Question

The value of ${}^{10}{\text{C}}_1+{}^{10}{C}_2+{}^{10}{C}_3+\dots +{}^{10}{C}_9$is
$$\begin{gathered} \text{A}\text{. }{\text{2}}^{10} \\ \text{B}\text{. }{2}^{11} \\ \text{C}.2^{10}-2 \\ \text{D}.2^{10}-1 \end{gathered}$$

Answer 1

Hint: - Here, we apply the formula of sum of binomial coefficient. Binomial coefficient is the number of ways to retrieve a set of p values from q different elements.

We will apply the formula of sum of binomial coefficient.

i.e. ${}^n{C}_0+{}^n{C}_1+{}^n{C}_2+\dots +{}^n{C}_n=2^n$

As given in question
${}^{10}{C}_1+{}^{10}{C}_2+{}^{10}{C}_3+\dots +{}^{10}{C}_9$

To use above formula we have to add and subtract ${}^{10}{\text{C}}_0$ and ${}^{10}{\text{C}}_{10}$.

Now, question becomes
${}^{10}{\text{C}}_0+{}^{10}{C}_1+{}^{10}{C}_2+\dots +{}^{10}{C}_{10}-{}^{10}{C}_0-{}^{10}{C}_{10}$
$\Rightarrow 2^{10}-2$
$\because {}^{10}{C}_0={}^{10}{C}_{10}=1$.
So option $\text{C}$ is the correct answer.

Note: - When you get these types of summation questions in binomial, you have to proceed as that question is set on any formula, or try to set a formula by addition or subtraction.