I'll just do the first part. There are just these 5 ways to break up 8 cars into a sum of 4 positive integers, without regard for order: 1. 1+1+1+5 = 8 2. 1+1+2+4 = 8 3. 1+1+3+3 = 8 4. 1+2+2+3 = 8 5. 2+2+2+2 = 8 Of course we must choose and order each of the 5 cases. Case 1: 5 cars in 1 container, 1 car in each of the other 3. We can choose the container to hold 5 cars in 4 ways. We can choose the 5 cars for that container C(8,5)=56 ways. The other 3 cars can be distributed to the other 3 containers in 3!=6 ways. That's (4)(56)(6) = 1344 ways Case 2: 4 cars in 1 container, 2 cars in another container, and 1 car in each of the other 2 containers. We can choose the container to hold 4 cars in 4 ways. We can choose the 4 cars for that container C(8,4)=70 ways. We can choose the container to hold 2 cars in 3 ways. We can choose the 2 cars for that container in C(4,2)=6 ways. The other 2 cars can be distributed to the other 2 containers in 2!=2 ways. That's (4)(70)(6)(2) = 3360 ways Case 3: 3 cars each in 2 containers, and 1 car in each of the other 2 containers. We can choose the 2 containers to hold 3 cars in C(4,2)=6 ways. We can choose the 3 cars for the container with the smaller number C(8,3)=56 ways. We can choose the cars for the container with the larger number C(5,3)=10 ways. The remaining 2 cars can be distributed to the other 2 containers in 2!=2 ways. That's (6)(56)(10)(2) = 6720 ways Case 4: 2 cars each in 2 containers, 3 cars in 1 container, and 1 car in 1 container We can choose the 2 containers to hold 2 cars in C(4,2)=6 ways. We can choose the 2 cars for the container with the smaller number C(8,2)=28 ways. We can choose the 2 cars for the container with the larger number C(6,2)=15 ways. We can choose the container to hold 3 cars in C(2,1)=2 ways. We can choose the 3 cars for that container in C(4,3)=4 ways. The container to hold only 1 car gets the 1 car left in 1 way. That's (6)(28)(15)(2)(4)(1)=20160 ways Case 5: Each container holds 2 cars each. We can choose the 2 cars for the container with the number 1 in C(8,2)=28 ways. We can choose the 2 cars for the container with the number 2 in C(6,2)=15 ways. We can choose the 2 cars for the container with the number 3 in C(4,2)=6 ways. We can choose the 2 cars for the container with the number 4 in C(2,2)=1 way. That's (28)(15)(6)(1)=2520 The total = 1344+3360+6720+20160+2520 = 34104 I haven't gone over this with a fine-tooth comb to see if I've made an error. I'll check it later, and correct it, or maybe another tutor will solve it. Edwin