Question

A set contains $\left( {2n + 1} \right)$ elements. If the number of subsets of this set which contains almost $n$ elements is $4096$, then the value of $n$ is ?

Answer 1

In order to find the value of $n$, list out the combinations possible from the subsets of the set of $\left( {2n + 1} \right)$ elements. Use the Binomial expansion rule to sort out the expansion to obtain the value equal to the combination of subsets. Compare it with $4096$, cancel the common terms and get the results.

Formula used:
1. ${\left( {1 + x} \right)^n}{ = ^n}{C_0}{x^0}{ + ^n}{C_1}{x^1}{ + ^n}{C_2}{x^2} + ............{ + ^n}{C_n}{x^n}$
2. $^n{C_r}{ = ^n}{C_{n - r}}$

Complete step by step answer:
We are given that a set contains $\left( {2n + 1} \right)$ elements in which the subsets have $n$ elements which must be starting from zero. So, the combination of subsets that can be formed are as follows: $^{2n + 1}{C_0}{ + ^{2n + 1}}{C_1}{ + ^{2n + 1}}{C_2} + ............{ + ^{2n + 1}}{C_n}$ which is given equal to $4096$, which can be written as:
$^{2n + 1}{C_0}{ + ^{2n + 1}}{C_1}{ + ^{2n + 1}}{C_2} + ............{ + ^{2n + 1}}{C_n} = 4096$ …….(1)

From the basic formula of Binomial expansion theorem, we know that: ${\left( {1 + x} \right)^n}{ = ^n}{C_0}{x^0}{ + ^n}{C_1}{x^1}{ + ^n}{C_2}{x^2} + ............{ + ^n}{C_n}{x^n}$
Since, we have $\left( {2n + 1} \right)$ elements instead of $n$ elements, so replacing $n$ with $\left( {2n + 1} \right)$ in the upper expansion, and we get:
${\left( {1 + x} \right)^{2n + 1}}{ = ^{2n + 1}}{C_0}{x^0}{ + ^{2n + 1}}{C_1}{x^1}{ + ^{2n + 1}}{C_2}{x^2} + ............{ + ^{2n + 1}}{C_{2n + 1}}{x^{2n + 1}}$
Now, taking the value of $x = 1$ in order to remove the variable from the equation and only $1$ because any power of $1$ gives $1$ only and also the numbers multiplied to $1$ returns the same value:
So, our equation becomes:
${\left( {1 + 1} \right)^{\left( {2n + 1} \right)}}{ = ^{2n + 1}}{C_0}{.1^0}{ + ^{2n + 1}}{C_1}{.1^1}{ + ^{2n + 1}}{C_2}{.1^2} + ............{ + ^{2n + 1}}{C_{2n + 1}}{.1^{2.1 + 1}}$
$\Rightarrow {\left( 2 \right)^{\left( {2n + 1} \right)}}{ = ^{2n + 1}}{C_0}{.1^0}{ + ^{2n + 1}}{C_1}{.1^1}{ + ^{2n + 1}}{C_2}{.1^2} + ............{ + ^{2n + 1}}{C_{2n + 1}}{.1^3}$ $\Rightarrow {\left( 2 \right)^{\left( {2n + 1} \right)}}{ = ^{2n + 1}}{C_0}{ + ^{2n + 1}}{C_1}{ + ^{2n + 1}}{C_2} + ............{ + ^{2n + 1}}{C_{2n + 1}}$ ……..(2)

Since, from the concepts of Permutation and combinations, we know that:
$^n{C_r}{ = ^n}{C_{n - r}}$
So, from this the equations obtained are:
$^{2n + 1}{C_0}{ = ^{2n + 1}}{C_{2n + 1}}$ , $^{2n + 1}{C_1}{ = ^{2n + 1}}{C_{2n}}$ , $^{2n + 1}{C_2}{ = ^{2n + 1}}{C_{2n - 1}}$ , and $^{2n + 1}{C_{n + 1}}{ = ^{2n + 1}}{C_n}$
Changing only one term of same terms in sequence in the equation (2), and we get:
${2^{2n + 1}} = {2.^{2n + 1}}{C_0} + {2.^{2n + 1}}{C_1} + {2.^{2n + 1}}{C_2} + ............ + {2.^{2n + 1}}{C_n}$
As on the right side there are two same values, so taking two common:
${2^{2n + 1}} = 2\left( {^{2n + 1}{C_0}{ + ^{2n + 1}}{C_1}{ + ^{2n + 1}}{C_2} + ............{ + ^{2n + 1}}{C_n}} \right)$
Dividing both the sides by $2$ and we get:
$\dfrac{{{2^{2n}}2}}{2} = \dfrac{2}{2}\left( {^{2n + 1}{C_0}{ + ^{2n + 1}}{C_1}{ + ^{2n + 1}}{C_2} + ............{ + ^{2n + 1}}{C_n}} \right)$
$\Rightarrow {2^{2n}} = \left( {^{2n + 1}{C_0}{ + ^{2n + 1}}{C_1}{ + ^{2n + 1}}{C_2} + ............{ + ^{2n + 1}}{C_n}} \right)$ ……..(3)

Since, the right-hand side matches with the equation (1):
So, from eq (1) and (3) we get:
${2^{2n}} = 4096$
The value of $4096$ is the same as $2$ for $12$ times, so it can be written as $4096 = {2^{12}}$.
Substituting this value in the above equation, and we get:
${2^{2n}} = {2^{12}}$
Since, the base is same so comparing the powers, and we get:
$2n = 12$
Dividing both the sides by $2$:
$\dfrac{{2n}}{2} = \dfrac{{12}}{2}$
$\therefore n = 6$

Therefore, the value of $n$ is $6$.

Note: We have taken $x = 1$ in order to get the base as $2$ because the value of $4096$ was to be expressed in terms of $2$. The rule applied in ${2^{2n + 1}} = {2^{2n}}{.2^1}$ is according to the law of radicals, which is ${a^p}.{a^q} = {a^{p + q}}$.Remember to use the binomial expansion rule for sorting out the expansion, any other methods for expansion cannot not be used.