A student council of 5 members is to be formed from a pool of 6 boys and 8 girls.\nIf Jason is one of the boys in the pool and must be included on the council, how many different councils can be formed?

Introduction / Context:
This question is about forming a committee with the extra condition that a specific person must be included. Such problems are common in combinatorics, and they help illustrate how to fix one member and then count combinations for the remaining positions.

Given Data / Assumptions:

• Total boys in the pool = 6 (including Jason).
• Total girls in the pool = 8.
• Total people in the pool = 14.
• Council size = 5 members.
• Jason must be on the council.
• Order of members within the council does not matter.

Concept / Approach:
Because Jason is required to be on the council, we first fix him as a member. After including Jason, we only need to choose the remaining 4 members from the remaining 13 people. Since the council is simply a group with no internal order, we use combinations C(n,r) to count the ways to choose these 4 additional members.

Step-by-Step Solution:
Total people in the pool = 14. Jason is always included, so he occupies one of the 5 council seats. Remaining people available = 14 - 1 = 13. Remaining seats to fill on the council = 5 - 1 = 4. Number of ways to choose 4 people from the remaining 13 = C(13,4). Compute C(13,4) = 13! / (4! * 9!). In product form, C(13,4) = (13 * 12 * 11 * 10) / (4 * 3 * 2 * 1). Compute numerator: 13 * 12 = 156, 11 * 10 = 110, then 156 * 110 = 17160. Compute denominator: 4 * 3 * 2 * 1 = 24. Now 17160 / 24 = 715.

Verification / Alternative check:
We can observe that C(13,4) should be a mid sized number since 13 is not very large. Also, C(13,4) equals C(13,9), reflecting symmetry in combinations. Known values around this range include C(10,4) = 210 and C(12,4) = 495, so 715 for C(13,4) fits the pattern of increasing values. Since Jason is fixed and we use all other choices only once, there is no double counting.

Why Other Options Are Wrong:

• 725: Slightly larger than C(13,4) and not equal to any standard combination from these numbers.
• 419: Too small; it misses a large portion of valid councils.
• 341: Also far from C(13,4) and does not correspond to any correct counting method here.

Common Pitfalls:
One common mistake is to forget to fix Jason first and instead compute C(14,5), which counts councils where Jason may or may not be present. Another is to think that Jason's fixed presence adds an extra factor like 5, which leads to overcounting. The safe approach is to treat Jason as already included and then apply the combinations formula to the remaining positions.

Final Answer:
The number of different student councils of 5 members that include Jason is 715.