How many ways are there to distribute 5 balls to 3 squares if each box must be filled with at least one ball if?

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Answer

How many ways are there to distribute 5 balls to 3 squares if each box must be filled with at least one ball if?
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Hint: There are five different balls and distributed among 3 persons such that each person gets at least one, then there are only two arrangements to distribute balls among people. The first one is 1,1,3 and second is 1,2,2. To find total ways just find the number of ways to arrange these two arrangements. There is also a short method, find all ways to distribute 5 balls and subtract the ways in which at least one person gets no balls.

Complete step-by-step answer:

There are two possible arrangements of balls 1,1,3 and 1,2,2 for distributed balls such that each person gets at least one ball.For arrangement 1,1,3 number of ways to distribute balls are $^5{C_1}{ \times ^4}{C_1}{ \times ^3}{C_3} = 5 \times 4 \times 1 = 20$ ways.Also, we can distribute 1,1,3 as 1,3,1 and 3,1,1.Then the total number of ways to distribute balls as 1,1,3 is 20x3=60.For arrangement 1,2,2 number of ways to distribute balls are\[^5{C_1}{ \times ^4}{C_2}{ \times ^2}{C_2} = 5 \times 3 \times 2 = 30\] ways.Also, we can distribute 1,2,2 as 2,1,2 and 2,2,1.Then the total number of ways to distribute balls as 1,2,2 is 30x3=90.So, the total number of ways to distribute balls such that each person gets at least one ball is 60+90+150.

Hence, the correct answer is option B.

Note: Total number of ways to distribute five different balls among 3 persons are ${3^5}$.

Number of ways to distribute five balls to 3 people at least one get no ball are $^3{C_1} \times ({2^5} - 1)$ ways.Then total number of ways to distribute balls such that each person get at least one ball is ${3^5}{ - ^3}{C_1} \times ({2^5} - 1) = 243 - 3*31$$ = 243 - 93 = 150$

(2) Since the boxes are indistinguishable, there are 5 different cases for arrangements of the number of balls in each box: $(5,0,0)$, $(4,1,0)$, $(3,2,0)$, $(3,1,1)$, or $(2,2,1)$.

$(5,0,0)$: There is only $1$ way to put all 5 balls in one box.

$(4,1,0)$: There are $\binom{5}{4} = 5$ choices for the 4 balls in one of the boxes.

$(3,2,0)$: There are $\binom{5}{3} = 10$ choices for the 3 balls in one of the boxes.

$(3,1,1)$: There are $\binom{5}{3} = 10$ choices for the 3 balls in one of the boxes, and we simply split the last two among the other indistinguishable boxes.

$(2,2,1)$: There are $\binom{5}{2} = 10$ options for one of the boxes with two balls, then $\binom{3}{2} = 3$ options for the second box with two balls, and one option remaining for the third. However since the boxes with two balls are indistinguishable, we are counting each pair of balls twice, and must divide by two. So there are $\dfrac{10 \times 3}{2} = 15$ arrangements of balls as $(2,2,1)$.

Thus the total number of arrangements for 3 indistinguishable boxes and 5 distinguishable balls is $1 + 5 + 10 + 10 + 15 = \boxed{41}$.

$\textbf{Alternate solution:}$ There are $3^5 = 243$ arrangements to put 5 distinguishable balls in 3 distinguishable boxes. Among these 243 arrangements, there is one case in our problem that is counted three times: if all 5 balls are placed in one box and the other two boxes both contain nothing. This leaves 240 other arrangements.

For every other case, the contents of each box is different, and so these cases are each counted $3! = 6$ times. Therefore there must be 40 of these cases, and we have $\boxed{41}$ cases total.