Question

A man has $10$ friends. In how many ways he can invite one or more of them to a party?

• A. $10!$
• B. $2^{10}$
• C. $10!-1$
• D. $2^{10}-1$

Answer 1

The correct option is D

$2^{10}-1$

Explanation for the correct option:

Step 1 Find the number of ways of inviting one or more friend to the party:

Number of ways in which he can invite $1$ friend $={}^{10}C_1$

similarly, for $2$friends, the number of ways is ${}^{10}C_1$ and for $10$ friends it is ${}^{10}C_{10}$

Since he is inviting one or more than one friend, the number of friends can range from $1$ to $10$.

so, the number of ways $={}^{10}C_1+{}^{10}C_2+{}^{10}C_3+{}^{10}C_4+{}^{10}C_5+{}^{10}C_6+{}^{10}C_7+{}^{10}C_8+{}^{10}C_9+{}^{10}C_{10}$

Step 2. add and subtract ${}^{10}C_0$ to it.

$=\left({}^{10}C_0+{}^{10}C_1+{}^{10}C_2+{}^{10}C_3+{}^{10}C_4+{}^{10}C_5+{}^{10}C_6+{}^{10}C_7+{}^{10}C_8+{}^{10}C_9+{}^{10}C_{10}\right)-{}^{10}C_0$

By using Binomial expansion, we get

$=2^{10}-{10}_{C_0}$

$={\mathbf{2}}^{\mathbf{10}}\mathbf{-}\mathbf{1}$

Hence, Option D is the correct answer.