Find number of ways in which 5 distinct balls in 3 identical boxes such that none of box is empty

Answer

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Hint: We designate the first box left as B1, the second box from the left as B2 and the third box from the left as B3. We ensure that no box is empty by putting one ball in each of them. We find the number of ways we can rest balls in three boxes. This is a problem based on the permutation and combinations. So we will be using the concept of this.

Complete step-by-step answer:

Let us denote the box from the first box left as B1, the second box from the left as B2 and the third box from the left as B3. In order to ensure that no box remains empty we put the first 1 ball in each of the boxes B1, B2, B3. So the number of balls left to be placed is $ 5-\left( 1+1+1 \right)=5-3=2 $ . We have to place 2 balls in the boxes B1, B2, B3. We can fill the box B1 with either 0, 1,2 balls . Similarly we can fill box B2 and box B3 with 0 or 1 or 2 balls. If we assume the number of balls we placed we place in boxes B1, B2. B3 respectively as $ {{x}_{1}},{{x}_{2}},{{x}_{3}} $ then all possible triplets such that $ {{x}_{1}}+{{x}_{2}}+{{x}_{3}}=2 $ are\[\left( 2,0,0 \right),\left( 0,2,0 \right),\left( 0,0,2 \right),\left( 1,1,0 \right),\left( 1,0,1 \right),\left( 0,1,1 \right)\]So there are 6 triplets and hence 6 numbers of ways to place 5 identical balls in 3 identical boxes. We can alternatively find number of triplets as the number of non-negative integral solution to the linear equation \[{{x}_{1}}+{{x}_{2}}+...+{{x}_{r}}=n\]which is given by $ {}^{n+r-1}{{C}_{n-1}} $ . Here in this problem $ {{x}_{1}}+{{x}_{2}}+{{x}_{3}}=2 $ we have $ n=2,r=3. $ So the number of non-negative integral solution is \[{}^{n+r-1}{{C}_{n-1}}={}^{2+3-1}{{C}_{2}}={}^{4}{{C}_{2}}=\dfrac{4!}{2!2!}=6\]

Note: We can directly find the answer by using the formula of number of positive integral solution to the equation $ {{x}_{1}}+{{x}_{2}}+...+{{x}_{r}}=n $ which is given by $ {}^{r-1}{{C}_{n-1}}. $ If the balls would not have been identical we would have selected 1 ball out of 5 in $ {}^{5}{{C}_{1}} $ ways , put them in boxes and then arranged them in $ 3! $ ways.

My approach :

I listed down the following cases :-

Case 1) 5 0 0 --> 5c5 = 1 way

Case 2) 4 1 0 --> 5c4 * 1c1 = 5 ways

Case 3) 3 2 0 --> 5c3 * 2c2 = 10 ways

Case 4) 3 1 1 --> 5c3 * 2c1 * 1c1 = 20 ways

Case 5) 2 2 1 --> 5c2 * 3c2 * 1c1= 30 ways

by adding all this I get 66 ways in which all possible combinations of selecting distinct objects is taken care of , and I feel that I don't need to arrange it further in boxes as all the boxes are identical

According to the solution provided for case 4 (i.e. 3,1,1) they have counted 10 ways and for case 5(i.e. 2,2,1) they have counted 15 ways , which makes their total to 41

can someone let me know where exactly am I going wrong with my approach ?