Answer

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**Hint:**

Observe that these are the terms of the binomial expression ${(a + b)^n}$ where $a = 1$, $b = 1$ and $n = 20$. We shall find the sum of the coefficients after using the formula. For further simplification use the concepts of combination.

**Complete step by step solution:**

We already know that, the binomial expansion of $a$ and $b$ raised to the power $n$ is given by

${(a + b)^n}{ = ^n}{C_0}{a^n}{b^0}{ + ^n}{C_1}{a^{n - 1}}{b^1} + ...{ + ^n}{C_{n - 1}}{a^1}{b^{n - 1}}{ + ^n}{C_n}{a^0}{b^n}$

Putting $a = 1$ , $b = 1$ and $n = 20$ in the above expression we get :

${(2)^n}{ = ^{20}}{C_0}{ + ^{20}}{C_1}{ + ^{20}}{C_2} + ...{ + ^{20}}{C_{20}}$ … (1)

Now, we have the sum of coefficients as the ${20}^{th}$ power of 2. We know that by the concept of combinations,$^n{C_r}{ = ^n}{C_{n - r}}$. So, using $^n{C_r}{ = ^n}{C_{n - r}}$ in equation (1), we get:

$

{(2)^{20}} = 2 \times {[^{20}}{C_0}{ + ^{20}}{C_1}{ + ^{20}}{C_2} + ...{ + ^{20}}{C_9}{ + ^{20}}{C_{10}}]{ - ^{20}}{C_{10}} \\

\Rightarrow {(2)^{20}}{ + ^{20}}{C_{10}} = 2 \times {[^{20}}{C_0}{ + ^{20}}{C_1}{ + ^{20}}{C_2} + ...{ + ^{20}}{C_9}{ + ^{20}}{C_{10}}] \\

\Rightarrow \dfrac{1}{2}[{(2)^{20}}{ + ^{20}}{C_{10}}] = {[^{20}}{C_0}{ + ^{20}}{C_1}{ + ^{20}}{C_2} + ...{ + ^{20}}{C_9}{ + ^{20}}{C_{10}}] \\

\Rightarrow {2^{19}} + \dfrac{1}{2}{(^{20}}{C_{10}}){ = ^{20}}{C_0}{ + ^{20}}{C_1}{ + ^{20}}{C_2} + ...{ + ^{20}}{C_9}{ + ^{20}}{C_{10}} \\

$

So, the sum of the given coefficients is ${2^{19}} + \dfrac{1}{2}{(^{20}}{C_{10}})$. This can be further simplified using the formula for combination $^n{C_r} = \dfrac{{n!}}{{r!(n - r)!}}$. Now, put the values $n = 20$ and $r = 10$ to simplify:

$

{2^{19}} + \dfrac{1}{2}{(^{20}}{C_{10}}) \\

\Rightarrow {2^{19}} + \dfrac{1}{2}.\dfrac{{20!}}{{10!(20 - 10)!}} \\

\Rightarrow {2^{19}} + \dfrac{1}{2}.\dfrac{{20!}}{{{{(10!)}^2}}} \\

$

So, the sum of expression is \[{2^{19}} - \dfrac{1}{2}.\dfrac{{20!}}{{{{(10!)}^2}}}\]

**Therefore, the correct answer is B.**

**Note:**

The concepts of permutations and combinations are vital while solving the problems related to binomial theorem. This question could have been attempted using the expansion of ${(1 + x)^n}$. Combination is used whenever we need to choose items.

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