+817 In how many ways can $4$ balls be placed in $8$ boxes if the balls and boxes are both distinguishable?
In how many ways can $4$ balls be placed in $8$ boxes if neither the balls nor the boxes are distinguishable?
In how many ways can $4$ balls be placed in $8$ boxes if the balls are indistinguishable, and the boxes are distinguishable?
In how many ways can $4$ balls be placed in $8$ boxes if the balls are distinguishable, and the boxes are indistinguishable?
+1533 1. If the balls and boxes are both distinguishable, then we can place the balls in the boxes in 84 ways.
2. If neither the balls nor the boxes are distinguishable, then we are simply putting 4 indistinguishable objects into 8 distinguishable boxes. This can be done in (48+4−1)=(411) ways.
3. If the balls are indistinguishable, and the boxes are distinguishable, then we can put the balls in the boxes in 8 ways. We can then permute the balls within each box in 4! ways. This gives us a total of 8⋅4!=192 ways.
4. If the balls are distinguishable, and the boxes are indistinguishable, then we can simply put the balls in the boxes in 4! ways.
Therefore, the answers are:
1.84=40962.(411)=3303.8⋅4!=1924.4!=24
