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10 Apr 2026, 04:10
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A
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
(hard)
Question Stats:
35% (02:25) correct
65%
(02:59)
wrong
based on 55
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A minibus has 3 rows of seats with 3 seats in each row. The six passengers who booked in advance have already taken their randomly assigned seats. A couple is next to board. What is the probability that there are two adjacent empty seats in the same row for the couple?
(A) \(\frac{13}{28}\)
(B) \(\frac{1}{2}\)
(C) \(\frac{15}{28}\)
(D) \(\frac{4}{7}\)
(E) \(\frac{9}{14}\)
(A) \(\frac{13}{28}\)
(B) \(\frac{1}{2}\)
(C) \(\frac{15}{28}\)
(D) \(\frac{4}{7}\)
(E) \(\frac{9}{14}\)
10 Apr 2026, 07:03
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A minibus has 3 rows of seats with 3 seats in each row. The six passengers who booked in advance have already taken their randomly assigned seats. A couple is next to board. What is the probability that there are two adjacent empty seats in the same row for the couple?
Determine how many ways there are for the six passengers who booked in advance to occupy seats.
9c6 = 9c3 = (9 × 8 × 7)/3! = 84
Determine how many ways two empty seats can be together in the same row with the six other seats taken.
There are two different ways for two empty seats to be together, three empty seats in a row and two empty seats in a row.
Three seats empty in one row:
3 ways
Two seats empty in the same row and the third occupied:
(1 way right 2 in row are empty + 1 way left 2 in row are empty) × 3 rows that can have 2 empty × 6c5 ways for the remaining 5 passengers to occupy seats in the other two rows = 2 × 3 × 6 = 36
3 + 36 = 39
39/84 = 13/28
(If you got 1/2, you probably overcounted by including the cases in which all three seats in a row were empty twice, as if two on the left in a row empty and the third also not taken by one of the other six passengers and two on the right in a row empty and the third also not taken by one of the other six passengers are different cases. Doing so produces three extra cases, meaning 42 favorable cases, rather than the correct 39.)
(A) \(\frac{13}{28}\)
(B) \(\frac{1}{2}\)
(C) \(\frac{15}{28}\)
(D) \(\frac{4}{7}\)
(E) \(\frac{9}{14}\)
Correct answer: A
Determine how many ways there are for the six passengers who booked in advance to occupy seats.
9c6 = 9c3 = (9 × 8 × 7)/3! = 84
Determine how many ways two empty seats can be together in the same row with the six other seats taken.
There are two different ways for two empty seats to be together, three empty seats in a row and two empty seats in a row.
Three seats empty in one row:
3 ways
Two seats empty in the same row and the third occupied:
(1 way right 2 in row are empty + 1 way left 2 in row are empty) × 3 rows that can have 2 empty × 6c5 ways for the remaining 5 passengers to occupy seats in the other two rows = 2 × 3 × 6 = 36
3 + 36 = 39
39/84 = 13/28
(If you got 1/2, you probably overcounted by including the cases in which all three seats in a row were empty twice, as if two on the left in a row empty and the third also not taken by one of the other six passengers and two on the right in a row empty and the third also not taken by one of the other six passengers are different cases. Doing so produces three extra cases, meaning 42 favorable cases, rather than the correct 39.)
(A) \(\frac{13}{28}\)
(B) \(\frac{1}{2}\)
(C) \(\frac{15}{28}\)
(D) \(\frac{4}{7}\)
(E) \(\frac{9}{14}\)
Correct answer: A
10 Apr 2026, 08:41
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Case 1:
Arranging them in blocks of 3 each. So essentially choosing 2 out of 3 block = 3C2 = 3 ways.
Case 2:
When only 1 pair of adjacent seats are available in each row.
To calculate this we first choose 3 in 1 block, 2 in the next block, 1 in the last block.
3C3*3C1*2C1*3C2*2 = 1*3*2*3*2 = 6*6 = 36 ways.
Total ways = 36+3 = 39 ways.
Total outcomes = 9C3 = 7*8*9/6 = 84.
39/84 = 13/28.
Answer: Option A
Arranging them in blocks of 3 each. So essentially choosing 2 out of 3 block = 3C2 = 3 ways.
Case 2:
When only 1 pair of adjacent seats are available in each row.
To calculate this we first choose 3 in 1 block, 2 in the next block, 1 in the last block.
3C3*3C1*2C1*3C2*2 = 1*3*2*3*2 = 6*6 = 36 ways.
Total ways = 36+3 = 39 ways.
Total outcomes = 9C3 = 7*8*9/6 = 84.
39/84 = 13/28.
Answer: Option A
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