Cards and Boxes

Since there are five even-numbered cards and five boxes, each box must get one and only one even-numbered card. Let's suppose this has been done; then the other five (odd-numbered) cards can be distributed among the five boxes in $5!$ ways. However, neither card 3 or card 9 can go into the same box as card 6; also card 5 cannot go into the same box as card 10. To eliminate these possibilities, we can use inclusion-exclusion: we subtract from $5!$ the number of ways of distributing the cards so either card 3 or card 9 does go into the same box as card 6, as well as the number of ways that card 5 goes into the same box as card 10; then we must add back in the number of ways of both of these happening.

• The number of ways of distributing the five odd-numbered cards so that 3 or 9 go into the box containing 6 is $\ \color{#3D99F6} 2 \times 4!$ .

• The number of ways of distributing the five odd-numbered cards so that 5 goes into the box containing 10 is $\ \color{#3D99F6} 1 \times 4!$ .

• The number of ways of distributing the five odd-numbered cards so that 3 or 9 go into the box containing 6 AND 5 goes into the box containing 10 is $\ \color{#3D99F6} 2 \times 1 \times 3!$ .

Thus the number of ways of distributing the five odd-numbered cards so that each box contains a coprime pair is

$5! - (2 \times 4!) - (1 \times 4!) + (2 \times 1 \times 3!) = 120 - 48 - 24 + 12 = \boxed{60}$ .

If a box contains two odd-number cards,then there must be other boxes containing two even-number cards(not coprime).
So,every box should contain an odd-number card and an even-number card.
Put the five odd-number cards in the five boxes.Since each box contains a card,I will call the box containing the number N "Box N".
I can't put 6 into Box 3 and 9,and I can't put 10 into Box 5,so consider 6 and 10 first.
If I put 6 in Box 1,then I can put 10 in Box 3,7,and 9.
If I put 6 in Box 5,then I can put 10 in Box 1,3,7,and 9.
If I put 6 in Box 7,then I can put 10 in Box 1,3,and 9.
There are $10$ ways to put 6 and 10 into two boxes.Since there are no limits for putting 2,4,8 in the boxes,putting them in the three other boxes will be $3!=6$ ways.
Hence,there are $6 \times 10=\boxed{60}$ ways to do this.