JAMB 2024 Mathematics · Question 21

In how many ways can a committee of 5 be selected from a group of 7 males and 3 females, if the committee must have one female?

Options

A.105ways
B.135 ways
C.210 ways
D.231 ways

Answer: A. 105ways

Explanation written with AI and checked against the answer key

1. Analyze the Question

We need to form a committee of 5 people from a total group of 10 people (7 males and 3 females). The condition is that the committee must have exactly one female.

2. Explain the Concept

This is a combination problem because the order in which the committee members are selected does not matter. The formula for combinations is:

^nC_r = \frac{n!}{r!(n-r)!}

To satisfy the condition of having exactly one female in a committee of 5:

We must choose 1 female from the 3 available females.
We must choose the remaining 4 members from the 7 available males (since $5 - 1 = 4$ ).

3. Show the Solution

Step 1: Choose 1 female from 3

^3C_1 = \frac{3!}{1!(3-1)!} = \frac{3}{1} = 3 \text{ ways}

Step 2: Choose 4 males from 7

^7C_4 = \frac{7!}{4!(7-4)!} = \frac{7 \times 6 \times 5 \times 4!}{4! \times 3 \times 2 \times 1}

^7C_4 = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = \frac{210}{6} = 35 \text{ ways}

Step 3: Calculate the total number of ways Since both actions must happen to form the committee, we multiply the results:

\text{Total ways} = (^3C_1) \times (^7C_4)

\text{Total ways} = 3 \times 35 = 105 \text{ ways}

4. Identify the Answer

Comparing our result to the given options, the correct choice is A.

5. Exam Tip

When a question says "must have one," it often implies "exactly one." If the question had said "at least one," you would have had to calculate the scenarios for 1, 2, and 3 females separately and add them together. Always read carefully to see if it implies "exactly" or "at least"!

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