Find the number of ways in which 6 distinct balls can be placed in 5 distinct boxes such that each box has at least 1 ball

Question

My approach :-

I first assumed all balls to be similar in nature , so that would give me 5 ways to distribute the balls in the boxes , which will be

2 1 1 1 1

1 2 1 1 1

1 1 2 1 1

1 1 1 2 1

1 1 1 1 2

basically that would be whole number solutions of a+b+c+d+e = 6 where all a,b,c,d,e >=1

Now since all the balls are distinct in nature I multiplied the 5 ways with 6! = giving me a total of 3600 ways , but the answer is given as 1800 ways , where am I going wrong ?

Answer 1

You have 5 possibilities to pick the box in which you want to have 2 balls. For this box, you have to pick 2 out of the 6 balls, so there are $6 \choose 2$ ways of doing that. Then you still have 4 balls and 4 boxes remaining, so the number of ways to place the remaining 4 balls amounts to $4 \cdot 3 \cdot 2 \cdot 1 = 4!$

Hence the total number of ways should be $$ N = 5 \cdot {6 \choose 2} \cdot 4! = 1800$$

Answer 2

The box with the two balls doesn't have them in a specific order. When you multiply with 6! you're implying that there is an order between the balls within the box with two balls whereas the order is unimportant in the context of the question. so to eliminate the cases with different ball orders in the box with two balls we must divide 3600 with 2 as there are two different ways that two balls can be put side by side.