1. How many ways...

How many ways can 17 identical balls be placed into 11 different boxes so that;

a) no box is empty
b) exactly on ebox is empty
c) at least on box is empty?

No need to evaluate any binomial coefficients.

2. Maybe this Wikipedia article can help. You need to count the number of 11-tuples of positive integers (for (a)) whose sum is 17. For (c), you could count the number of 11-tuples of natural numbers (zero allowed) and from there subtract (a).

4. See if this helps and then post here:

a) First you put one ball into each box. There is only one way to do that since choosing balls makes no sence because they are all identical. This is equal to choosing 11 boxes out of 11 boxes, hence one way to do it. This leaves you with six balls you have to place in eleven boxes. Number you're looking for is equal to the number of ways to place six balls in 11 boxes.

b) First choose one box that will remain empty. This can be done in 11 ways. Reasoning after that is similar to the above with 10 instead of 11 boxes.

c) For each k ranging from 1 to 10 choose k out of 11 boxes that will remain empty. After that things become similar to b).

5. For question c.
$\displaystyle \binom{17+11-1}{17}-\binom{6+11-1}{6}$.

Do you see why that is the answer?

6. Originally Posted by Plato
For question c.
$\displaystyle \binom{17+11-1}{17}-\binom{6+11-1}{6}$.

Do you see why that is the answer?
yes, i see. first you had \ $binom{17+11-1}{17}$ which represents all the ways to choose 11 balls from 17 boxes. that caters for all the possibilities.
now you have minius the number of ways that at least one box is empty. what you did was the the number of ways no box is empty.

did i explain it correctly?

7. Well put it this way: Subtract the number of ways that no box is empty from the total number of ways.
That is the number of ways that at least one box is empty.