A decision committee of 5 members is to be formed from a group of 4 actors, 3 directors and 2 producers. In how many distinct ways can we form a committee consisting of exactly 2 actors, 2 directors and 1 producer?

Introduction / Context:
This question checks understanding of combinations in selection problems. We are selecting a fixed size committee from different categories of people: actors, directors and producers. The composition of the committee is restricted, which makes it a standard application of combinations where order does not matter but category counts do matter.

Given Data / Assumptions:

• There are 4 actors, 3 directors and 2 producers in total.
• A committee of exactly 5 members has to be formed.
• The committee must contain exactly 2 actors, exactly 2 directors and exactly 1 producer.
• The order in which members are selected does not matter, only the group composition matters.

Concept / Approach:
We use combinations because we are choosing people without regard to order. For each category, we count the number of ways to choose the required number of people using nCr. Then, since the selections from actors, directors and producers are independent, the total number of committees equals the product of the combination counts from each category. This is a typical multiplication principle application in combinatorics.

Step-by-Step Solution:
Step 1: Choose 2 actors from 4 actors. The number of ways is 4C2.Step 2: Compute 4C2 = 4! / (2! * 2!) = 6.Step 3: Choose 2 directors from 3 directors. The number of ways is 3C2.Step 4: Compute 3C2 = 3! / (2! * 1!) = 3.Step 5: Choose 1 producer from 2 producers. The number of ways is 2C1 = 2.Step 6: Multiply the independent choices. Total committees = 4C2 * 3C2 * 2C1 = 6 * 3 * 2 = 36.

Verification / Alternative check:
An alternative is to think in terms of listing small illustrative subsets. For instance, suppose actors are A1 to A4, directors are D1 to D3 and producers are P1, P2. Any valid committee must contain 2 labels from A set, 2 from D set and 1 from P set. If you start listing a few such committees and count the number of choices at each stage, you will see that for each pair of actors and directors, there are exactly 2 choices of producers. That is, for each of the 6 * 3 = 18 actor director combinations, you can pair P1 or P2, giving 18 * 2 = 36 committees. This matches the earlier calculation.

Why Other Options Are Wrong:

• 18: This counts only 4C2 * 3C2 and ignores the choices for producers.
• 24: This could arise from a miscalculation, such as 4C2 * 3C1 * 2C1, which uses the wrong director count.
• 32: This may come from misusing powers of 2 or guessing rather than applying combinations.

Common Pitfalls:
A frequent error is to treat each committee as an ordered arrangement and use permutations instead of combinations. Another mistake is to forget that selections from each group are independent, leading to incorrect addition instead of multiplication of counts. Some students also misread the requirement and pick the wrong number of directors or actors, which changes the combination values. Carefully mapping each phrase of the question to a combination calculation is essential.

Final Answer:
The total number of different committees that can be formed is 36 distinct ways.