A boy has five blue marbles, four green marbles and three red marbles. In
how many ways can he arrange four of them in a row, if the marbles of any
one colour are indistinguishable?
There are enough blue marbles to have BBBB.
There are enough green marbles to have GGGG.
If there were enough red marbles to have RRRR, then the answer would
be achieved this way:
3 choices of color for the 1st marble,
3 choices of color for the 2nd marble,
3 choices of color for the 3rd marble, and
3 choices of color for the 4th marble.
That would be 3∙3∙3∙3 = 34 = 81 ways.
However there are NOT enough red marbles to have RRRR.
So RRRR is the ONLY case of the 81 that we can't have,
because none of the others can contain more than 3 R's.
So we subtract 1 from the 81:
Answer: 80
Edwin