Discrete Math, counting and probability (white balls and black balls)? Can anyone help me with this please?
An urn contains ten white balls numbered from 1 to 10, and ten black balls numbered
from 1 to 10. A sample of 5 balls is chosen from the urn.
(a) How many dierent samples are there?
(b) How many samples in (a) have at least one white ball?
(c) How many samples in (a) have the property that the sum of the numbers on the balls
is even?
(d) How many samples in (a) have the property that the product of the numbers on the
balls is even?
(e) How many samples in (a) have the property that the numbers on the balls are distinct?
 A: (a), (b) Your calculations for the first two are right. There are $\binom{20}{5}$ possible samples. 
For at least one white, your alternative procedure is more efficient. The answer $\binom{20}{5}-\binom{10}{5}$ is right. 
(c) The sum is even if we have an odd number of evens (for then we have an even number of odds). So we want $1$, $3$, or $5$ evens. The number of ways to choose $1$ even and $4$ odd is $\binom{10}{1}\binom{10}{4}$. Write down also the number of ways to get $3$ even and $2$ odd, also $5$ even and $0$ odd, and add. 
But we do not need to add! For we get an odd sum if we get $1$ odd and $4$ even, or $3$ odd and $3$ even, or $5$ odd. If you write down the number of ways to do this, you will get exactly the same expression as we got for even sum. So the number with even sum is $\frac{1}{2}\binom{20}{5}$.
(d) There are $\binom{10}{5}$ ways to choose all odd. The rest will give even product. 
(e) There are $\binom{10}{5}$ ways to choose $5$ distinct numbers from $10$. For each such choice, look at the numbers you have. Each can be "represented" by a black or a white, giving a total of $\binom{10}{5}2^5$. 
A: Hints on how to go about each:
a. Consider the possible breakdowns and how many of each can you have?  For example, how many ways could you draw 5 white balls, 4 white and 1 black, 3 white and 2 black, ...
b. There are a couple of ways to go here.  First, is the consider all the possible ways to select 5 items from a set of 20 and then remove the all black balls cases that would give you the at least one white ball case for a removal way to count this.  The flip would be to sum all but the all black ball case in a.
c. This requires a bit more work to consider as you'd want to look at various odd and even cases.  While there is the requirement for at least one even, the other 4 balls have to pair up for the sum to be even.
d. This requires something a bit easier in that now you're looking for any even number amongst the 5 balls.
e. This requires you consider which duplicate cases can exist to remove from the overall set in a or else be very careful in constructing a counting case for all the balls to be distinct.
