5 keys are randomly placed among 3 places designated as p1,p2 and p3 Find the probability of the following events
1- place p1 is empty
2- place p1 and only p1 is empty
3- exactly 1 place is empty
4- At least 1 place is empty
5- No place is empty
6- 2 places are empty
7- 3 places are empty
8- either p1 or p2 is empty

Repeat the above experiment with n keys and three places Check you general expression against your answer to above experiment
The persons p1 p2 p3 are among those 10 persons who have placed their cheques on a counter.Each person take a cheque random from the cheques on the counter. What is the probability that at least one either p1 p2 or p3 will get his own cheque?

1. Place p1 is empty:

This combines the cases where you have the distributions:
0 5 0
0 4 1
0 3 2
0 2 3
0 1 4
0 0 5

Can you get each probability?

2. Place p1 and only p1 is empty

This is simpler, and includes:
0 4 1
0 3 2
0 2 3
0 1 4

3. Exactly 1 place is empty:

0 4 1
0 3 2
0 2 3
0 1 4
4 0 1
3 0 2
2 0 3
1 0 4
4 1 0
3 2 0
2 3 0
1 4 0

4. At least 1 place is empty:

0 0 5
0 5 0
5 0 0
0 4 1
0 3 2
0 2 3
0 1 4
4 0 1
3 0 2
2 0 3
1 0 4
4 1 0
3 2 0
2 3 0
1 4 0

5. No place is empty:

1 1 3
1 2 2
1 3 1
2 1 2
2 2 1
3 1 1

6. Two empty places:

0 0 5
0 5 0
5 0 0

7. Can't you seriously not figure this out?

8. Either p1 or p2 is empty:

0 0 5
0 5 0
5 0 0
0 4 1
0 3 2
0 2 3
0 1 4
4 0 1
3 0 2
2 0 3
1 0 4

Assuming the keys are identical and the boxes are different.

If we are distributing r identical objects into n different boxes, then there are

C(r+n-1,r) ways

In this case, C(5+3-1,5)=21 ways

I would think keys are distinct, though, but perhaps not. It is not specified.

i.e. If no space is to be empty, then place one key in each of the 3 boxes. Then, there are only a matter of distributing the 2 remaining keys in the 3 boxes in C(2+3-1,2)=6 ways. Which is what U has in his post. Thus, the probability no box is empty would be 6/21=2/7.

If the keys are distinct, that is another matter.