Eight staff members (a, b, c, d, e, f, g, h) each have one scooter to be moved f

Question

Eight staff members (a, b, c, d, e, f, g, h) each have one scooter to be moved from one office building to another. Unfortunately the removalists took the scooters away before the staff members had a chance to put labels with delivery instructions on them. The removalists know the eight room locations for delivery, so will deliver one scooter per room. Give all your answers as exact fractions.

(a) In how many ways can the scooters be delivered to the eight offices?

(b) What is the probability that ‘a’ and ‘b’ each get their correct scooters?

(c) What is the probability that ‘a’ and ‘b’ get their correct scooters, but that ‘c’ does not?

(d) What type of (statistical) event is the outcome in part (c) compared to that in part (b)?

(e) Consider now the situation where the eight staff share their new offices in pairs, being (a,b) , (c,d) , (e,f) , (g,h). The removalists know to deliver two scooters to each of the four new offices. What is the probability that ‘a’ and ‘b’ get their correct scooters?

(f) Consider now the situation where the four staff are to share two new offices, with the allocations being (a,b,c,d) , (e,f,g,h). The removalists know to deliver four scooters to each of the two new offices. What is the probability that ‘a’ and ‘b’ get their correct scooters?

Explanation / Answer

(a) The first scooter can be delivered to 8 offices, the second scooter can be delivered to 7 offices, and so on.

So the total number of possible ways in which the 8 scooters can be delivered to the 8 offices = 8x7x6x5x4x3x2x1 = 8! = 40,320

(b) Since 'a' and 'b' get their correct scooters, we have 6 other scooters to allot among 6 offices. This can be done in 6! = 6x5x4x3x2x1 = 720 ways

the probability that ‘a’ and ‘b’ each get their correct scooters = 720 / 40320 = 0.0179

(c) With 'a' and 'b' have been fixed.

There are 6 scooters left to be delivered to 6 offices. There are 5 choices for c among the 6 possible, because one of them is the correct scooter. So the scooter to c's office has 5 choices. The rest of the 5 scooters can be delivered to 5 offices in 5! ways.

So the required number of ways = 5 x 5! = 600

the probability that ‘a’ and ‘b’ get their correct scooters, but that ‘c’ does not = 600 / 40320 = 0.0149

(d) The outcome in part (c) is a subset of outcomes in part (b)

(e) The two scooters of a and b can be allotted to them in 2 ways and the remaining 6 scooters can be allotted among the 6 staff members in 6! ways.

The number of ways in which a and b get their correct scooters in pair = 2 x 6! = 1440

Probability that  a and b get their correct scooters = 1440 / 40320 = 0.0357

(f) the number of ways in which 4 scooters can be selected for the first group = C[8,4] = 8! / (4! x 4!) = 70

with a's and b's scooters in the first group, rest of the 2 scooters for the first group can be selected in C[6,2] = 15 ways

Probability that a and b get their correct scooters = 15/70 = 0.2143

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