Question

a number of children are posing for a group picture. there are five 8 - year - olds and four 9 - year - olds posing in the picture. how many arrangements are possible where the 8 - year - olds and 9 - year - olds alternate in the lineup? 9! 5!×4! $\frac{9!}{4!5!}$ 5! + 4!

Explanation:

Step1: Determine the starting - point

Since there are 5 eight - year - olds and 4 nine - year - olds, the lineup must start with an eight - year - old.

Step2: Calculate permutations of each group

The number of ways to arrange the 5 eight - year - olds among themselves is given by the permutation formula \(n!\), where \(n = 5\), so there are \(5!\) ways. The number of ways to arrange the 4 nine - year - olds among themselves is \(4!\) ways.

Step3: Use the multiplication principle

By the multiplication principle of counting, the total number of arrangements where they alternate is the product of the number of arrangements of eight - year - olds and nine - year - olds. So the total number of arrangements is \(5!\times4!\).

Answer:

\(5!\times4!\)