Question

The value of \[^{30}{C_0}^{30}{C_{10}}{ - ^{30}}{C_1}^{30}{C_{11}}{ + ^{30}}{C_2}^{30}{C_{12}} - .......{ + ^{30}}{C_{10}}^{30}{C_{20}}\] is
(A) \[^{30}{C_{10}}\]
(B) \[^{30}{C_{15}}\]
(C) \[^{30}{C_{13}}\]
(D) \[^{60}{C_{30}}\]

Answer 1

Hint: Here we will apply the Binomial Expansion to solve the given problem. A combination is the number of ways we can combine things, when the order does not matter.
The binomial theorem (or binomial expansion) describes the algebraic expansion of powers of a binomial. According to the theorem, it is possible to expand the polynomial \[{(x + y)^n}\].

Formula used: Combination rule: \[^n{C_r}{ = ^n}{C_{n - r}}\]
Binomial expansion:
\[{(x + y)^n}{ = ^n}{C_0}{x^n}{ + ^n}{C_1}{x^{n - 1}}{y^1}{ + ^n}{C_2}{x^{n - 2}}{y^2} + ......{ + ^n}{C_{n - 1}}x{y^{n - 1}}{ + ^n}{C_n}{y^n}\]
Putting \[x = 1,y = x\], we get,
\[{(1 + x)^n}{ = ^n}{C_0}{ + ^n}{C_1}{x^1}{ + ^n}{C_2}{x^2} + ......{ + ^n}{C_{n - 1}}{x^{n - 1}}{ + ^n}{C_n}{x^n}\]
Putting \[x = 1,y = - x\], we get,
\[{(1 - x)^n}{ = ^n}{C_0}{ + ^n}{C_1}{\left( { - x} \right)^1}{ + ^n}{C_2}{\left( { - x} \right)^2} + ......{ + ^n}{C_{n - 1}}{\left( { - x} \right)^{n - 1}}{ + ^n}{C_n}{\left( { - x} \right)^n}\]
Solving, \[{(1 - x)^n}{ = ^n}{C_0}{ - ^n}{C_1}{x^1}{ + ^n}{C_2}{x^2} + ...... + {\left( { - 1} \right)^{n - 1}}^n{C_{n - 1}}{x^{n - 1}} + {\left( { - 1} \right)^n}^n{C_n}{x^n}\]

