I have a probability problem which is simplified as following, I need to check if my solution is true:

If there is a bag contains 20 of small present boxes. Three men try to put presents in these boxes consecutively. The first man chooses four boxes randomly and put a chocolate bar inside each one and put them back to the bag, then the second one chooses four boxes and put a coin inside each one and put them back to the bag. Finally, the third man selects four boxes and put a jelly bean inside each one and put them back to the bag. A child (who know the number of boxes that contain presents) comes later and chooses the same number of boxes randomly from the bag:

1) Assuming that no box contains more than one present, calculate the probability of selecting all boxes with presents.

2) Assuming that these boxes can have more than one present, calculate the probability of selecting all boxes with presents.

Solution of first question:

The child will choose (4*3) boxes with following probabilities:

P(1) =12/20
P(2) = 11/19

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P(12) = 1/9

Therefore, P = P(1)*P(2)...P(12)= (4*3)!(20-4*3)!/20!, as a result for (m) number of men, (x) number of presents each man put and (n) number of total boxes the general form of (P) is:

P = (mx)!(n-mx)!/n!.

Solution of the second question:

I think to calculate the probability $P_c$ of choosing the same box by the different men to be inserted by their present then subtract this from the term (mx) in P.

P_c is calculated as advised in http://math.stackexchange.com/a/1835968/346940:

The probability that some two men chooses the same box is the complement of the probability that no two men chooses the same box.

The number of ways to choose the three men's boxess such that they are all different is

\binom{20}{4}\binom{16}{4}\binom{12}{4} = 4845 * 1820 * 495

The number of ways to choose the same boxess with no restriction is

\binom{20}{4}\binom{20}{4}\binom{20}{4}= 1140 * 4845* 4845* 4845

The probability of the boxess being different is therefore the first expression above, divided by the second expression:

P_c =4845* 1820* 495/4845*4845*4845=900900/23474025


Terefore, P_d = 1 - P_c and the solution is:

P_2= (mx - P_d)!(n-(mx-P_d))!/n!

I am quite sure about the first question solution, I would like to check the second one please.