Question:
The sum $^{20}{C_0}{ + ^{20}}{C_1}{ + ^{20}}{C_2}...{ + ^{20}}{C_{10}}$ is equal to
A. $20! + \dfrac{{20!}}{{2{{(10!)}^2}}}$
B. \[{2^{19}} - \dfrac{1}{2}.\dfrac{{20!}}{{{{(10!)}^2}}}\]
C. ${2^{19}}{ + ^{20}}{C_{10}}$
D. None of these

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.