Complete step-by-step answer:
We need to find out the value of \[^{30}{C_0}^{30}{C_{10}}{ - ^{30}}{C_1}^{30}{C_{11}}{ + ^{30}}{C_2}^{30}{C_{12}} - .......{ + ^{30}}{C_{10}}^{30}{C_{20}}\].
Now we can use the binomial expansion putting \[x = 1,y = x,n = 30\] we get,
\[ \Rightarrow {(1 + x)^{30}}{ = ^{30}}{C_0}{ + ^{30}}{C_1}{x^1}{ + ^{30}}{C_2}{x^2} + ......{ + ^{30}}{C_{30 - 1}}{x^{30 - 1}}{ + ^{30}}{C_{30}}{x^{30}}\]
Solving we get,
\[ \Rightarrow {(1 + x)^{30}}{ = ^{30}}{C_0}{ + ^{30}}{C_1}{x^1}{ + ^{30}}{C_2}{x^2} + ......{ + ^{30}}{C_{29}}{x^{29}}{ + ^{30}}{C_{30}}{x^{30}} \ldots \ldots \ldots \left( i \right)\]
Again, we can use the binomial expansion putting \[x = 1,y = - x,n = 30\]we get,
\[ \Rightarrow {(1 - x)^{30}}{ = ^{30}}{C_0} + \left( { - 1} \right){ \times ^{30}}{C_1}{x^1} + {\left( { - 1} \right)^2}{ \times ^{30}}{C_2}{x^2} + ...... + {\left( { - 1} \right)^{30 - 1}}{ \times ^{30}}{C_{30 - 1}}{x^{30 - 1}} + {\left( { - 1} \right)^{30}}{ \times ^{30}}{C_{30}}{x^{30}}\]Solving we get,
\[ \Rightarrow {(1 - x)^{30}}{ = ^{30}}{C_0} + \left( { - 1} \right){ \times ^{30}}{C_1}{x^1} + {\left( { - 1} \right)^2}{ \times ^{30}}{C_2}{x^2} + ...... + {\left( { - 1} \right)^{29}}{ \times ^{30}}{C_{29}}{x^{29}} + {\left( { - 1} \right)^{30}}{ \times ^{30}}{C_{30}}{x^{30}}\]
\[ \Rightarrow {(1 - x)^{30}}{ = ^{30}}{C_0}{ - ^{30}}{C_1}{x^1}{ + ^{30}}{C_2}{x^2} + ......{ - ^{30}}{C_{30 - 1}}{x^{30 - 1}}{ + ^{30}}{C_{30}}{x^{30}} \ldots \ldots ..\left( {ii} \right)\]
Now multiplying (i) and (ii) we get,
\[
   \Rightarrow {(1 + x)^{30}} \times {(1 - x)^{30}} \\
  {\text{ }} = \left\{ {^{30}{C_0}{ + ^{30}}{C_1}{x^1}{ + ^{30}}{C_2}{x^2} + ......{ + ^{30}}{C_{29}}{x^{29}}{ + ^{30}}{C_{30}}{x^{30}}} \right\} \times \left\{ {^{30}{C_0}{ - ^{30}}{C_1}{x^1}{ + ^{30}}{C_2}{x^2} + ......{ - ^{30}}{C_{30 - 1}}{x^{30 - 1}}{ + ^{30}}{C_{30}}{x^{30}}} \right\} \\
 \]
\[
   \Rightarrow {\left\{ {(1 + x)(1 - x)} \right\}^{30}} \\
  {\text{ }} = \left\{ {^{30}{C_0}{ + ^{30}}{C_1}{x^1}{ + ^{30}}{C_2}{x^2} + ......{ + ^{30}}{C_{29}}{x^{29}}{ + ^{30}}{C_{30}}{x^{30}}} \right\} \times \left\{ {^{30}{C_0}{ - ^{30}}{C_1}{x^1}{ + ^{30}}{C_2}{x^2} + ......{ - ^{30}}{C_{30 - 1}}{x^{30 - 1}}{ + ^{30}}{C_{30}}{x^{30}}} \right\} \\
 \]
Solving the L.H.S we get,
\[ \Rightarrow {\left\{ {(1 + x)(1 - x)} \right\}^{30}} = {(1 - {x^2})^{30}}\]
Hence, we have to calculate the multiplication for the RHS,
\[
   \Rightarrow {(1 - {x^2})^{30}} \\
  {\text{ }} = \left\{ {^{30}{C_0}{ + ^{30}}{C_1}{x^1}{ + ^{30}}{C_2}{x^2} + ......{ + ^{30}}{C_{29}}{x^{29}}{ + ^{30}}{C_{30}}{x^{30}}} \right\} \times \left\{ {^{30}{C_0}{ - ^{30}}{C_1}{x^1}{ + ^{30}}{C_2}{x^2} + ......{ - ^{30}}{C_{30 - 1}}{x^{30 - 1}}{ + ^{30}}{C_{30}}{x^{30}}} \right\} \ldots \ldots \left( {iii} \right) \\
