Question

Find the value of: ${}^{20}{{C}_{0}}-{}^{20}{{C}_{1}}+{}^{20}{{C}_{2}}-.......+{}^{20}{{C}_{18}}=?$
A. 1
B. 0
C. 19
D. 20

Answer 1

Hint : To solve the question given above, we will first find out about the binomial expansion of any term and then we will apply the formula for binomial expansion of ${{\left( a+b \right)}^{n}}$ . In this formula, we will put a = x, b = -1 and n = 20. After writing the expansion in terms of x, we will put x = 1 in the expansion. From this expansion, we will get the required value of ${}^{20}{{C}_{0}}-{}^{20}{{C}_{1}}+{}^{20}{{C}_{2}}-.......+{}^{20}{{C}_{18}}$.

Complete step-by-step answer :
Before we start to solve the question given above, we must first know what is a binomial expansion and what will be the binomial expansion of ${{\left( a+b \right)}^{n}}$ . The binomial expansion describes the algebraic expansion of powers of a binomial. In other words, binomial expansion is the expansion of any power ${{\left( a+b \right)}^{n}}$ of a binomial $\left( a+b \right)$ as a certain sum of products ${{a}^{i}}{{b}^{j}}$ , where both i and j are integers. The binomial expansion of ${{\left( a+b \right)}^{n}}$ is given as shown below:
${{\left( a+b \right)}^{n}}={}^{n}{{C}_{0}}{{a}^{n}}{{b}^{0}}+{}^{n}{{C}_{1}}{{a}^{n-1}}{{b}^{1}}+{}^{n}{{C}_{2}}{{a}^{n-2}}{{b}^{2}}+......+{}^{n}{{C}_{n-1}}{{a}^{1}}{{b}^{n-1}}+{}^{n}{{C}_{n}}{{a}^{0}}{{b}^{n}}..............\left( 1 \right)$
Now, we will assume that the value of the term given in question is I. Thus, we have following equation:
$I={}^{20}{{C}_{0}}-{}^{20}{{C}_{1}}+{}^{20}{{C}_{2}}-.......+{}^{20}{{C}_{18}}.............\left( 2 \right)$
Now, we will put a = x, b = -1 and n = 20 in equation (1). Thus, we will get following equation:
$\begin{aligned} & {{\left( x-1 \right)}^{20}}={}^{20}{{C}_{0}}{{x}^{20}}{{\left( -1 \right)}^{0}}+{}^{20}{{C}_{1}}{{x}^{19}}{{\left( -1 \right)}^{1}}+.......+{}^{20}{{C}_{19}}{{x}^{1}}{{\left( -1 \right)}^{19}}+{}^{20}{{C}_{20}}{{x}^{0}}{{\left( -1 \right)}^{20}} \\\ & \Rightarrow {{\left( x-1 \right)}^{20}}={}^{20}{{C}_{0}}{{x}^{20}}-{}^{20}{{C}_{1}}{{x}^{19}}+.......+{}^{20}{{C}_{18}}{{x}^{2}}-{}^{20}{{C}_{19}}{{x}^{1}}+{}^{20}{{C}_{20}}.............\left( 3 \right) \\\ \end{aligned}$
Now, we will put x = 1 in equation (3). Thus, we will get following equation:
$\begin{aligned} & \Rightarrow {{\left( 1-1 \right)}^{20}}={}^{20}{{C}_{0}}{{\left( 1 \right)}^{20}}-{}^{20}{{C}_{1}}{{\left( 1 \right)}^{19}}+.......+{}^{20}{{C}_{18}}{{\left( 1 \right)}^{2}}-{}^{20}{{C}_{19}}{{\left( 1 \right)}^{1}}+{}^{20}{{C}_{20}} \\\ & \Rightarrow {{0}^{20}}={}^{20}{{C}_{0}}-{}^{20}{{C}_{1}}+.......+{}^{20}{{C}_{18}}-{}^{20}{{C}_{19}}+{}^{20}{{C}_{20}} \\\ & \Rightarrow \left( {}^{20}{{C}_{0}}-{}^{20}{{C}_{1}}+{}^{20}{{C}_{2}}.......+{}^{20}{{C}_{18}} \right)-\left( {}^{20}{{C}_{19}} \right)+\left( {}^{20}{{C}_{20}} \right)=0............\left( 4 \right) \\\ \end{aligned}$
From (2) and (4), we have,
$\begin{aligned} & \Rightarrow I-{}^{20}{{C}_{19}}+{}^{20}{{C}_{20}}=0 \\\ & \Rightarrow I={}^{20}{{C}_{19}}-{}^{20}{{C}_{20}} \\\ \end{aligned}$
Now, we know that we can write ${}^{n}{{C}_{r}}\ as\ {}^{n}{{C}_{n-r}}$ . On applying this identity in the above equation, we will get,

& \Rightarrow I={}^{20}{{C}_{20-19}}-{}^{20}{{C}_{20-20}} \\\ & \Rightarrow I={}^{20}{{C}_{1}}-{}^{20}{{C}_{0}} \\\ & \Rightarrow I=\dfrac{20}{1}-1 \\\ & \Rightarrow I=20-1 \\\ & \Rightarrow I=19................\left( 5 \right) \\\ \end{aligned}$$ From (2) and (5), we can say that, $${}^{20}{{C}_{0}}-{}^{20}{{C}_{1}}+{}^{20}{{C}_{2}}-.......+{}^{20}{{C}_{18}}=19$$ **So, the correct answer is “Option C”.** **Note** : We can also solve the question given above in an alternate way as shown: We know that, $ \begin{aligned} & \Rightarrow {}^{n}{{C}_{0}}-{}^{n}{{C}_{1}}+{}^{n}{{C}_{2}}-{}^{n}{{C}_{3}}.......+{{\left( -1 \right)}^{n}}{}^{n}{{C}_{n}}=0 \\\ & \Rightarrow {}^{n}{{C}_{0}}-{}^{n}{{C}_{1}}+{}^{n}{{C}_{2}}.......+{}^{n}{{C}_{n-2}}=0-{{\left( -1 \right)}^{n-1}}{}^{n}{{C}_{n-1}}-{{\left( -1 \right)}^{n}}{}^{n}{{C}_{n}} \\\ \end{aligned} $ On putting n = 20, we will get following equation: $$\begin{aligned} & \Rightarrow {}^{20}{{C}_{0}}-{}^{20}{{C}_{1}}+{}^{20}{{C}_{2}}-.......+{}^{20}{{C}_{18}}=0-{{\left( -1 \right)}^{19}}{}^{20}{{C}_{19}}-{{\left( -1 \right)}^{20}}{}^{20}{{C}_{20}} \\\ & \Rightarrow {}^{20}{{C}_{0}}-{}^{20}{{C}_{1}}+{}^{20}{{C}_{2}}-.......+{}^{20}{{C}_{18}}=20-1=19 \\\ \end{aligned}$$