Question

From 6 gentlemen and 4 ladies a committee of 5 is to be formed. Find the number of ways if
(a) The committee is to include at least one lady.
(b) There is no restriction about its formation.
A. 240 , 255
B. 242 , 256
C. 246 , 252
D. None of these

Answer 1

Hint: In this particular question , we have to proceed by first selecting 5 members with no restrictions. Then subtract the number of ways no lady is selected with that amount to get the number of ways of selecting the committee with at least 1 lady.

Complete step-by-step answer:
Number of men = 6
Number of women = 4
Total members = 6 + 4 = 10
If there is no restriction , then the number of ways to select 5 members will be $^{10}{C_5} = \dfrac{{10!}}{{5!\left( {10 - 5} \right)!}} = \dfrac{{10!}}{{5! \times 5!}} = \dfrac{{10 \times 9 \times 8 \times 7 \times 6 \times 5!}}{{5 \times 4 \times 3 \times 2 \times 1 \times 5!}} = 18 \times 14 = 252$ ways ( b )
 ( Since$^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$ )
Now selecting the committee with at least 1 lady = Total number of ways of selecting 5 members without restriction – Number of ways no lady is selected
$ \Rightarrow $ Number of ways no lady is selected ( all men are selected ) = $^6{C_5}
 = \dfrac{{6!}}{{5! \times \left( {6 - 5} \right)!}} = \dfrac{{6!}}{{5!}} = 6$ ways
Therefore selecting the committee with at least 1 lady = 252 – 6 = 246 ways ( a )
Hence the correct answer is C. ( 246 , 252 )

Note: Remember to recall the formula and concept of selection of people using Permutations and Combinations to solve such types of questions. Note that selection of at least one lady could also be done by adding the total number of ways selecting 1 , 2 , 3 , 4 and 5 ladies.