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20 Feb 2026, 03:44
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D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
65%
(hard)
Question Stats:
53% (01:56) correct
47%
(01:42)
wrong
based on 104
sessions
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Noah and Mia are with five friends at the theatre. In how many ways can they be seated in seven adjacent seats so that there is one person per seat and exactly one friend sits between Noah and Mia?
A. 120
B. 240
C. 600
D. 1200
E. 1440
A. 120
B. 240
C. 600
D. 1200
E. 1440
20 Feb 2026, 04:08
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kevincan
Let the 7 friends be Noah, Mia, and A,B,C,D,E.
There are 7 places, let the positions be 1,2,3,4,5,6 and 7.
Selecting three successive places For Noah , Mia and other person, can be
(1,2,3)
(2,3,4)
(3,4,5)
(4,5,6)
(5,6,7)
So, we know there are 5 positions.
Noah and Mia can interchange positions, so we get a 2!.
And, selecting one among A,B,C,D,E for the position between Noah and Mia. 5C1 = 5.
And the remaining 4 can be arranged in 4! Ways.
Total ways = 5 * 2! * 5 * 4! = 1200 ways
Option D
20 Feb 2026, 05:21
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Great question — this is a classic seat-arrangement problem with a constraint, and it's one where students consistently lose points by undercounting or overcounting. Let me break it down carefully.
Key concept being tested: Constrained Permutations
The setup: 7 people (Noah, Mia, and 5 friends — call them A, B, C, D, E) need to be seated in 7 adjacent seats, with the constraint that exactly one friend sits between Noah and Mia.
Step 1: Find the number of valid position-sets for the Noah–[friend]–Mia block.
We need three consecutive seats where Noah is in seat 1, one friend is in seat 2, and Mia is in seat 3 (or vice versa). The three-seat window can start at position 1, 2, 3, 4, or 5 in a row of 7:
(1,2,3), (2,3,4), (3,4,5), (4,5,6), (5,6,7) → 5 position windows
Step 2: Arrange Noah and Mia within the block.
Noah can be on the left with Mia on the right, or Mia on the left with Noah on the right. That's 2 ways.
Step 3: Choose which friend sits in the middle.
Any 1 of the 5 remaining friends can be the person sitting between them → 5C1 = 5 ways
Step 4: Arrange the remaining 4 friends in the other 4 seats.
The remaining 4 people can sit in any order → 4! = 24 ways
Step 5: Multiply everything together.
5 × 2 × 5 × 24 = 1,200
Answer: D
Common trap: Many students treat the Noah–friend–Mia group as a single block and then arrange 5 units in a row (giving 5! = 120 as the block positions), forgetting that the block itself is 3 seats wide within a fixed 7-seat row — not a circular or flexible arrangement. The block can only start in 5 specific positions, not slide freely.
Takeaway: When you have a "sandwich" constraint (X is between A and B), count the valid windows for the trio explicitly, then multiply by the internal arrangements of the trio and the external arrangements of everyone else.
Key concept being tested: Constrained Permutations
The setup: 7 people (Noah, Mia, and 5 friends — call them A, B, C, D, E) need to be seated in 7 adjacent seats, with the constraint that exactly one friend sits between Noah and Mia.
Step 1: Find the number of valid position-sets for the Noah–[friend]–Mia block.
We need three consecutive seats where Noah is in seat 1, one friend is in seat 2, and Mia is in seat 3 (or vice versa). The three-seat window can start at position 1, 2, 3, 4, or 5 in a row of 7:
(1,2,3), (2,3,4), (3,4,5), (4,5,6), (5,6,7) → 5 position windows
Step 2: Arrange Noah and Mia within the block.
Noah can be on the left with Mia on the right, or Mia on the left with Noah on the right. That's 2 ways.
Step 3: Choose which friend sits in the middle.
Any 1 of the 5 remaining friends can be the person sitting between them → 5C1 = 5 ways
Step 4: Arrange the remaining 4 friends in the other 4 seats.
The remaining 4 people can sit in any order → 4! = 24 ways
Step 5: Multiply everything together.
5 × 2 × 5 × 24 = 1,200
Answer: D
Common trap: Many students treat the Noah–friend–Mia group as a single block and then arrange 5 units in a row (giving 5! = 120 as the block positions), forgetting that the block itself is 3 seats wide within a fixed 7-seat row — not a circular or flexible arrangement. The block can only start in 5 specific positions, not slide freely.
Takeaway: When you have a "sandwich" constraint (X is between A and B), count the valid windows for the trio explicitly, then multiply by the internal arrangements of the trio and the external arrangements of everyone else.
