An ice-cream café sells a mega sundae in which the customer will order 5 scoops of ice-cream. The staff will arrange the 5 scoops on a plate in either a circular or row manner. The café offers exactly 10 distinct flavors of ice-cream daily.

1. Find the number of ways to choose 5 distinct flavours from the 10 available.

Number of ways to choose 5 flavours = $^{10}C_5 = 252$

2. Multiply by the number of ways to arrange the 5 chosen flavours in a circle to find the total arrangements.

Number of circular arrangements for 5 items = $(5-1)! = 4! = 24$

$\text{Total number of ways} = 252 \times 24 = 6048$

1. Determine the possible distributions of the 5 scoops among the 3 chosen flavours.

Since exactly 3 flavours (vanilla, chocolate, banana) are chosen for all 5 scoops and each flavour must be present at least once, the possible combinations of scoop counts are:

• $3$ of one flavour, and $1$ of each of the other two (i.e. $3-1-1$)
• $2$ of two flavours, and $1$ of the third (i.e. $2-2-1$)

2. Calculate the number of row arrangements for the 3-1-1 distribution.

There are $\binom{3}{1} = 3$ ways to choose which of the $3$ flavours gets $3$ scoops.
For each choice, the number of ways to arrange the 5 scoops in a row (with 3 identical items) is $\frac{5!}{3!1!1!} = 20$.
Number of arrangements = $3 \times 20 = 60$.

3. Calculate the number of row arrangements for the 2-2-1 distribution.

There are $\binom{3}{1} = 3$ ways to choose which of the $3$ flavours gets exactly $1$ scoop.
For each choice, the number of ways to arrange the 5 scoops in a row (with two sets of 2 identical items) is $\frac{5!}{2!2!1!} = 30$.
Number of arrangements = $3 \times 30 = 90$.

4. Find the total number of arrangements by adding the possibilities from both cases.

Total number of arrangements = $60 + 90 = 150$.