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Question

Answers

$

{\text{A}}{\text{. }}{{\text{2}}^{10}} \\

{\text{B}}{\text{. }}{2^{11}} \\

{\text{C}}.{\text{ }}{2^{10}} - 2 \\

{\text{D}}.{\text{ }}{2^{10}} - 1 \\

$

Answer

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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. ${}^n{C_0} + {}^n{C_1} + {}^n{C_2} + \ldots + {}^n{C_n} = {2^n}$

As given in question

${}^{10}{C_1} + {}^{10}{C_2} + {}^{10}{C_3} + \ldots + {}^{10}{C_9}$

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

Now, question becomes

${}^{10}{{\text{C}}_0} + {}^{10}{C_1} + {}^{10}{C_2} + \ldots + {}^{10}{C_{10}} - {}^{10}{C_0} - {}^{10}{C_{10}}$

$ \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.

We will apply the formula of sum of binomial coefficient.

i.e. ${}^n{C_0} + {}^n{C_1} + {}^n{C_2} + \ldots + {}^n{C_n} = {2^n}$

As given in question

${}^{10}{C_1} + {}^{10}{C_2} + {}^{10}{C_3} + \ldots + {}^{10}{C_9}$

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

Now, question becomes

${}^{10}{{\text{C}}_0} + {}^{10}{C_1} + {}^{10}{C_2} + \ldots + {}^{10}{C_{10}} - {}^{10}{C_0} - {}^{10}{C_{10}}$

$ \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.