# Distributing $4$ indistinguishable black marbles and $6$ distinguishable coloured marbles into $5$ distinguishable boxes.

Prove that there are exactly $$8100$$ different ways of distributing $$4$$ indistinguishable black marbles and $$6$$ distinguishable coloured marbles ( none of them black) into $$5$$ distinguishable boxes in such a way that each box contains exactly $$2$$ marbles.

I have done problems involving indistinguishable balls and indistinguishable/distinguishable boxes, distinguishable balls and indistinguishable/distinguishable boxes.

I am confused about how to handle the situation when both indistinguishable and distinguishable balls are given at the same time.

Just distribute the distinguishable balls as a product of two multinomial coefficients,
[laying down a pattern]$$\;\times\;$$ [permuting it]
and forget about filler indistinguishable black balls, thus

$$2-2-2-0-0\; pattern:\; \Large\binom6{2,2,2,0,0}\binom5{3,2} = 900$$

$$2-2-1-1-0\;pattern:\; \Large\binom6{2,2,1,1,0}\binom5{2,2,1} =5400$$

$$2-1-1-1-1\;pattern:\; \Large\binom6{2,1,1,1,1}\binom5{1,4} = 1800$$

Count ways to assign black balls to the boxes – partitioned by how many boxes contain two black balls – then count ways to arrange distinguishable balls among the remaining spaces — remembering that ball placement inside a box is not ordered.

Eg: The count of ways to assign two boxes with double-black, and arrange the distinguishable balls into pairs among the remaining three boxes, is: $$\dfrac{5!}{2!\,3!}\cdotp\dfrac{6!}{2!^3}$$

Do similarly for the remaining cases.

We can break this problem down into simply arranging the distinguishable marbles in 5 distinguishable boxes(such that no box gets more than 2 marbles), and then using the indistinguishable ones a "filler" of sorts.

We can do this by considering that the distinguishable marbles can be arranged in 3 ways,

1. $$4$$ boxes with one marble, $$1$$ with two marbles: This can be done in $$\binom51\frac{6!}{2!}=1800$$ ways.
2. $$2$$ boxes with one marble, $$2$$ with two marbles and 1 with none: $$\binom5{2,2,1}\frac{6!}{2!2!}=5400$$ ways.
3. $$3$$ boxes with two marbles, $$2$$ with none: $$\binom{5}{3}\frac{6!}{2!2!2!}=900$$ ways.

Thus, we have a total of $$1800+5400+900=\bbox[5px,border:2px solid #C0A000]{8100\text{ ways.}}$$