Question

A committee of 5 is to be formed from 9 ladies and 8 men, if the committee commands a lady majority, then the number of ways this can be done is
(a) 2352
(b) 1008
(c) 3486
(d) 3360

Answer 1

Hint: Here we need a committee of 5 members with the majority of ladies. So, we will first find the possible combination of the number of ladies with the number of men to form the required committee. Then for each way, we will find the number of ways in which the number of ladies can be selected from all ladies and multiply it by the number of ways in which the number of men can be selected from all men. At last, we will add all the possibilities and get the final answer. We will use the combination for solving this sum. The way of selecting r items out of n is given by \[^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}.\]

Complete step-by-step answer:
Here we are given the total number of ladies as 9 and the total number of men as 8. We need to select a committee of 5 members in which the ladies should be in majority. Hence, the possible ways could be
(i) Selecting all 5 members as ladies
(ii) Selecting 4 ladies and one man
(iii) Selecting 3 ladies and two men
Now let us calculate the total ways a committee can be selected for each type of selection. We will use the combination for this. The ways of selecting r items out of n is given by \[^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}.\]
(i) We have to select all members as ladies. Hence, we need to find ways of selecting 5 ladies out of total ladies which is 9. Hence, the total ways will be
\[^{9}{{C}_{5}}=\dfrac{9!}{5!\left( 9-5 \right)!}=\dfrac{9!}{5!\times 4!}\]
\[{{\Rightarrow }^{9}}{{C}_{5}}=\dfrac{9\times 8\times 7\times 6\times 5!}{5!\times 4\times 3\times 2}\]
\[{{\Rightarrow }^{9}}{{C}_{5}}=3\times 7\times 6\]
\[{{\Rightarrow }^{9}}{{C}_{5}}=126\]
So, for selecting all 5 ladies, the total ways are 126.
(ii) We need to select 4 ladies out of 9 ladies and one man out of 8 men. Hence, the number of ways becomes equal to \[^{9}{{C}_{4}}\times {{\text{ }}^{8}}{{C}_{1}}.\]
\[{{\Rightarrow }^{9}}{{C}_{4}}\times {{\text{ }}^{8}}{{C}_{1}}=\dfrac{9!}{4!\left( 9-4 \right)!}\times \dfrac{8!}{1!\left( 8-1 \right)!}\]
\[{{\Rightarrow }^{9}}{{C}_{4}}\times {{\text{ }}^{8}}{{C}_{1}}=\dfrac{9\times 8\times 7\times 6\times 5!}{4\times 3\times 2\times 1\times 5!}\times \dfrac{8\times 7!}{7!}\]
\[{{\Rightarrow }^{9}}{{C}_{4}}\times {{\text{ }}^{8}}{{C}_{1}}=126\times 8\]
\[{{\Rightarrow }^{9}}{{C}_{4}}\times {{\text{ }}^{8}}{{C}_{1}}=1008\]
Hence, the ways of selecting 4 ladies and 1 man are 1008.
(iii) We need to select 3 ladies out of 9 ladies and two men out of 8 men. Hence the number of ways becomes equal to \[^{9}{{C}_{3}}\times {{\text{ }}^{8}}{{C}_{2}}.\]
\[{{\Rightarrow }^{9}}{{C}_{3}}\times {{\text{ }}^{8}}{{C}_{2}}=\dfrac{9!}{3!\left( 9-3 \right)!}\times \dfrac{8!}{2!\left( 8-2 \right)!}\]
\[{{\Rightarrow }^{9}}{{C}_{3}}\times {{\text{ }}^{8}}{{C}_{2}}=\dfrac{9\times 8\times 7\times 6!}{3\times 2!6!}\times \dfrac{8\times 7\times 6!}{2\times 6!}\]
\[{{\Rightarrow }^{9}}{{C}_{3}}\times {{\text{ }}^{8}}{{C}_{2}}=84\times 28\]
\[{{\Rightarrow }^{9}}{{C}_{3}}\times {{\text{ }}^{8}}{{C}_{2}}=2352\]
Hence, the ways of selecting 3 ladies and 2 men are 2352.
Now, we need to find the total ways which could be (i) or (ii) or (iii). Hence, the total ways would be found by adding them. So, the total number of ways is
\[\Rightarrow 126+1008+2352=3486\]
Hence, the total ways of selecting the committee of 5 members with the majority of ladies are 3486.

So, the correct answer is “Option (C)”.

Note: Students should make sure that every possibility is covered while taking permutation. They can use a trick that for ‘and’ we use multiplication and for ‘or’ we use addition. Here, we have multiplied the number of ladies by the number of ways of men because both were happening simultaneously whereas all ways are added because any one of them could be applied.