DrMaths wrote:Out of 5 positions, 1 to 5, the 2 empty chairs can be arranged, for example, at positions 1 and 2, written as 12 below.
Using this notation, here is a list of the empty seat positions: 12,13,14,15,23,24,25,34,35,45 = 10 combinations.
That answer is not provided in the question.
The result in blue must be multiplied by the number of ways the 3 people can be arranged in the occupied seats (3!):
10 * 3! = 60.
We could also proceed as follows:
From the 5 seats, the number of ways to choose 3 to be occupied = 5C3 = (5*4*3)/(3*2*1) = 10.
Within the occupied seats, the number of ways to arrange the 3 people = 3! = 6.
To combine the options above, we multiply:
10*6 = 60.
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