C_n} = {2^n}
  3. Exponential rule: 1an=a−n\dfrac{1}{{{a^n}}} = {a^{ - n}}
  4. Exponential rule: am⋅an=am+n{a^m} \cdot {a^n} = {a^{m + n}}

Complete step by step solution:
We are given a set which contains (2n+1)\left( {2n + 1} \right) elements. We will find the number of subsets of the set which contains at most nn elements.
The number of subsets of the set which contain at most nn elements is given by 2n+1C0+2n+1C1+2n+1C2+2n+1C3+.....+2n+1Cn=K{}^{2n + 1}{C_0} + {}^{2n + 1}{C_1} + {}^{2n + 1}{C_2} + {}^{2n + 1}{C_3} + ..... + {}^{2n + 1}{C_n} = K
Multiplying by 22 on both the sides of the above equation, we get
⇒2K=2(2n+1C0+2n+1C1+2n+1C2+2n+1C3+.....+2n+1Cn)\Rightarrow 2K = 2\left( {{}^{2n + 1}{C_0} + {}^{2n + 1}{C_1} + {}^{2n + 1}{C_2} + {}^{2n + 1}{C_3} + ..... + {}^{2n + 1}{C_n}} \right)
By using the property of Binomial Coefficients, we get
⇒2K=(2n+1C0+2n+1C1+2n+1C2+2n+1C3+.....+2n+1Cn+2n+1Cn+1+.......+2n+1C2n+2n+1C2n+1)\Rightarrow 2K = \left( {{}^{2n + 1}{C_0} + {}^{2n + 1}{C_1} + {}^{2n + 1}{C_2} + {}^{2n + 1}{C_3} + ..... + {}^{2n + 1}{C_n} + {}^{2n + 1}{C_{n + 1}} + ....... + {}^{2n + 1}{C_{2n}} + {}^{2n + 1}{C_{2n + 1}}} \right)
By adding the first and last terms successively, we get
⇒2K=(2n+1C0+2n+1C2n+1)+(2n+1C1+2n+1C2n)+(2n+1C2+2n+1C2n−1)+......+(2n+1Cn+2n+1C2n+1)\Rightarrow 2K = \left( {{}^{2n + 1}{C_0} + {}^{2n + 1}{C_{2n + 1}}} \right) + \left( {{}^{2n + 1}{C_1} + {}^{2n + 1}{C_{2n}}} \right) + \left( {{}^{2n + 1}{C_2} + {}^{2n + 1}{C_{2n - 1}}} \right) + ...... + \left( {{}^{2n + 1}{C_n} + {}^{2n + 1}{C_{2n + 1}}} \right)
By using the property of binomial theorem nC0+nC1+nC2+........+nCn=2n{}^n{C_0} + {}^n{C_1} + {}^n{C_2} + ........ + {}^n{C_n} = {2^n}, we get
⇒2K=22n+1\Rightarrow 2K = {2^{2n + 1}}
By rewriting the equation, we get
⇒K=22n+12\Rightarrow K = \dfrac{{{2^{2n + 1}}}}{2}
By using the exponential rule 1an=a−n\dfrac{1}{{{a^n}}} = {a^{ - n}}, we get
⇒K=22n+1⋅2−1\Rightarrow K = {2^{2n + 1}} \cdot {2^{ - 1}}
Using the exponential rule am⋅an=am+n{a^m} \cdot {a^n} = {a^{m + n}}, we get
⇒K=22n+1−1\Rightarrow K = {2^{2n + 1 - 1}}
⇒K=22n\Rightarrow K = {2^{2n}}
Therefore, the number of subsets of the set which contains at most nn elements is 22n{2^{2n}}.

Thus, option (D) is the correct answer.

Note:
We know that the binomial coefficient uses the concept of combinations. In Binomial expansion, the number of terms is greater by 1 than the power of the binomial expansion. The sum of the exponents of a Binomial expansion is always mm which is the power of the binomial expansion. We should also use the trigonometric formula and ratios while solving the binomial expansion for the variable mm. Binomial coefficients are the integers which are coefficients in the Binomial Theorem.