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Question: A set contains \(2n+1\) elements. The number of subsets of this set containing more than \(n\) eleme...

A set contains 2n+12n+1 elements. The number of subsets of this set containing more than nn elements is equal to
A. 2n−1{{2}^{n-1}}
B. 2n{{2}^{n}}
C. 2n+1{{2}^{n+1}}
D. 22n{{2}^{2n}}

Explanation

Solution

We express the summation form in the combination form. Therefore, number of ways the choosing can be done for nn to 2n+12n+1 elements out of 2n+12n+1 elements is 2n+1Cr,r=n(1)2n+1{}^{2n+1}{{C}_{r}},r=n\left( 1 \right)2n+1. We take the sum and use the theorems ∑r=0nnCr=2n\sum\limits_{r=0}^{n}{{}^{n}{{C}_{r}}}={{2}^{n}} and nCr=nCn−r{}^{n}{{C}_{r}}={}^{n}{{C}_{n-r}}.

Complete answer: A set contains 2n+12n+1 elements. We need to find the number of subsets of this set containing more than nn elements.
So, we have to choose the number of elements we are taking.
We can take from nn to 2n+12n+1.
The choosing can be done in combination.
So, number of ways the choosing can be done for nn to 2n+12n+1 number of elements out of 2n+12n+1 elements is 2n+1Cn,2n+1Cn+1,2n+1Cn+2,2n+1Cn+3,.........,2n+1C2n+1{}^{2n+1}{{C}_{n}},{}^{2n+1}{{C}_{n+1}},{}^{2n+1}{{C}_{n+2}},{}^{2n+1}{{C}_{n+3}},.........,{}^{2n+1}{{C}_{2n+1}}.
Total will be 2n+1Cn+2n+1Cn+1+2n+1Cn+2+.........+2n+1C2n+1=∑r=n2n+12n+1Cr{}^{2n+1}{{C}_{n}}+{}^{2n+1}{{C}_{n+1}}+{}^{2n+1}{{C}_{n+2}}+.........+{}^{2n+1}{{C}_{2n+1}}=\sum\limits_{r=n}^{2n+1}{{}^{2n+1}{{C}_{r}}}.
We know the binomial theorem (1+x)n=∑r=0nnCrxr{{\left( 1+x \right)}^{n}}=\sum\limits_{r=0}^{n}{{}^{n}{{C}_{r}}{{x}^{r}}}. We put the value of x=1x=1.
We get ∑r=0nnCr=2n\sum\limits_{r=0}^{n}{{}^{n}{{C}_{r}}}={{2}^{n}}. We also know that nCr=nCn−r{}^{n}{{C}_{r}}={}^{n}{{C}_{n-r}}.
We now can express ∑r=02n+12n+1Cr=2n+1C0+2n+1C1+.........+2n+1C2n+2n+1C2n+1\sum\limits_{r=0}^{2n+1}{{}^{2n+1}{{C}_{r}}}={}^{2n+1}{{C}_{0}}+{}^{2n+1}{{C}_{1}}+.........+{}^{2n+1}{{C}_{2n}}+{}^{2n+1}{{C}_{2n+1}}.

& {}^{2n+1}{{C}_{0}}+{}^{2n+1}{{C}_{1}}+.....+{}^{2n+1}{{C}_{n}}+{}^{2n+1}{{C}_{n+1}}+{}^{2n+1}{{C}_{n+2}}+....+{}^{2n+1}{{C}_{2n}}+{}^{2n+1}{{C}_{2n+1}} \\\ & ={}^{2n+1}{{C}_{2n+1}}+{}^{2n+1}{{C}_{2n}}+.....+{}^{2n+1}{{C}_{n+1}}+{}^{2n+1}{{C}_{n+1}}+{}^{2n+1}{{C}_{n+2}}+....+{}^{2n+1}{{C}_{2n}}+{}^{2n+1}{{C}_{2n+1}} \\\ & =2\left( {}^{2n+1}{{C}_{n+1}}+{}^{2n+1}{{C}_{n+2}}+....+{}^{2n+1}{{C}_{2n}}+{}^{2n+1}{{C}_{2n+1}} \right) \\\ \end{aligned}$$ We replace the values for $n$ with $2n+1$ in $$\sum\limits_{r=0}^{n}{{}^{n}{{C}_{r}}}={{2}^{n}}$$ to get $$\sum\limits_{r=0}^{2n+1}{{}^{2n+1}{{C}_{r}}}={{2}^{2n+1}}$$. So, $$2\left( {}^{2n+1}{{C}_{n+1}}+{}^{2n+1}{{C}_{n+2}}+....+{}^{2n+1}{{C}_{2n}}+{}^{2n+1}{{C}_{2n+1}} \right)=\sum\limits_{r=0}^{2n+1}{{}^{2n+1}{{C}_{r}}}={{2}^{2n+1}}$$. This gives $${}^{2n+1}{{C}_{n+1}}+{}^{2n+1}{{C}_{n+2}}+....+{}^{2n+1}{{C}_{2n}}+{}^{2n+1}{{C}_{2n+1}}=\dfrac{{{2}^{2n+1}}}{2}={{2}^{2n}}$$ **Therefore the correct option is D.** **Note:** There are some constraints in the form of ${}^{n}{{C}_{r}}=\dfrac{n!}{r!\times \left( n-r \right)!}$. The general conditions are $n\ge r\ge 0;n\ne 0$. Also, we need to remember the fact that the notion of choosing r objects out of n objects is exactly equal to the notion of choosing $\left( n-r \right)$ objects out of n objects. The mathematical expression is ${}^{n}{{C}_{r}}=\dfrac{n!}{r!\times \left( n-r \right)!}={}^{n}{{C}_{n-r}}$.