From a group of 5 men and 11 women, in how many ways can a team of 11 be…

When a selection is split into mutually exclusive cases — here, by how many men are on the team — the total count is the SUM of each case's combinations (the addition principle), and each case is counted using the combination formula C(n, r) = n! / (r! (n − r)!), which counts unordered selections of r items from n distinct items.

Since the team of 11 may contain at most 3 men, the valid cases are exactly 0, 1, 2, or 3 men — each paired with enough women to complete the team of 11 (women available: 11):

1. 0 men, 11 women: C(5, 0) × C(11, 11) = 1 × 1 = 1

2. 1 man, 10 women: C(5, 1) × C(11, 10) = 5 × 11 = 55

3. 2 men, 9 women: C(5, 2) × C(11, 9) = 10 × 55 = 550

4. 3 men, 8 women: C(5, 3) × C(11, 8) = 10 × 165 = 1650

Adding the four cases: 1 + 55 + 550 + 1650 = 2256.

Cross-check: without the “at most 3 men” restriction, the team could also have 4 men (C(5,4) × C(11,7) = 5 × 330 = 1650) or 5 men (C(5,5) × C(11,6) = 1 × 462 = 462). Adding all six cases gives 2256 + 1650 + 462 = 4368, which matches the unrestricted count C(16, 5) = 4368 (choosing which 5 of the 16 people are left out of the team of 11) — confirming the case-wise total is consistent.

So the number of ways to select the team is 2256.