 \]
By applying Binomial expansion in the L.H.S, that is substituting the $x$ as ${x^2}$ in binomial expansion of \[{(1 - x)^{30}}\] we get,
\[
   \Rightarrow {(1 - {x^2})^{30}} \\
  { = ^{30}}{C_0} + {\left( { - 1} \right)^1}{ \times ^{30}}{C_1}{\left( {{x^2}} \right)^1} + {\left( { - 1} \right)^2}{ \times ^{30}}{C_2}{\left( {{x^2}} \right)^2} + ....... + {\left( { - 1} \right)^{10}}{ \times ^{30}}{C_{10}}{\left( {{x^2}} \right)^{10}} + .... + {\left( { - 1} \right)^{30 - 1}}{ \times ^{30}}{C_{30 - 1}}{\left( {{x^2}} \right)^{30 - 1}} + {\left( { - 1} \right)^{30}}{ \times ^{30}}{C_{30}}{\left( {{x^2}} \right)^{30}} \\
 \]
Simplifying (add and subtract the terms in power) we get,
\[ \Rightarrow {(1 - {x^2})^{30}}{ = ^{30}}{C_0}{ - ^{30}}{C_1}{x^2}{ + ^{30}}{C_2}{x^4} + ...{ + ^{30}}{C_{10}}{\left( {{x^2}} \right)^{10}} - .....{ - ^{30}}{C_{29}}{\left( {{x^2}} \right)^{29}}{ + ^{30}}{C_{30}}{\left( {{x^2}} \right)^{30}}\]
Multiplying the power values to simplify,
\[ \Rightarrow {(1 - {x^2})^{30}}{ = ^{30}}{C_0}{ - ^{30}}{C_1}{x^2}{ + ^{30}}{C_2}{x^4} + ...{ + ^{30}}{C_{10}}{x^{20}} - .....{ - ^{30}}{C_{29}}{x^{58}}{ + ^{30}}{C_{30}}{x^{60}}\]
Putting the above expression of L.H.S in (iii), we get,
\[
  { \Rightarrow ^{30}}{C_0}{ - ^{30}}{C_1}{x^2}{ + ^{30}}{C_2}{x^4} + ...{ + ^{30}}{C_{10}}{x^{20}} - .....{ - ^{30}}{C_{29}}{x^{58}}{ + ^{30}}{C_{30}}{x^{60}} \\
  {\text{ }} = \left\{ {^{30}{C_0}{ + ^{30}}{C_1}{x^1}{ + ^{30}}{C_2}{x^2} + ......{ + ^{30}}{C_{29}}{x^{29}}{ + ^{30}}{C_{30}}{x^{30}}} \right\} \times \left\{ {^{30}{C_0}{ - ^{30}}{C_1}{x^1}{ + ^{30}}{C_2}{x^2} + ......{ - ^{30}}{C_{30 - 1}}{x^{30 - 1}}{ + ^{30}}{C_{30}}{x^{30}}} \right\} \\
 \]
Now, we need to find out the value of
\[^{30}{C_0}^{30}{C_{10}}{ - ^{30}}{C_1}^{30}{C_{11}}{ + ^{30}}{C_2}^{30}{C_{12}} - .......{ + ^{30}}{C_{10}}^{30}{C_{20}}\]
We can write it as,
\[{ \Rightarrow ^{30}}{C_0}^{30}{C_{20}}{ - ^{30}}{C_1}^{30}{C_{19}}{ + ^{30}}{C_2}^{30}{C_{18}} - .......{ + ^{30}}{C_{10}}^{30}{C_{10}}\]
Since we know from combination rule,
\[{ \Rightarrow ^n}{C_r}{ = ^n}{C_{n - r}}\]
Comparing the coefficient of \[{x^{20}}\] from both sides we get,
\[{\therefore ^{30}}{C_{10}}{ = ^{30}}{C_0}{ \times ^{30}}{C_{20}}{ - ^{30}}{C_1}{ \times ^{30}}{C_{19}}{ + ^{30}}{C_2}^{30}{C_{18}} - .......{ + ^{30}}{C_{10}}^{30}{C_{10}}\]

Hence (A) is the correct option.

Note: In mathematics, a combination is a selection of items from a collection, such that the order of selection does not matter. For example, given three fruits, say an apple, an orange and a pear, there are three combinations of two that can be drawn from this set: an apple and a pear; an apple and an orange; or a pear and an orange.