Discussion
A committee of 4 people is to be appointed from 3 officers of the production department, 4 officers of the purchase department, 2 officers of the sales department and one chartered accountant. What is the probability of forming the committee that there must be one from each category ?
Answer: D
The probability is \(\frac{8}{70}\).
There are \(3 + 4 + 2 + 1 = 10\) people. One from each of the four categories can be chosen in \(3 \times 4 \times 2 \times 1\) ways.
\[ \begin{aligned} \text{total} &= \binom{10}{4} = 210 \\ \text{favourable} &= 3 \times 4 \times 2 \times 1 = 24 \\ P &= \frac{24}{210} = \frac{8}{70} \end{aligned} \]
\(\frac{8}{70}\) is \(\frac{4}{35}\), left unreduced to match the options. The other options correspond to 6, 12 and 18 favourable committees out of 210, all undercounts of the 24 ways to pick one person from each category.
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