How many ways can $4$ marbles selected from $2$ black and $4$ white marbles be arranged in a row?

Question

There are $2$ black marbles and $4$ white marbles.

How many ways can $4$ of these marbles be arranged in a row?

Answer 1

There are three cases to consider, depending on the number of black marbles that are used:

1. Four white marbles, no black marbles: There is one such sequence since the white marbles must occupy all four positions.
2. Three white marbles and one black marble: There are four positions the black marble could occupy. The remaining positions must be filled with white marbles, so there are four such sequences.
3. Two white marbles and two black marbles: Such a sequence is completely determined by choosing which two of the four positions will be occupied by the black marbles. The remaining positions must be filled with the white marbles. Hence, there are $\binom{4}{2} = 6$ such sequences.

The number of arrangements of four marbles is $1 + 4 + 6 = 11$. We can express the answer in the form $$\binom{4}{0} + \binom{4}{1} + \binom{4}{2}$$ where $\binom{4}{k}$ counts the number of ways of choosing which of the four positions will be occupied by black marbles when, respectively, $0$, $1$, or $2$ black marbles are used.

Answer 2

Consider three cases. First, when all 4 white marbles are taken, these can be arranged in one way. The second case, with 3 white and 1 black, has $4!/3!=4$ ways. The third, with 2 white and 2 black, has $4!/(2!2!)=6$ ways. And the answer is $1+4+6=11$.