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A 5-member panel is to be chosen from 6 economists and 4 historians and then seated in a row, subject to the rule that no 2 historians may sit next to each other. The panel must contain exactly 2 historians and 3 economists. In how many ways can the panel be both chosen and seated?
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Correct answer
C
First select the members: choose 3 economists from 6 and 2 historians from 4, C(6,3) × C(4,2) = 20 × 6 = 120 selections. Now seat 3 economists and 2 historians in a row with no two historians adjacent.
Seat the 3 economists first in 3! = 6 orders. They create 4 gaps (before, between, between, after) into which the 2 historians go, one per gap, and the historians are placed and ordered in P(4,2) = 4 × 3 = 12 ways. Seatings per selection = 6 × 12 = 72.
Total = 120 × 72 = 8640.
Verification by total-minus-adjacent: the total seatings of 5 distinct people is 5! = 120. The seatings with the 2 historians adjacent (glue them into one block) are 4! × 2! = 24 × 2 = 48, so the valid count is 120 − 48 = 72, which matches the 72 found from the gap method. So 120 × 72 = 8640 is confirmed.
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