Question

Find the number of ways in which a committee of 6 members can be formed from 8 gentlemen and 4 ladies so that the committee contains at least 3 ladies.
A. \[252\]
B. \[672\]
C. \[444\]
D. \[420\]

Answer 1

Hint: From the given question, extract the required data. Then, using the given condition (at least 3 ladies) calculate the possible number of ways to form a committee. After that, calculate the possible number of ways to form a committee when there are 4 ladies present in that committee. In the end, add both possible number of ways and get the required answer.



Formula Used:The combination formula: \[{}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}\]
Factorial: \[n! = 1 \times 2 \times ... \times n = n\left( {n - 1} \right)!\]



Complete step by step solution:Given:
Number of ladies: 4
Number of gentlemen: 8
Number of members in committee: 6 (at least 3 ladies)

We have to form a committee of 6 and in each committee formed there must be at least 3 ladies.
There are only two possible ways:
Case 1: 3 ladies and 3 gentlemen
In this case, we have to select 3 ladies from 4 ladies and 3 gentlemen out of 8 gentlemen.
So, the possible number of ways are:
\[{}^4{C_3} \times {}^8{C_3} = \dfrac{{4!}}{{3!\left( {4 - 3} \right)!}} \times \dfrac{{8!}}{{3!\left( {8 - 3} \right)!}}\]
\[ \Rightarrow {}^4{C_3} \times {}^8{C_3} = \dfrac{{4 \times 3!}}{{3!1!}} \times \dfrac{{8 \times 7 \times 6 \times 5!}}{{3!5!}}\]
\[ \Rightarrow {}^4{C_3} \times {}^8{C_3} = 4 \times \dfrac{{8 \times 7 \times 6}}{{3!}}\]
\[ \Rightarrow {}^4{C_3} \times {}^8{C_3} = 4 \times 8 \times 7\]
\[ \Rightarrow {}^4{C_3} \times {}^8{C_3} = 224\] \[.....\left( 1 \right)\]

Case 2: 4 ladies and 2 gentlemen
In this case, we have to select all ladies and 2 gentlemen out of 8 gentlemen.
So, the possible number of ways are:
\[{}^4{C_4} \times {}^8{C_2} = 1 \times \dfrac{{8!}}{{2!\left( {8 - 2} \right)!}}\]
\[ \Rightarrow {}^4{C_4} \times {}^8{C_2} = \dfrac{{8 \times 7 \times 6!}}{{2!6!}}\]
\[ \Rightarrow {}^4{C_4} \times {}^8{C_2} = \dfrac{{8 \times 7}}{2}\]
\[ \Rightarrow {}^4{C_4} \times {}^8{C_2} = 28\] \[.....\left( 2 \right)\]

So, the total number of ways to form a committee are:
\[\left( {{}^4{C_3} \times {}^8{C_3}} \right) + \left( {{}^4{C_4} \times {}^8{C_2}} \right) = 224 + 28\]
\[ \Rightarrow \left( {{}^4{C_3} \times {}^8{C_3}} \right) + \left( {{}^4{C_4} \times {}^8{C_2}} \right) = 252\]



Option ‘A’ is correct



Note: Students often get confused with the combination and permutation formula. So, remember the following formulas.
The combination formula: \[{}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}\]
The permutation formula: \[{}^n{P_r} = \dfrac{{n!}}{{\left( {n - r} \right)!}}\]