Forming 3-boy groups with an exclusion: From boys {A, B, C, D, E}, how many groups of 3 can be formed if C and D must not be in the same group?

Introduction / Context:This is combinations with an exclusion constraint: count all 3-person groups, then subtract the forbidden ones that include both C and D together.

Given Data / Assumptions:

• Five distinct boys: A, B, C, D, E.
• We need groups (order does not matter) of size 3.
• Constraint: C and D cannot both appear in a chosen group.

Concept / Approach:

• Total groups = C(5,3).
• Forbidden groups contain {C, D} plus one of the remaining three.
• Allowed = total − forbidden.

Step-by-Step Solution:

Verification / Alternative check:List groups and strike those containing both C and D; seven remain. This matches the calculation.

Why Other Options Are Wrong:

• 8 double-counts or mis-subtracts forbidden groups.
• 5 and 6 undercount the allowed combinations.

Common Pitfalls:

• Treating ordered arrangements instead of combinations.

Final Answer:7