Question

Suppose you are doing Christmas photo arrangements with a family of five adults and seven children. The smallest child is Tiny Tim.

(a) Assume all the adults are in the back row and all the children are in the front row. How many ways are there to arrange the family in this case?

(b) All else being equal, what is the probability that a randomly chosen such arrangement has Tiny Tim on the far left of the row of children?

Answer 1

(a) By the rule of permutations the number of ways of arranging n things/people all taken together = n!

Here we have 5 adults who are in the back row, and they can be arranged in 5! = 120 ways

There are 7 children who are in the front row and they can be arranged in 7! = 5040 ways

Therefore the total number of arrangements = 120 * 5040 = 604800

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(b) Probability = Favorable Outcomes / Total Outcomes

Total outcomes as found in (a) = 604800

Favorable outcomes has Tiny Tim on the far left.

Since his place his fixed. The remaining 6 children can be arrange in 6! = 720 ways.

For adults the number of ways of arranging them = 5! = 120

Therefore the total Favorable outcomes = 120 * 720 = 86400

The required probability = 86400 / 604800 = 1 / 7 = 0.1429

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