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Question: The value of \[{}^n{C_0} - {}^n{C_1} + {}^n{C_2} - ...... + {( - 1)^{{n^n}}}{C_n}\]is: (A) 1 ...

The value of nC0−nC1+nC2−......+(−1)nnCn{}^n{C_0} - {}^n{C_1} + {}^n{C_2} - ...... + {( - 1)^{{n^n}}}{C_n}is:
(A) 1
(B) 0
(c) 2
(c) n

Explanation

Solution

Hint- Proceed the solution with the help of binomial expansion. The binomial expansion of (1−x)n=nC0+nC1(−x)1+nC2(−x)2+..........+nCn(−x)n{(1 - x)^n} = {}^n{C_0} + {}^n{C_1}{( - x)^1} + {}^n{C_2}{( - x)^2} + .......... + {}^n{C_n}{( - x)^n}. If however we remove the variable x then we can get the series of binomial coefficients.

Complete step-by-step solution -
The binomial expansion of (1−x)n=nC0+nC1(−x)1+nC2(−x)2+..........+nCn(−x)n{(1 - x)^n} = {}^n{C_0} + {}^n{C_1}{( - x)^1} + {}^n{C_2}{( - x)^2} + .......... + {}^n{C_n}{( - x)^n}.
Put x=1 in this expansion.
Since, we know that (1−x)n=nC0+nC1(−x)1+nC2(−x)2+..........+nCn(−x)n{(1 - x)^n} = {}^n{C_0} + {}^n{C_1}{( - x)^1} + {}^n{C_2}{( - x)^2} + .......... + {}^n{C_n}{( - x)^n}
Putting x=1 on both sides,
(1−1)n=nC0+nC1(−1)1+nC2(−1)2+..........+nCn(−1)n{(1 - 1)^n} = {}^n{C_0} + {}^n{C_1}{( - 1)^1} + {}^n{C_2}{( - 1)^2} + .......... + {}^n{C_n}{( - 1)^n} ⇒0=nC0−nC1+nC2−..........+(−1)nnCn \Rightarrow 0 = {}^n{C_0} - {}^n{C_1} + {}^n{C_2} - .......... + {( - 1)^n}{}^n{C_n}
⇒nC0−nC1+nC2−..........+(−1)nnCn=0\Rightarrow {}^n{C_0} - {}^n{C_1} + {}^n{C_2} - .......... + {( - 1)^n}{}^n{C_n} = 0

Note- Binomial theorem is a fast method of expanding a binomial expression that has been raised to some large power.
we got the expansion of (1−x)n{(1 - x)^n} using the expansion of
(1+x)n={(1 + x)^n} = nC0+nC1(x)1+nC2(x)2+..........+nCn(x)n{}^n{C_0} + {}^n{C_1}{(x)^1} + {}^n{C_2}{(x)^2} + .......... + {}^n{C_n}{(x)^n}, as (1−x)n{(1 - x)^n}= (1+(−x))n{(1 + ( - x))^n}.
The expansion can also be written as (1+x)n={(1 + x)^n} = 1+n1!x+n(n−1)2!x2+n(n−1)(n−2)3!x3+........1 + \dfrac{n}{{1!}}x + \dfrac{{n(n - 1)}}{{2!}}{x^2} + \dfrac{{n(n - 1)(n - 2)}}{{3!}}{x^3} + ........