A school committee of 5 students is to be formed from a class of 12 students. If John must be on the committee, in how many different committees can be formed under this condition?

Introduction / Context:
This is a committee selection problem with a simple condition: one particular student, John, must be included. The presence of a mandatory member slightly reduces the problem to choosing the remaining members from the remaining pool. These types of constraints are very common in exam questions related to combinations and selection with conditions.

Given Data / Assumptions:

• Total students in the class: 12.
• We must form a committee of 5 students.
• John is one of the 12 students and must be included in every valid committee.
• All students are distinct individuals.
• Committees are sets; the order of members does not matter.

Concept / Approach:
Since John must be in the committee, we fix him as one member. That leaves 4 more positions to be filled. The remaining available students are the other 11 students (everyone except John). Thus the problem reduces to choosing 4 students from these remaining 11 without any further restrictions. That is a combination problem: 11C4.

Step-by-Step Solution:
Step 1: Reserve one place on the committee for John. Step 2: Remaining positions to fill = 5 - 1 = 4. Step 3: Remaining students available for these 4 positions = 12 - 1 = 11. Step 4: Since order does not matter, the number of ways to choose the remaining 4 members is 11C4. Step 5: 11C4 = 11! / (4! * 7!). Step 6: We can compute this if needed, but the expression 11C4 already correctly represents the count. Step 7: Therefore, the desired number of committees is 11C4.

Verification / Alternative check:
We can compute 11C4 numerically: 11! / (4! * 7!) = (11 * 10 * 9 * 8) / (4 * 3 * 2 * 1) = 792. This confirms that there are 792 valid committees. The symbolic option 11C4 is correctly the combination expression that gives this number, so using it as the answer is appropriate when options are in terms of combinatorial notation.

Why Other Options Are Wrong:
11P4: This counts ordered selections, which is not relevant for committees where order does not matter. 11P5: This would correspond to ordered selections of 5 from 11, and it also ignores the fixed presence of John. 11C5: This would mean choosing 5 students from 11 without including John, which does not reflect the requirement that John must be on the committee. Only 11C4 correctly represents choosing the remaining 4 members among 11 when John is already included.

Common Pitfalls:
Some students forget to account for the fact that John is fixed and still use 12C5, which ignores the condition. Others misinterpret the notation and pick a permutation expression like 11P4, which would treat different orders of the same committee as distinct, overcounting heavily. Always carefully read conditions about required members and whether the order of selection matters.

Final Answer:
The number of different committees of 5 students that include John is 11C4 (numerically equal to 792).