Let P be the probability that no card gets placed into a box with the same label.

P=P_0=1-\left(P_4+P_3+P_2+P_1\right) where P_nrepresents the probability that **exactly n cards** are placed in a box with the same label.

Finding P_4, etc is easier than finding P_0.

Finding P_4: All 4 cards are placed in a box with the same label. This can happen exactly once. So, **P_4=\frac{1}{24}**.

Finding P_3: Exactly 3 cards are placed in a box with the same label. This implies that the 4th card should also be placed in a box with the same label. Hence, it is not possible for exactly 3 cards to be placed in a box with the same label. So, P_3=\frac{0}{24}

Finding P_2: Exactly 2 cards are placed in a box with the same label. The remaining 2 cards should not be placed in a box with the same label.

There are 6 ways of picking the 2 numbers to place in a box with the same label. For each of the 6 ways, there is exactly one way in which the remaining 2 cards are placed into a box of different label. So, P_2=\frac{6}{24}.

Finding P_1: Exactly 1 card must be placed in a box with the same label. There are 4 ways of picking this 1 card.

For each such combination, each of the remaining 3 cards have to be placed in a box of different card. Let the 3 cards be named A, B and C. Pick the card A and place it in the box named B. Now, pick the card that does not correspond to the box in which A is placed - in this case C. This can be placed only in box A; and the last card B can be placed in box C. One combination is (A, B); (C, A) and (B, C). Similarly, if A is placed in box C, the combination is (A, C); (B, A) and (C, B)

For each combination, there are 2 ways in which all remaining cards are placed in boxes with a different label.

So, P_1=\frac{\left(4\cdot2\right)}{24}

Hence, P_0=1-\left(\frac{1}{24}+\frac{0}{24}+\frac{6}{24}+\frac{8}{24}\right)=1-\frac{15}{24}=\frac{9}{24}