Question

A certain law firm consists of 4 senior partners and 6 junior partners.

How many different groups of 3 partners can be formed in which at least one member of the group is a senior partner?

A) 48 b) 100 c) 120 d) 288 e) 600

Answer 1

Answer: b) 100

Explanation:

Given : Number of senior partners =4

Number of junior partners. = 6

Total partners = 4+ 6= 10

Total Number of ways to choose 3 partners
`=^(10)C_3=(10!)/(3!(10-3)!)`

`[\because\ ^nC_r=(n!)/(r!(n-r)!)]`

`=(10*9*8*7!)/(7!*6)=120`

i.e. Total Number of ways to choose 3 partners =120

The number of ways that none of 3 partners are seniors =tex]=^{6}C_3=\dfrac{6!}{3!(6-3)!}=20[/tex]

Now , the different groups of 3 partners can be formed in which at least one member of the group is a senior partner

= Total Number of ways to choose 3 partners- Number of ways that none of 3 partners are seniors

= 120-20=100

Hence, the correct answer is b) 100.