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

Hint: First we will find the combination formula to find the ways to invite one friend or more of them. Then by using the binomial theorem we will find the value of the sum of the combination.

Formula Used:
The number of ways to select \[r\] objects from \[n\] objects is \[^n{C_r}\].
Binomial expansion:
\[{\left( {1 + x} \right)^n}{ = ^n}{C_0}{ + ^n}{C_1}x{ + ^n}{C_2}{x^2} + \cdots { + ^n}{C_n}{x^n}\]

Complete step by step solution:
The man can invite 1 friend out of 10 friends.
The number of to invite 1 friend is \[^{10}{C_1}\].
The man can invite 2 friends out of 10 friends.
The number of to invite 2 friends is \[^{10}{C_2}\].
The man can invite 3 friends out of 10 friends.
The number of to invite 3 friends is \[^{10}{C_3}\].
Similarly,
The man can invite 10 friends out of 10 friends.
The number of to invite 10 friends is \[^{10}{C_{10}}\].
The total number of ways to invite them to the party is
\[^{10}{C_1}{ + ^{10}}{C_2}{ + ^{10}}{C_3} + \cdots { + ^{10}}{C_{10}}\]
We know that,
\[{\left( {1 + x} \right)^n}{ = ^n}{C_0}{ + ^n}{C_1}x{ + ^n}{C_2}{x^2} + \cdots { + ^n}{C_n}{x^n}\]
Substitute \[n = 10\] and \[x = 1\]
 \[{\left( {1 + 1} \right)^{10}}{ = ^{10}}{C_0}{ + ^{10}}{C_1}{ + ^{10}}{C_2} + \cdots { + ^{10}}{C_{10}}\]
\[ \Rightarrow {\left( {1 + 1} \right)^{10}}{ = ^{10}}{C_0}{ + ^{10}}{C_1}{ + ^{10}}{C_2} + \cdots { + ^{10}}{C_{10}}\]
\[ \Rightarrow {2^{10}}{ = ^{10}}{C_0}{ + ^{10}}{C_1}{ + ^{10}}{C_2} + \cdots { + ^{10}}{C_{10}}\] …….(i)
We know that \[^{10}{C_0} = 1\]
Substitute the value of \[^{10}C\] in equation (i)
\[ \Rightarrow {2^{10}} = 1{ + ^{10}}{C_1}{ + ^{10}}{C_2} + \cdots { + ^{10}}{C_{10}}\]
Subtract 1 from each side of the equation:
\[ \Rightarrow {2^{10}} - 1 = 1{ + ^{10}}{C_1}{ + ^{10}}{C_2} + \cdots { + ^{10}}{C_{10}} - 1\]
\[ \Rightarrow {2^{10}} - 1{ = ^{10}}{C_1}{ + ^{10}}{C_2} + \cdots { + ^{10}}{C_{10}}\]

Hence the correct option is D.

Note: Students often apply the concept factorial. If the question related to the arrangement of seats for 10 friends, then we can find it by using the factorial. But the question is related to inviting one or more people. So, we apply the concept combination.