# Number of ways in which balls are distributed

In how many ways can we distribute 5 different balls into 4 different boxes, given that order does not matter inside the boxes and empty boxes are not allowed?

My attempt

First I chose $$4$$ balls out of $$5$$ and arranged them for the $$4$$ boxes: $$\binom 54 \times 4!.$$

Then for the remaining ball I can choose any of the $$4$$ boxes. Multiplying them, we get $$480$$, which is double the answer given as correct answer. Why I am wrong? And how can I solve the problem if the order matters inside the boxes?

• Cases will repeat in your way.. Use inclusion-exclusion principle Mar 15, 2020 at 17:33

For any allowed distribution we have just one box with $$2$$ balls and the others with $$1$$ ball each. We choose such box in $$4$$ ways. Now multiply such number by the number of permutations of the $$5$$ balls, i.e. $$5!$$ and finally divide the result by $$2$$ because in the box with two balls order does not matter. Hence the result is $$\frac{4\cdot 5!}{2}=240.$$

With your attempt you count a disposition more than once.

Named $$A,B,C,D$$ the boxes and $$a,b,c,d,e$$ the balls you count (for example) two times the combination: $$a,e\in A$$, $$b\in B$$, $$c\in C$$, $$d\in D$$.

The first time you choose the set $$\{a,b,c,d\}$$, put $$a\in A, b\in B, c\in C, d\in D$$, then you put $$e\in A$$;

The second time you chose the set $$\{e,b,c,d\}$$, put $$e\in A, b\in B, c\in C, d\in D$$, then you put $$a\in A$$.

Since I do not see how to fix this count, I suggest another approach:

First, choose the box which will contain $$2$$ balls in $$\binom{4}{1}$$, second choose the two balls you will put in the box chosen in $$\binom{5}{2}$$, then choose how to put the last three balls in last three boxes in $$3!$$.

So the answer should be: $$\binom{4}{1} \cdot \binom{5}{2} \cdot 3! = 240$$

Using inclusion-exclusion principle,

$$4^5 - ^4C_1 3^5 + ^4C_2 2^5 - ^4C_3 1^5$$

Subtract cases when all 5 objects go in 3 boxes (exclusion) then include when objects go in 2 boxes, because they are subtracted more than the number of times such case comes. Then again exclude when all objects go in one box.

Two methods:

Method 1: There is only one case possible 2,1,1,1

So, first we will go with unnamed distribution of 5 distinct things and then we will arrange them in 4 places.

$$\frac{5!}{(2!)(1!)^3(3!)}4!= 240$$

Method 2:

Distribution of identical objects then arranging them. Solve x+y=5 and arrange the balls in the boxes

$$=C(_{3-1}^{5-1})\frac{5!}{2!}=240$$