1. ## Combinations

A box contains ten bricks identical except for color. Three bricks are red, two are white, two are yellow, two are blue and one is black. In how many ways can three distinguishable bricks be:
(a) taken from the box
(b) arranged in a row?

I can't hit on any way to find (a), pairs of same colored bricks are possible as long as a differently colored brick is in the middle, but other than this I don't know what to do. And for (b) I don't know either.
I really am horrible at this permutations and combinations, so any help will be greatly appreciated.
Thanks

2. Originally Posted by arze
A box contains ten bricks identical except for color. Three bricks are red, two are white, two are yellow, two are blue and one is black. In how many ways can three distinguishable bricks be:
(a) taken from the box
(b) arranged in a row?

I can't hit on any way to find (a), pairs of same colored bricks are possible as long as a differently colored brick is in the middle, but other than this I don't know what to do. And for (b) I don't know either.
I really am horrible at this permutations and combinations, so any help will be greatly appreciated.
Thanks
For [a] - Plz clarify if pick for e.g. 2 blue and 1 red a valid outcome. Note the 2 blues are not distinguisable, as required by the question.

Somehow, I think, question is not very clear. Or maybe I have not understood it.

3. Originally Posted by arze
A box contains ten bricks identical except for color. Three bricks are red, two are white, two are yellow, two are blue and one is black. In how many ways can three distinguishable bricks be:
(a) taken from the box
(b) arranged in a row?
Short of using a generating function the only other way is to take up the various cases.
There is one way that all three can be the same color: red.
How many ways can all three be different colors?
How many ways can only two be of different colors?
So make a list and count.

Using a generating function we expand $(1 + x + x^2 + x^3 )(1 + x + x^2 )^3 (1 + x)$.
Doing so we see that the coefficient of $x^3$ is 27.
That is the number of ways to select three bricks.

To do part (b), count the arrangements in each of the three cases.
There is one way to arrange all red.
There are $3!$ ways to arrange three different colors.
There are $\frac{3!}{2!}$ ways to arrange two of one colors and one different color.

4. number of ways all are different would be $^5C_3=10$?
and number of ways only two of different colors $1\times^5C_2=10$?
I haven't learnt about generating functions.