From 6 programmers and 4 typists, an office wants to recruit 5 people. What is the number of ways this can be done so as to recruit at least one typist?

Concept:

The notation C(n, r) is the number of combinations/groups of n different things taking r at a time and is given by: \(C\left( {n,\;r} \right) = \frac{{{\rm{n}}!}}{{{\rm{r}}!\left( {{\rm{n\;}} - {\rm{\;r}}} \right)!}}\)

Calculation:

Given: 6 programmers and 4 typists, an office wants to recruit 5 people.

Here, n = 10 and r = 5.

Case 1: Group of 5 with 1 typist

1 typist can be chosen from 4 in C(4, 1) = 4 ways

4 programmers can be chosen from 6 in C(6, 4) = 15 ways

⇒ No. of ways to form a group of 5 with 1 typist = 4 × 15 = 60.

Case 2: Group of 5 with 2 typists

2 typist can be chosen from 4 in C(4, 2) = 6 ways

3 programmers can be chosen from 6 in C(6, 3) = 20 ways

⇒ No. of ways to form a group of 5 with 2 typist = 6 × 20 = 120.

Case 3: Group of 5 with 3 typists

3 typist can be chosen from 4 in C(4, 3) = 4 ways

2 programmers can be chosen from 6 in C(6, 2) = 15 ways

⇒ No. of ways to form a group of 5 with 2 typist = 4 × 15 = 60.

Case 4: Group of 5 with 4 typists

4 typist can be chosen from 4 in C(4, 4) = 1 way

1 programmers can be chosen from 6 in C(6, 1) = 6 ways

⇒ No. of ways to form a group of 5 with 2 typist = 1 × 6 = 6.