Good thinking, but 6C3 =
20 overcounts because
not every selection of 3 people for the front row is valid!Simple example to see why:Label people
1 (shortest) to
6 (tallest).
Say you pick {4, 5, 6} for the front row.
Then {1, 2, 3} must go to the back row.
Arranged by height:
Quote:
Back row: 1 — 2 — 3
Front row: 4 — 5 — 6
Now check: Is each back person taller than the front person?
- Column 1: Is 1 > 4?
NO!This selection breaks the rule, so it doesn't count.
Key insight: The 3 tallest people cannot ALL be in the front row — there'd be no one tall enough to stand behind them!Many of your 20 selections have this same problem. When you carefully count only the valid ones, you get
5.
Answer: A (5)MG1105
Bunuel Basically the only option we have is to choose the three people who will stand in the front row. All other permutation will be governed by the height order and hence is 1 (since there can only be one way to line up people as per their height unless the height is same). So the answer should be 6C3, right? Just wanted to confirm if this is the correct approach.