There are 41 students in a class, and the number of girls is one more than the number of boys. A team of four students is to be formed such that all four team members are not of the same gender and the numbers of girls and boys in the team are not equal. In how many different ways can such a team of four be formed?

Introduction / Context:
This problem combines combinatorics with constraints based on gender composition. We must form a four student team from a class where the counts of boys and girls are closely related. However, the team is restricted: it cannot be all boys, it cannot be all girls, and it also cannot have an equal number of boys and girls. This forces us to carefully enumerate the allowed gender combinations and count the selections for each case.

Given Data / Assumptions:

• Total students = 41.
• Number of girls is one more than the number of boys.
• Let number of boys be B and number of girls be G.
• B + G = 41 and G = B + 1.
• Team size = 4 students.
• Disallowed teams: all four from same gender, and teams with equal numbers of boys and girls.

Concept / Approach:
First, we solve for the numbers of boys and girls. Then we list all possible gender splits for a team of four and remove those that violate the conditions. The valid splits are those where there is at least one of each gender and the counts are not equal. After identifying valid splits, we compute the number of teams for each split using combinations and then sum the results.

Step-by-Step Solution:
Step 1: Solve for the number of boys and girls. B + G = 41 and G = B + 1. Substitute G: B + (B + 1) = 41, so 2B + 1 = 41, giving 2B = 40 and B = 20. Thus, G = B + 1 = 21. Step 2: List all possible gender splits for a team of 4. Possible splits: 4 boys 0 girls, 3 boys 1 girl, 2 boys 2 girls, 1 boy 3 girls, 0 boys 4 girls. Conditions: not all same gender and not equal numbers of boys and girls. Therefore, allowed splits are 3 boys 1 girl and 1 boy 3 girls. Step 3: Count teams with 3 boys and 1 girl. Choices for boys: C(20, 3). Choices for girls: C(21, 1). Number of teams = C(20, 3) * C(21, 1) = 1140 * 21 = 23940. Step 4: Count teams with 1 boy and 3 girls. Choices for boys: C(20, 1) = 20. Choices for girls: C(21, 3). C(21, 3) = 1330. Number of teams = 20 * 1330 = 26600. Step 5: Add both valid cases. Total valid teams = 23940 + 26600 = 50540.

Verification / Alternative check:
We can quickly check the disallowed cases. Teams of 4 boys: C(20, 4). Teams of 4 girls: C(21, 4). Teams of 2 boys and 2 girls: C(20, 2) * C(21, 2). None of these should be included. Counting only 3 boys 1 girl and 1 boy 3 girls, as we have done, matches the logical constraints. The final total 50540 is consistent with the combinatorial calculations.

Why Other Options Are Wrong:
The values 49450, 46587 and 52487 do not equal the sum of the two allowed case counts and likely include some disallowed configurations or arithmetic errors. The neat number 48000 is also incorrect and does not match the precise combination calculations for the valid gender splits.

Common Pitfalls:
Many learners forget to exclude teams with equal numbers of boys and girls or accidentally allow all boy or all girl teams. Others miscompute C(21, 3) or C(20, 3), leading to small errors that change the final total. Systematically listing all possible splits, marking which are allowed and which are disallowed, and then applying the combination formula carefully helps avoid such mistakes.

Final Answer:
The team of four students can be formed in 50540 different ways under the given gender constraints.