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June 2006 p6 q4
2878
The diagram shows the seating plan for passengers in a minibus, which has 17 seats arranged in 4 rows. The back row has 5 seats and the other 3 rows have 2 seats on each side. 11 passengers get on the minibus.
Of the 11 passengers, 5 are unmarried and the other 6 consist of 3 married couples.
In how many ways can 5 of the 11 passengers on the bus be chosen if there must be 2 married couples and 1 other person, who may or may not be married?
Solution
To solve this problem, we need to select 2 married couples and 1 additional person from the remaining passengers.
First, choose 2 married couples from the 3 available couples. The number of ways to do this is given by the combination formula:
\(\binom{3}{2} = 3\)
After choosing 2 married couples, there are 7 passengers left (5 unmarried and 2 from the remaining couple). We need to choose 1 additional person from these 7 passengers:
\(\binom{7}{1} = 7\)
Therefore, the total number of ways to choose the passengers is: