Distributing 3 distinct balls in 5 distinct boxes such that a box can be empty

balls-in-binscombinatoricssolution-verification

My approach :-

I made the following cases :-

let the balls be A,B,C and Boxes be B1, B2, B3, B4, B5

Case 1) 1 1 1 0 0 —> 3c1 * 2c1 * 1c1 (These 6 cases in my opinion are representing a particular order in which the balls are being filled in the 5 boxes: ABC, ACB, BAC, BCA, CAB, CBA ), for example here BCA which means that B ball is filled first in a particular box , followed by C ball which further is followed by A in end, how shall I permute each of these 6 cases into the boxes?

What I thought here for example in the case of ABC (A being filled first , then B , followed by C in end) , I tried to make the following cases :

 B1        B2       B3       B4        B5  

 A         B        C         0         0  -> (the case in which A is first placed in Box 1 followed by B ball in box 2 and C ball in box 3) 

 A         B        0         C         0 -> the case where A is filled first in box 1 followed by B in box 2 and then C in box 4)

 A         B        0         0         C 

 0         A        B         C         0

 0         A        B         0         C

 0         A        0         B         C

 B         A        C         0         0 -> **the case where B is filled first in box 1 followed by A in box 2 and then C in box 3)**
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Am I doing it correctly ?

Case 2) 2 1 0 0 0 —> 3c2 * 1c1 would give me the order in which the balls would be filled like here I would have 3 cases {AB,C},{AC,B}, {BC, A}, how shall I permute each of these 3 cases into the boxes ?

Case 3) and finally the case of 3 0 0 0 0 —> 3c3 would give me 1 case of {A,B,C} , how to permute this 1 case ?

I am struggling to understand after the initial ordering of balls , how shall I permute them further into the boxes ?

Best Answer

What matters here is which ball is placed in which box. You have to select which box will receive which ball.

Method 1: There are five ways to place each ball, so the three distinct balls can be placed in five distinct boxes in $5^3 = 125$ ways.

Method 2: We consider cases.

Case 1: Each ball is placed in a separate box.

There are five ways to place ball A in one of the boxes, which leaves four ways to place ball B in a different box. That, in turn, leaves three ways to place ball C in one of the remaining boxes. Hence, as Henry indicated in the comments, there are $5 \cdot 4 \cdot 3 = 60$ such distributions.

Case 2: Exactly two balls are placed in one box, with the other ball placed in a separate box.

There are five ways to select the box which will receive two balls, $\binom{3}{2}$ ways to select which two balls will be placed in that box, and four ways to select which of the remaining boxes will receive the remaining ball. There are $$5 \cdot \binom{3}{2} \cdot 4 = 5 \cdot 3 \cdot 4 = 60$$ such distributions.

Case 3: All five balls are placed in the same box.

There are five boxes in which all the balls could be placed, so there are $5$ such distributions.

Total: Since the three cases are mutually exclusive and exhaustive, we can obtain the total number of distributions by adding the above cases, which yields $60 + 60 + 5 = 125$, as above.

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