A research team of 6 people is to be formed from 10 chemists, 5 politicians, 8 economists and 15 biologists. In how many different ways can such a team be chosen if the team must contain at least 5 chemists?

Introduction / Context:
This problem is a classic example of forming a committee with a restriction on the composition of its members. We are given several different categories of specialists and are asked to count the number of ways to form a research team of fixed size when at least a minimum number of chemists must be included. Such questions check the ability to break a counting problem into cases and use combinations correctly while satisfying a constraint like “at least k of a particular type”.

Given Data / Assumptions:

• Total chemists available = 10.
• Total politicians available = 5.
• Total economists available = 8.
• Total biologists available = 15.
• Team size required = 6 people.
• At least 5 chemists must be on the team.
• All individuals are distinct and each person can be selected at most once.

Concept / Approach:
The phrase “at least 5 chemists” means the team can have either exactly 5 chemists or exactly 6 chemists. We handle each possible case separately and then add the counts, because the cases are mutually exclusive but collectively exhaustive. Within each case, the members are chosen using combinations since the order in which people are picked does not matter; the team is an unordered group. The other non chemist members come from the remaining categories combined as a single pool of distinct people.

Step-by-Step Solution:
Step 1: Total non chemist pool size = politicians + economists + biologists = 5 + 8 + 15 = 28. Case 1: Exactly 5 chemists and 1 non chemist. Choose 5 chemists from 10: C(10, 5). Choose 1 non chemist from 28: C(28, 1). Number of teams in Case 1 = C(10, 5) * C(28, 1). C(10, 5) = 252 and C(28, 1) = 28, so Case 1 count = 252 * 28 = 7056. Case 2: Exactly 6 chemists and 0 non chemists. Choose 6 chemists from 10: C(10, 6). C(10, 6) = 210. Total number of valid teams = Case 1 + Case 2 = 7056 + 210 = 7266.

Verification / Alternative check:
Because the team size is 6 and we need at least 5 chemists, there are only two possibilities: (5 chemists, 1 non chemist) or (6 chemists, 0 non chemists). There is no other composition that satisfies the condition. The combination values C(10, 5) and C(10, 6) can be checked manually using factorial definitions. Adding these two non overlapping cases gives 7266, which is consistent with combinatorial reasoning and the problem statement.

Why Other Options Are Wrong:
Values like 6379 or 6400 do not correspond to any clean combination expression and suggest arithmetic or counting mistakes. The value 7350 is larger than the correct answer and might come from including teams with fewer than 5 chemists. The remaining distractor 7000 is also not equal to C(10, 5) * 28 plus C(10, 6). None of these options correctly model both allowed cases and therefore they are incorrect.

Common Pitfalls:
Learners often misinterpret “at least 5 chemists” as “exactly 5 chemists” and forget the case with 6 chemists. Another common error is to separate the non chemists by category and overcomplicate the selection rather than treating all non chemists as one pool. Some also mistakenly use permutations instead of combinations even though the team is unordered. Clearly writing out the cases and using combinations for each avoids these mistakes.

Final Answer:
The number of different 6 person research teams that can be formed with at least 5 chemists is 7266.