Question

From six gentlemen and four ladies, a committee of five is to be formed. Number of ways in which this can be done if the committee is to include at least one lady is:
A. 252
B. 246
C. 248
D. 250

Answer 1

Hint: Here, using the given condition(at least one lady should be there) we will have four possible ways to form a committee with at least one lady. After that we will add them. Doing this will get the answer.

Complete step-by-step answer:
According to the question we have to make a committee of 5 and in each committee formed there must be at least one lady. There are 6 gentlemen and 4 ladies.
We can select 5 members for the committee and include at least one lady in the following four ways as done below:
(1) 1 lady and 4 gentlemen
(2) 2 ladies and 3 gentlemen
(3) 3 ladies and 2 gentlemen
(4) 4 ladies and 1 gentlemen
So, for the first condition we have to select 1 lady from 4 ladies and 4 gentlemen, therefore we can do:
$^4{C_1}{ \times ^6}{C_4} = \dfrac{{4!}}{{1!(3!)}} \times \dfrac{{6!}}{{4!(2!)}} = \dfrac{{4 \times 3 \times 2 \times 1}}{{3 \times 2 \times 1}} \times \dfrac{{6 \times 5 \times 4 \times 3 \times 2 \times 1}}{{2 \times 4 \times 3 \times 2 \times 1}} = 60$ ….(1)
For the second condition we have to select 2 ladies from 4 ladies and 3 gentlemen, therefore we can do:
$^4{C_2}{ \times ^6}{C_3} = \dfrac{{4!}}{{2!(2!)}} \times \dfrac{{6!}}{{3!(3!)}} = \dfrac{{4 \times 3 \times 2 \times 1}}{{2 \times 2}} \times \dfrac{{6 \times 5 \times 4 \times 3 \times 2 \times 1}}{{3 \times 2 \times 1 \times 3 \times 2 \times 1}} = 120$ ……(2)
For the third condition we have to select 3 ladies from 4 ladies and 2 gentlemen, therefore we can do:
$^4{C_3}{ \times ^6}{C_2} = \dfrac{{4!}}{{3!(1!)}} \times \dfrac{{6!}}{{4!(2!)}} = \dfrac{{4 \times 3 \times 2 \times 1}}{{3 \times 2 \times 1}} \times \dfrac{{6 \times 5 \times 4 \times 3 \times 2 \times 1}}{{4 \times 3 \times 2 \times 1 \times 2 \times 1}} = 60$……(3)
For the fourth condition we have to select all 4 ladies from 4 ladies and 1 gentleman, therefore we can do:
$^4{C_4}{ \times ^6}{C_1} = \dfrac{{4!}}{{4!}} \times \dfrac{{6!}}{{5!}} = \dfrac{{4 \times 3 \times 2 \times 1}}{{4 \times 3 \times 2 \times 1}} \times \dfrac{{6 \times 5 \times 4 \times 3 \times 2 \times 1}}{{5 \times 4 \times 3 \times 2 \times 1}} = 6$ ……(4)
On adding (1), (2), (3) & (4) we get the total number of ways of forming committee are:
60 + 120 + 60 + 6 = 246
Hence, the required number of committees is 246.
Hence the correct option is B.

Note: Whenever you face such types of problems you have to think of the number of ways then according to the condition provided you have to select people with the help of combinations as it is used for selections. This way will take you to the right answer.