If it had asked about 6 distinct balls would the answer be 147?

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This is $\color{red}{\text{ “indistinguishable objects into indistinguishable boxes” }}$ problem which is a standard combinatorial problem. There is no simple closed formula for the number of ways to distribute $n$ indistinguishable objects into $j$ indistinguishable boxes.

So, We will enumerate all the ways to distribute.

Best way is to go in a sequence, covering all possibilities, So, that we do not overcount, we do not undercount.

Case 1 : $\color{blue}{\text{If only one bin is used : }}$

(6,0,0) i.e. Only 1 way (Since All balls have to be put in this single bin that is used ; Since all bins are identical, so, it doesn’t matter which bin we use)

Case 2 : $\color{blue}{\text{If two bins are used : }}$

The distribution can be done as any of the following :

(5,1) (which means 5 identical balls in one bin, 1 ball in another bin, and the third bin is unused)

(4,2) (which means 4 identical balls in one bin, 2 balls in another bin, and the third bin is unused)

(3,3) (which means 3 identical balls in one bin, 3 balls in another bin, and the third bin is unused)

i.e. 3 ways to distribute 6 identical balls into 3 identical bins if exactly two of the bins are used.

Case 3 : $\color{blue}{\text{If three bins are used : }}$

The distribution can be done as any of the following :

(4,1,1) (which means 4 identical balls in one bin, 1 ball in another bin, and 1 ball in the third bin)

(3,2,1) (which means 3 identical balls in one bin, 2 ball in another bin, and 1 ball in the third bin)

(2,2,2) (which means 2 identical balls in one bin, 2 ball in another bin, and 2 ball in the third bin)

i.e. 3 ways to distribute 6 identical balls into 3 identical bins if all three of the bins are used.

So, total we have 7 ways to distribute 6 identical balls in 3 identical bins.

NOTE that we cannot distribute as (2,2,1,1) because only three bins are available, not four.

So, We will enumerate all the ways to distribute.

Best way is to go in a sequence, covering all possibilities, So, that we do not overcount, we do not undercount.

Case 1 : $\color{blue}{\text{If only one bin is used : }}$

(6,0,0) i.e. Only 1 way (Since All balls have to be put in this single bin that is used ; Since all bins are identical, so, it doesn’t matter which bin we use)

Case 2 : $\color{blue}{\text{If two bins are used : }}$

The distribution can be done as any of the following :

(5,1) (which means 5 identical balls in one bin, 1 ball in another bin, and the third bin is unused)

(4,2) (which means 4 identical balls in one bin, 2 balls in another bin, and the third bin is unused)

(3,3) (which means 3 identical balls in one bin, 3 balls in another bin, and the third bin is unused)

i.e. 3 ways to distribute 6 identical balls into 3 identical bins if exactly two of the bins are used.

Case 3 : $\color{blue}{\text{If three bins are used : }}$

The distribution can be done as any of the following :

(4,1,1) (which means 4 identical balls in one bin, 1 ball in another bin, and 1 ball in the third bin)

(3,2,1) (which means 3 identical balls in one bin, 2 ball in another bin, and 1 ball in the third bin)

(2,2,2) (which means 2 identical balls in one bin, 2 ball in another bin, and 2 ball in the third bin)

i.e. 3 ways to distribute 6 identical balls into 3 identical bins if all three of the bins are used.

So, total we have 7 ways to distribute 6 identical balls in 3 identical bins.

NOTE that we cannot distribute as (2,2,1,1) because only three bins are available, not four.

1 vote

In the given question ball as well as Bin are similar means looking similar

so here in given question

six identical Ball and three identical Bins

so lets case first when all Bins are not empty

Bin Bin Bin

4 | 1 | 1 |

3 | 2 | 1 |

2 | 2 | 2 |

case second when only one bins is empty

Bin Bin Bin

5 | 1 | empty |

4 | 2 | empty |

3 | 3 | empty |

lets case third when only two bins are empty

Bin Bin Bin

6 | empty | empty |

so overall arrangement which is looking different is 3+3+1=7