1. In this case, we are essentially looking for the number of ways to distribute $4$ indistinguishable balls into $3$ distinguishable boxes with no empty boxes. This is equivalent to finding the number of ways to place $2$ dividers (to separate the balls into $3$ groups) among a total of $4$ balls and $2$ dividers. This can be calculated using combinations, yielding:
$$\binom{4+2}{2}=15$$
2. In this case, we are looking for the number of ways to distribute $4$ indistinguishable balls into $3$ indistinguishable boxes with no empty boxes. This is essentially the same as finding the number of ways to write $4$ as a sum of $3$ non-negative integers, which is a classic stars and bars problem. The answer is the number of ways to arrange $4$ identical balls and $2$ identical dividers, giving:
$$\binom{4+2}{2}=15$$
3. In this case, we are looking for the number of ways to distribute $4$ distinguishable balls into $3$ distinguishable boxes with no empty boxes. This is equivalent to the number of ways to partition a set of $4$ elements into $3$ non-empty subsets, which is a special case of Stirling numbers of the second kind. The answer is:
$$3!\times S(4,3)=3!\times6=36$$
4. In this case, we are looking for the number of ways to distribute $4$ distinguishable balls into $3$ indistinguishable boxes with no empty boxes. This is equivalent to the number of ways to partition a set of $4$ elements into $3$ non-empty subsets, as in the previous case. The answer is the same:
$$3!\times S(4,3)=3!\times6=36$$
In how many ways can $4$ balls be placed in $3$ boxes if the balls and boxes are both distinguishable, and no box can be empty?
In how many ways can $4$ balls be placed in $3$ boxes if neither the balls nor the boxes are distinguishable, and no box can be empty?
In how many ways can $4$ balls be placed in $3$ boxes if the balls are indistinguishable, and the boxes are distinguishable, and no box can be empty?
In how many ways can $4$ balls be placed in $3$ boxes if the balls are distinguishable, and the boxes are indistinguishable, and no box can be empty?
1 answer