
In how many ways can a committee consisting of one or more members be formed out of 12 members of the Municipal Corporation?
A. 4095
B. 5095
C. 4905
D. 4090
Answer
161.1k+ views
Hint: We have to make a committee using 12 members. The committee is consisting of one member or more members. We have to find the number of ways to make committee consisting of one or more members using combination formula. Then add the number of ways to get the required answer.
Formula Used:Combination formula:
\[{}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}\]
Complete step by step solution:The number of members of the Municipal Corporation is 12.
We can make a committee using one or more members of the Municipal Corporation.
Case 1: Making committee using 1 member
The number of ways to make a committee consisting 1 member is \[{}^{12}{C_1}\].
Case 2: Making committee using 2 members
The number of ways to make a committee consisting 2 members is \[{}^{12}{C_2}\]
Case 3: Making committee using 3 members
The number of ways to make a committee consisting 3 members is \[{}^{12}{C_3}\].
Case 4: Making committee using 4 members
The number of ways to make a committee consisting 4 members is \[{}^{12}{C_4}\].
Case 5: Making committee using 5 members
The number of ways to make a committee consisting 5 members is \[{}^{12}{C_5}\].
Case 6: Making committee using 6 members
The number of ways to make a committee consisting 6 members is \[{}^{12}{C_6}\].
Case 7: Making committee using 7 members
The number of ways to make a committee consisting 7 members is \[{}^{12}{C_7}\].
Case 8: Making committee using 8 members
The number of ways to make a committee consisting 8 members is \[{}^{12}{C_8}\].
Case 9: Making committee using 9 members
The number of ways to make a committee consisting 9 members is \[{}^{12}{C_9}\].
Case 10: Making committee using 10 members
The number of ways to make a committee consisting 10 members is \[{}^{12}{C_{10}}\].
Case 11: Making committee using 11 members
The number of ways to make a committee consisting 11 members is \[{}^{12}{C_{11}}\].
Case 12: Making committee using 12 members
The number of ways to make a committee consisting 12 members is \[{}^{12}{C_{12}}\].
We know that,
\[{\left( {1 + x} \right)^{12}} = {}^{12}{C_0} + {}^{12}{C_1}x + {}^{12}{C_2}{x^2} + {}^{12}{C_3}{x^3} + {}^{12}{C_4}{x^4} + {}^{12}{C_5}{x^5} + {}^{12}{C_6}{x^6} + {}^{12}{C_7}{x^7} + \cdots + {}^{12}{C_{11}}{x^{11}} + {}^{12}{C_{12}}{x^{12}}\]
Putting x = 1:
\[{2^{12}} = {}^{12}{C_0} + {}^{12}{C_1} + {}^{12}{C_2} + {}^{12}{C_3} + {}^{12}{C_4} + {}^{12}{C_5} + {}^{12}{C_6} + {}^{12}{C_7} + \cdots + {}^{12}{C_{11}} + {}^{12}{C_{12}}\]
\[ \Rightarrow {2^{12}} = 1 + {}^{12}{C_1} + {}^{12}{C_2} + {}^{12}{C_3} + {}^{12}{C_4} + {}^{12}{C_5} + {}^{12}{C_6} + {}^{12}{C_7} + \cdots + {}^{12}{C_{11}} + {}^{12}{C_{12}}\]
\[ \Rightarrow {2^{12}} - 1 = {}^{12}{C_1} + {}^{12}{C_2} + {}^{12}{C_3} + {}^{12}{C_4} + {}^{12}{C_5} + {}^{12}{C_6} + {}^{12}{C_7} + \cdots + {}^{12}{C_{11}} + {}^{12}{C_{12}}\]
Total number of ways to make committee is \[{}^{12}{C_1} + {}^{12}{C_2} + {}^{12}{C_3} + {}^{12}{C_4} + {}^{12}{C_5} + {}^{12}{C_6} + {}^{12}{C_7} + \cdots + {}^{12}{C_{11}} + {}^{12}{C_{12}}\]
\[ = {2^{12}} - 1\]
\[ = 4096 - 1\]
\[ = 4095\]
Option ‘A’ is correct
Note: Students often make mistake to solve given question. They used permutation formula to solve the given question. But the order of selection does not matter thus we will apply combination formula to solve it.
