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Question: Question: The sum \(^{20}{C_0}{ + ^{20}}{C_1}{ + ^{20}}{C_2}...{ + ^{20}}{C_{10}}\) is equal to ...

Question:
The sum 20C0+20C1+20C2...+20C10^{20}{C_0}{ + ^{20}}{C_1}{ + ^{20}}{C_2}...{ + ^{20}}{C_{10}} is equal to
A. 20!+20!2(10!)220! + \dfrac{{20!}}{{2{{(10!)}^2}}}
B. 219−12.20!(10!)2{2^{19}} - \dfrac{1}{2}.\dfrac{{20!}}{{{{(10!)}^2}}}
C. 219+20C10{2^{19}}{ + ^{20}}{C_{10}}
D. None of these

Explanation

Solution

Observe that these are the terms of the binomial expression (a+b)n{(a + b)^n} where a=1a = 1, b=1b = 1 and n=20n = 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 aa and bb raised to the power nn is given by
(a+b)n=nC0anb0+nC1an−1b1+...+nCn−1a1bn−1+nCna0bn{(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=1a = 1 , b=1b = 1 and n=20n = 20 in the above expression we get :
(2)n=20C0+20C1+20C2+...+20C20{(2)^n}{ = ^{20}}{C_0}{ + ^{20}}{C_1}{ + ^{20}}{C_2} + ...{ + ^{20}}{C_{20}} … (1)
Now, we have the sum of coefficients as the 20th{20}^{th} power of 2. We know that by the concept of combinations,nCr=nCn−r^n{C_r}{ = ^n}{C_{n - r}}. So, using nCr=nCn−r^n{C_r}{ = ^n}{C_{n - r}} in equation (1), we get:
(2)20=2×[20C0+20C1+20C2+...+20C9+20C10]−20C10 ⇒(2)20+20C10=2×[20C0+20C1+20C2+...+20C9+20C10] ⇒12[(2)20+20C10]=[20C0+20C1+20C2+...+20C9+20C10] ⇒219+12(20C10)=20C0+20C1+20C2+...+20C9+20C10  {(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 219+12(20C10){2^{19}} + \dfrac{1}{2}{(^{20}}{C_{10}}). This can be further simplified using the formula for combination nCr=n!r!(n−r)!^n{C_r} = \dfrac{{n!}}{{r!(n - r)!}}. Now, put the values n=20n = 20 and r=10r = 10 to simplify:
219+12(20C10) ⇒219+12.20!10!(20−10)! ⇒219+12.20!(10!)2  {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 219−12.20!(10!)2{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{(1 + x)^n}. Combination is used whenever we need to choose items.