Asked by Lily
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?
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?
Answers
Answered by
GPT 3.5
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:
$$3!\times S(4,3)=3!\times6=36$$
$$\binom{4+2}{2}=15$$
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:
$$3!\times S(4,3)=3!\times6=36$$