Formula Used:Combination formula:
\[{}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}\]
Complete step by step solution:The number of members of the Municipal Corporation is 12.
We can make a committee using one or more members of the Municipal Corporation.
Case 1: Making committee using 1 member
The number of ways to make a committee consisting 1 member is \[{}^{12}{C_1}\].
Case 2: Making committee using 2 members
The number of ways to make a committee consisting 2 members is \[{}^{12}{C_2}\]
Case 3: Making committee using 3 members
The number of ways to make a committee consisting 3 members is \[{}^{12}{C_3}\].
Case 4: Making committee using 4 members
The number of ways to make a committee consisting 4 members is \[{}^{12}{C_4}\].
Case 5: Making committee using 5 members
The number of ways to make a committee consisting 5 members is \[{}^{12}{C_5}\].
Case 6: Making committee using 6 members
The number of ways to make a committee consisting 6 members is \[{}^{12}{C_6}\].
Case 7: Making committee using 7 members
The number of ways to make a committee consisting 7 members is \[{}^{12}{C_7}\].
Case 8: Making committee using 8 members
The number of ways to make a committee consisting 8 members is \[{}^{12}{C_8}\].
Case 9: Making committee using 9 members
The number of ways to make a committee consisting 9 members is \[{}^{12}{C_9}\].
Case 10: Making committee using 10 members
The number of ways to make a committee consisting 10 members is \[{}^{12}{C_{10}}\].
Case 11: Making committee using 11 members
The number of ways to make a committee consisting 11 members is \[{}^{12}{C_{11}}\].
Case 12: Making committee using 12 members
The number of ways to make a committee consisting 12 members is \[{}^{12}{C_{12}}\].
We know that,
\[{\left( {1 + x} \right)^{12}} = {}^{12}{C_0} + {}^{12}{C_1}x + {}^{12}{C_2}{x^2} + {}^{12}{C_3}{x^3} + {}^{12}{C_4}{x^4} + {}^{12}{C_5}{x^5} + {}^{12}{C_6}{x^6} + {}^{12}{C_7}{x^7} + \cdots + {}^{12}{C_{11}}{x^{11}} + {}^{12}{C_{12}}{x^{12}}\]
Putting x = 1:
\[{2^{12}} = {}^{12}{C_0} + {}^{12}{C_1} + {}^{12}{C_2} + {}^{12}{C_3} + {}^{12}{C_4} + {}^{12}{C_5} + {}^{12}{C_6} + {}^{12}{C_7} + \cdots + {}^{12}{C_{11}} + {}^{12}{C_{12}}\]
\[ \Rightarrow {2^{12}} = 1 + {}^{12}{C_1} + {}^{12}{C_2} + {}^{12}{C_3} + {}^{12}{C_4} + {}^{12}{C_5} + {}^{12}{C_6} + {}^{12}{C_7} + \cdots + {}^{12}{C_{11}} + {}^{12}{C_{12}}\]
\[ \Rightarrow {2^{12}} - 1 = {}^{12}{C_1} + {}^{12}{C_2} + {}^{12}{C_3} + {}^{12}{C_4} + {}^{12}{C_5} + {}^{12}{C_6} + {}^{12}{C_7} + \cdots + {}^{12}{C_{11}} + {}^{12}{C_{12}}\]
Total number of ways to make committee is \[{}^{12}{C_1} + {}^{12}{C_2} + {}^{12}{C_3} + {}^{12}{C_4} + {}^{12}{C_5} + {}^{12}{C_6} + {}^{12}{C_7} + \cdots + {}^{12}{C_{11}} + {}^{12}{C_{12}}\]
\[ = {2^{12}} - 1\]
\[ = 4096 - 1\]
\[ = 4095\]
Option ‘A’ is correct
Note: Students often make mistake to solve given question. They used permutation formula to solve the given question. But the order of selection does not matter thus we will apply combination formula to solve it.
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