In the Service Club there are 14 juniors and 23 seniors.\nThe club wants to send 4 representatives to a state conference, with exactly 2 juniors and 2 seniors.\nHow many different groups of representatives are possible?

Introduction / Context:
This problem is about forming a small team from two categories of members while respecting a fixed composition. Such questions test understanding of combinations and how to multiply combination counts from different groups when the selections are independent.

Given Data / Assumptions:

• Number of juniors in the club = 14.
• Number of seniors in the club = 23.
• Total people available = 14 + 23 = 37.
• We must choose exactly 4 representatives.
• Required composition: 2 juniors and 2 seniors.
• Order of representatives in the group does not matter.

Concept / Approach:
The selection can be broken into two independent choices: choosing juniors and choosing seniors. For each group we use combinations because we only care about which people are chosen, not the order in which they are picked. The total number of valid groups is then the product of the number of ways to choose juniors and the number of ways to choose seniors.

Step-by-Step Solution:
Number of ways to choose 2 juniors from 14 juniors = C(14,2). Compute C(14,2) = 14! / (2! * 12!) = (14 * 13) / 2 = 91. Number of ways to choose 2 seniors from 23 seniors = C(23,2). Compute C(23,2) = 23! / (2! * 21!) = (23 * 22) / 2 = 253. Total number of valid representative groups = C(14,2) * C(23,2) = 91 * 253. Compute 91 * 253: first 91 * 200 = 18200, then 91 * 50 = 4550 and 91 * 3 = 273. Add 18200 + 4550 + 273 = 23023.

Verification / Alternative check:
The result should be substantially larger than either 91 or 253, because we are combining choices from two groups. An approximate product 100 * 250 would give about 25000, so 23023 is a reasonable exact value. There is no overlap between the decisions about juniors and seniors, and each valid combination of juniors and seniors defines a unique group of 4 representatives, so the multiplication method is correct.

Why Other Options Are Wrong:

• 23024: Off by 1 from the correct answer, usually from a small arithmetic slip.
• 24023: Larger than the true product of 91 and 253, so it cannot be correct.
• 25690: Much larger and does not match any natural combination product from the given numbers.

Common Pitfalls:
A typical mistake is to use C(37,4), which counts all 4 person committees without regard to the required mix of juniors and seniors. Another error is to miscompute C(14,2) or C(23,2) by forgetting to divide by 2 or by mishandling the multiplication. Writing out the combination formulas clearly and computing the products in stages helps to avoid these errors.

Final Answer:
The number of different 4 person groups containing exactly 2 juniors and 2 seniors is 23023.