Evan has 15 toy bricks of which 5 are red, 4 are blue, 4 are yellow and 2 are green. The bricks of the same colour are identical.

1. Determine the number of ways to choose 2 different colours from the 4 available colours, as bricks of the same colour are identical.

$\text{Number of ways} = \binom{4}{2} = 6$

1. Calculate the number of ways to choose 3 identical bricks and 3 different bricks.

For 3 identical bricks, only Red, Blue, and Yellow have at least 3 bricks available.
Number of ways $= 3$

For 3 different bricks, we choose 3 colours out of the 4 available colours.
Number of ways $= \binom{4}{3} = 4$

2. Calculate the number of ways to choose 2 identical bricks and 1 different brick.

All 4 colours have at least 2 bricks, so there are 4 choices for the colour of the 2 identical bricks.
The remaining 1 brick can be chosen from the other 3 colours.
Number of ways $= 4 \times 3 = 12$

3. Calculate the total number of ways by summing the possibilities from all cases.

$\text{Total number of ways} = 3 + 4 + 12 = 19$

1. List all possible combinations of 4 bricks that have a total value of $8.

Let $r$, $b$, $y$, and $g$ be the number of red, blue, yellow, and green bricks respectively.
We need $r + b + y + g = 4$ and $3r + 2b + y = 8$.
By systematic listing, the possible combinations of 4 bricks are:

• 2 Red, 1 Blue, 0 Yellow, 1 Green (Value: $2(\char36{}3) + 1(\char36{}2) + 0(\char36{}1) + 1(\char36{}0) = \char36{}8$)
• 2 Red, 0 Blue, 2 Yellow, 0 Green (Value: $2(\char36{}3) + 0(\char36{}2) + 2(\char36{}1) + 0(\char36{}0) = \char36{}8$)
• 1 Red, 2 Blue, 1 Yellow, 0 Green (Value: $1(\char36{}3) + 2(\char36{}2) + 1(\char36{}1) + 0(\char36{}0) = \char36{}8$)
• 0 Red, 4 Blue, 0 Yellow, 0 Green (Value: $0(\char36{}3) + 4(\char36{}2) + 0(\char36{}1) + 0(\char36{}0) = \char36{}8$)

2. Calculate the number of ways to choose the distinct bricks for each valid combination.

Since the bricks of the same colour are non-identical, we use combinations (${}^nC_r$) to find the ways to pick the specific bricks:

• 2 Red, 1 Blue, 1 Green: $\binom{5}{2} \times \binom{4}{1} \times \binom{2}{1} = 10 \times 4 \times 2 = 80$
• 2 Red, 2 Yellow: $\binom{5}{2} \times \binom{4}{2} = 10 \times 6 = 60$
• 1 Red, 2 Blue, 1 Yellow: $\binom{5}{1} \times \binom{4}{2} \times \binom{4}{1} = 5 \times 6 \times 4 = 120$
• 4 Blue: $\binom{4}{4} = 1$

3. Sum the number of ways from all valid cases.

Total number of ways $= 80 + 60 + 120 + 1 = 261$