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Question: The value of \[^{30}{C_0}^{30}{C_{10}}{ - ^{30}}{C_1}^{30}{C_{11}}{ + ^{30}}{C_2}^{30}{C_{12}} - ......

The value of 30C030C10−30C130C11+30C230C12−.......+30C1030C20^{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) 30C10^{30}{C_{10}}
(B) 30C15^{30}{C_{15}}
(C) 30C13^{30}{C_{13}}
(D) 60C30^{60}{C_{30}}

Explanation

Solution

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{(x + y)^n}.

Formula used: Combination rule: nCr=nCn−r^n{C_r}{ = ^n}{C_{n - r}}
Binomial expansion:
(x+y)n=nC0xn+nC1xn−1y1+nC2xn−2y2+......+nCn−1xyn−1+nCnyn{(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=xx = 1,y = x, we get,
(1+x)n=nC0+nC1x1+nC2x2+......+nCn−1xn−1+nCnxn{(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=−xx = 1,y = - x, we get,
(1−x)n=nC0+nC1(−x)1+nC2(−x)2+......+nCn−1(−x)n−1+nCn(−x)n{(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=nC0−nC1x1+nC2x2+......+(−1)n−1nCn−1xn−1+(−1)nnCnxn{(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 30C030C10−30C130C11+30C230C12−.......+30C1030C20^{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=30x = 1,y = x,n = 30 we get,
⇒(1+x)30=30C0+30C1x1+30C2x2+......+30C30−1x30−1+30C30x30\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,
⇒(1+x)30=30C0+30C1x1+30C2x2+......+30C29x29+30C30x30………(i)\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=30x = 1,y = - x,n = 30we get,
⇒(1−x)30=30C0+(−1)×30C1x1+(−1)2×30C2x2+......+(−1)30−1×30C30−1x30−1+(−1)30×30C30x30\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,
⇒(1−x)30=30C0+(−1)×30C1x1+(−1)2×30C2x2+......+(−1)29×30C29x29+(−1)30×30C30x30\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}}
⇒(1−x)30=30C0−30C1x1+30C2x2+......−30C30−1x30−1+30C30x30……..(ii)\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 xx as x2{x^2} in binomial expansion of (1−x)30{(1 - x)^{30}} we get,

⇒(1−x2)30 =30C0+(−1)1×30C1(x2)1+(−1)2×30C2(x2)2+.......+(−1)10×30C10(x2)10+....+(−1)30−1×30C30−1(x2)30−1+(−1)30×30C30(x2)30  \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,
⇒(1−x2)30=30C0−30C1x2+30C2x4+...+30C10(x2)10−.....−30C29(x2)29+30C30(x2)30\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,
⇒(1−x2)30=30C0−30C1x2+30C2x4+...+30C10x20−.....−30C29x58+30C30x60\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
30C030C10−30C130C11+30C230C12−.......+30C1030C20^{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,
⇒30C030C20−30C130C19+30C230C18−.......+30C1030C10{ \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,
⇒nCr=nCn−r{ \Rightarrow ^n}{C_r}{ = ^n}{C_{n - r}}
Comparing the coefficient of x20{x^{20}} from both sides we get,
∴30C10=30C0×30C20−30C1×30C19+30C230C18−.......+30C1030C10{\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.