Committee and Seating Counts With Restrictions
A student council has 9 members: 5 seniors and 4 juniors. They must form a 4-person fundraising committee and, separately, arrange all 9 members in a row for a photograph. Two rules apply. Rule 1: the committee must contain at least one junior. Rule 2: in the photograph, no two juniors may stand next to each other. A student proposes computing the committee count as C(9,4) - C(5,4) and the photo count as 9! - (arrangements with juniors together), and a classmate objects that the second subtraction overcounts because pairs of juniors can be adjacent in more than one way. The council wants exact counts and a clear justification.
1. Compute the number of valid 4-person committees under Rule 1, and verify the count by a second method.
Read the worked answer
Method 1 (complement): total committees C(9,4) = 126. Committees with no junior contain only the 5 seniors: C(5,4) = 5. Valid committees = 126 - 5 = 121.
Method 2 (direct): count by number of juniors j = 1, 2, 3, 4, choosing j juniors from 4 and 4 - j seniors from 5:
- j = 1: C(4,1)·C(5,3) = 4 · 10 = 40
- j = 2: C(4,2)·C(5,2) = 6 · 10 = 60
- j = 3: C(4,3)·C(5,1) = 4 · 5 = 20
- j = 4: C(4,4)·C(5,0) = 1 · 1 = 1 Total = 40 + 60 + 20 + 1 = 121.
Both methods give 121, confirming the count. The complement method is shorter because 'at least one junior' has a simple negation ('no juniors').
2. Compute the number of photograph arrangements satisfying Rule 2, and explain why the classmate is right that subtracting arrangements with juniors together overcounts.
Read the worked answer
Use the gap method. Arrange the 5 seniors first: 5! = 120 ways. This creates 6 gaps (before the first senior, between consecutive seniors, after the last). Place the 4 juniors into 4 of these 6 gaps, at most one per gap, so no two juniors are adjacent: choose the gaps in C(6,4) = 15 ways and arrange the 4 juniors within them in 4! = 24 ways.
Total = 5! · C(6,4) · 4! = 120 · 15 · 24 = 43,200.
Why the subtraction overcounts: 'juniors together' is not a single event when there are 4 juniors. If we subtract arrangements where at least one specified pair is adjacent, an arrangement with two separate adjacent pairs is subtracted twice, and one with three juniors in a block is subtracted three times. Inclusion-exclusion would be needed to correct this, making the direct gap method far cleaner.
Check by complement idea: total arrangements 9! = 362,880; valid ones 43,200; invalid ones 319,680. The gap method avoids the inclusion-exclusion trap entirely and gives the exact count directly.