Question

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

Answer 1

Hint: We express the summation form in the combination form. Therefore, number of ways the choosing can be done for $n$ to $2n+1$ elements out of $2n+1$ elements is ${}^{2n+1}{{C}_{r}},r=n\left( 1 \right)2n+1$. We take the sum and use the theorems \[\sum\limits_{r=0}^{n}{{}^{n}{{C}_{r}}}={{2}^{n}}\] and \[{}^{n}{{C}_{r}}={}^{n}{{C}_{n-r}}\].

Complete answer: A set contains $2n+1$ elements. We need to find the number of subsets of this set containing more than $n$ elements.
So, we have to choose the number of elements we are taking.
We can take from $n$ to $2n+1$.
The choosing can be done in combination.
So, number of ways the choosing can be done for $n$ to $2n+1$ number of elements out of $2n+1$ elements is ${}^{2n+1}{{C}_{n}},{}^{2n+1}{{C}_{n+1}},{}^{2n+1}{{C}_{n+2}},{}^{2n+1}{{C}_{n+3}},.........,{}^{2n+1}{{C}_{2n+1}}$.
Total will be \[{}^{2n+1}{{C}_{n}}+{}^{2n+1}{{C}_{n+1}}+{}^{2n+1}{{C}_{n+2}}+.........+{}^{2n+1}{{C}_{2n+1}}=\sum\limits_{r=n}^{2n+1}{{}^{2n+1}{{C}_{r}}}\].
We know the binomial theorem \[{{\left( 1+x \right)}^{n}}=\sum\limits_{r=0}^{n}{{}^{n}{{C}_{r}}{{x}^{r}}}\]. We put the value of $x=1$.
We get \[\sum\limits_{r=0}^{n}{{}^{n}{{C}_{r}}}={{2}^{n}}\]. We also know that \[{}^{n}{{C}_{r}}={}^{n}{{C}_{n-r}}\].
We now can express \[\sum\limits_{r=0}^{2n+1}{{}^{2n+1}{{C}_{r}}}={}^{2n+1}{{C}_{0}}+{}^{2n+1}{{C}_{1}}+.........+{}^{2n+1}{{C}_{2n}}+{}^{2n+1}{{C}_{2n+1}}\].
\[\begin{align}
  & {}^{2n+1}{{C}_{0}}+{}^{2n+1}{{C}_{1}}+.....+{}^{2n+1}{{C}_{n}}+{}^{2n+1}{{C}_{n+1}}+{}^{2n+1}{{C}_{n+2}}+....+{}^{2n+1}{{C}_{2n}}+{}^{2n+1}{{C}_{2n+1}} \\
 & ={}^{2n+1}{{C}_{2n+1}}+{}^{2n+1}{{C}_{2n}}+.....+{}^{2n+1}{{C}_{n+1}}+{}^{2n+1}{{C}_{n+1}}+{}^{2n+1}{{C}_{n+2}}+....+{}^{2n+1}{{C}_{2n}}+{}^{2n+1}{{C}_{2n+1}} \\
 & =2\left( {}^{2n+1}{{C}_{n+1}}+{}^{2n+1}{{C}_{n+2}}+....+{}^{2n+1}{{C}_{2n}}+{}^{2n+1}{{C}_{2n+1}} \right) \\
\end{align}\]
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 ${}^{n}{{C}_{r}}=\dfrac{n!}{r!\times \left( n-r \right)!}={}^{n}{{C}_{n-r}}$.