There are 7 men and 10 women in a selection pool for a committee. The committee will have three distinct positions President, Vice President and Treasurer. In how many different ways can the committee be formed if exactly two of the three positions are occupied by men and the third position is occupied by a woman?

Introduction / Context:
This problem combines combinations and permutations when forming a small committee with designated positions. Unlike a team where order does not matter, here the roles of President, Vice President and Treasurer are distinct, so different assignments of the same people to different positions count as different committees. The condition that exactly two of these three positions must be filled by men forces us to consider the gender composition and the role assignments carefully.

Given Data / Assumptions:

• Total men available = 7.
• Total women available = 10.
• Three distinct positions: President, Vice President, Treasurer.
• Exactly two positions must be held by men and one by a woman.
• All individuals are distinct, and one person can hold only one position.

Concept / Approach:
We must first choose which two men and which one woman will serve, and then assign these three chosen people to the specific positions. Choosing the people is a combinations step because order is not yet relevant. Assigning them to named positions is a permutations step because each different assignment to the roles produces a new arrangement. The total number of valid committees is the product of these two stages.

Step-by-Step Solution:
Step 1: Choose 2 men out of 7. Number of ways = C(7, 2) = 21. Step 2: Choose 1 woman out of 10. Number of ways = C(10, 1) = 10. Step 3: Now there are 3 chosen people (2 men and 1 woman). Assign them to the three distinct posts. Number of ways to assign 3 distinct people to 3 positions = 3! = 6. Step 4: Multiply all stages together. Total number of valid committees = C(7, 2) * C(10, 1) * 3!. Total = 21 * 10 * 6 = 1260.

Verification / Alternative check:
An alternative viewpoint is to first select a woman for one of the three posts and then fill the remaining two posts with men. However, because the three posts are symmetric at the beginning, the method above already accounts for all possibilities with a clean combination followed by a permutation. Checking each step numerically confirms that 21 * 10 = 210 ways to choose the people, and for each such choice there are 6 different assignments to positions, giving 210 * 6 = 1260 arrangements.

Why Other Options Are Wrong:
The values 1200, 1240, 1620 and 840 do not match the correct combination and permutation breakdown. For example, 840 might arise if the assignment to roles is undercounted, while 1620 might come from miscounting the number of ways to choose the men or women. None of these values equals C(7, 2) * C(10, 1) * 3!, so they are incorrect.

Common Pitfalls:
Students often forget to account for the fact that the three positions are distinct and mistakenly use only combinations for all steps. Others may accidentally choose 2 men and 1 woman but then assign only some of them to roles, ignoring permutations. Another error is accidentally allowing 3 men or 3 women, which violates the “exactly two men” condition. Explicitly separating the selection of people from the assignment of posts prevents these mistakes.

Final Answer:
The committee with exactly two men and one woman in the three distinct posts can be formed in 1260 different ways.