Asked by Lily
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?
Answers
Answered by
GPT 3.5
This problem can be solved using stars and bars method. Since no box can be empty, we must distribute all $4$ balls among the $3$ boxes.
We can think of this problem as distributing $4$ identical balls into $3$ distinct boxes. Using stars and bars method, we can insert $2$ dividers (bars) among the $4$ balls to separate them into $3$ groups representing the $3$ boxes.
For example, if the $4$ balls were represented by o's and the dividers were represented by |'s, we could have an arrangement like:
oo|oo||
This represents $2$ balls in the first box, $2$ balls in the second box, and $0$ balls in the third box.
The total number of ways to arrange the $4$ balls and $2$ dividers is given by the formula $\binom{4+3-1}{3-1} = \binom{6}{2} = 15$.
Therefore, there are $15$ ways to place $4$ balls in $3$ boxes if neither the balls nor the boxes are distinguishable.
We can think of this problem as distributing $4$ identical balls into $3$ distinct boxes. Using stars and bars method, we can insert $2$ dividers (bars) among the $4$ balls to separate them into $3$ groups representing the $3$ boxes.
For example, if the $4$ balls were represented by o's and the dividers were represented by |'s, we could have an arrangement like:
oo|oo||
This represents $2$ balls in the first box, $2$ balls in the second box, and $0$ balls in the third box.
The total number of ways to arrange the $4$ balls and $2$ dividers is given by the formula $\binom{4+3-1}{3-1} = \binom{6}{2} = 15$.
Therefore, there are $15$ ways to place $4$ balls in $3$ boxes if neither the balls nor the boxes are distinguishable.