In how many different ways can you distribute 3 apples, 3 oranges and 2 bananas to 8 children, if each child receives one fruit?
my work: 3 apples implies choosing 3 children from eight where the order of selection does not matter. That gives 8x7x6/3x2x1=56 options.
3 oranges the equals 3 choices from 5 where the order doesn't matter = 5x4x3/2x1=30 options.
And when we have distributed those, the remaining 2 children get the bananas - no further ootions.
Total possibilities = 56 x 30 = 1680 possibilities.
There is a problem when you calculate the number of ways you can distribute $3$ oranges among $5$ people. You calculate this as being
$$ \frac{5 \times 4 \times 3}{2 \times 1} = 30.$$
This is incorrect. The correct number of ways you can distribute $3$ oranges among $5$ people is $$ \frac{5\times 4 \times 3}{3 \times 2 \times 1} = 10$$
Combining this with the first part you did correctly we get the final answer $56 \times 10 = 560.$
Remember that in general the number of ways of choosing $k \leq n $ objects among $n$ is
$$ {n \choose k}:=\frac{n !}{k! (n-k)!}.$$