426 views
Please explain how to solve the following:

There are 3 boxes and 5 balls.
a. Distribute into [1,1,3]
a.1. Boxes are different and balls are different
a.2. Boxes are undistinguishable and balls are different
b. Distribute into [2,2,1]
b.1. Boxes are different and balls are different
b.2. Boxes are undistinguishable and balls are different

I know the formula n!/(q1!.q2!.q3!). But not sure on how to deal when boxes are different and when boxes are of same type.

Ok, you can solve all this stuff even you don't know any formulae.

First case : [1 , 1 , 3 ]

1 :   balls  and boxes are different

let B1 , B2 , B3 be boxes , and b1,b2,b3,b4,b5  be balls

Now you put three balls either in B1 or in B2 or in B3

In B1 then,  left two balls be x and y ,then you may put x into B2 and y into B3 or  x into B3 and y into B2

In B2 then,  left two balls be x and y ,then you may put x into B1 and y into B3 or  x into B3 and y into B1

In B3 then,  left two balls be x and y ,then you may put x into B1 and y into B2 or  x into B2 and y into B1

6 cases , Now you can select three balls out of 5 in ,  C( 5 , 3 ) : 10 ways.

Total ways : 6 * 10 : 60

2 : Boxes are same , balls are different

Now you have no choice for boxes , because they are same, only you can focus on balls , either you put three balls in first box

or second or third , and left two balls in left two boxes , all may be same , (because boxes are same )

Hence you have to consider only one case :  Put three balls in one and left two balls in left two boxes.

Now you can select three balls out of 5 in ,  C( 5 , 3 ) : 10 ways

Number of ways to distribute [1 ,1 , 3 ]  in same boxes : 10.

This same technique can be used for second example also , You can try out yourself.

### 1 comment

Thank you for your detailed solution

Answer to question no 1 :

A) 1) . Given objects to be distinct and boxes distinct [ Ordered parition problem ] :

So here we have 5 objects and we have to partition that one gets 3 and other 2 gets 1 each..Say this partition be [ A B C ] ..But as A B and C are distinct so their permutation is also necessary ..But there is a repetition as 2 groups get 1 object each..

So no of ways of distribution  =  ( 5! / [ (3! * (1!)2) ] )  *  ( 3! / 2! )

=  120 / 2

=  60

A) 2) . Given objects to be distinct and boxes identical [Unordered partition] :

So here we have A B C partitions as identical so no concept of permutations of partitions..

So no of ways of distribution = ( 5! / 3! * 2!)   =  10 as 2 bags are identical

Similarly we can solve B) part also..

Thank you Habib. Could you please check if the following formulas are correct?

1. In unordered partition of n objects such that group sizes are [q1, q2,..., qt]:

No of distributions = n!/((q1!.q2!...qt!)*(no of groups of size q1)!*(no of groups of size q2)*...*(no of groups of size qt))

2. In ordered partition of n objects such that group sizes are [q1, q2,..., qt]:
n!*(total no of groups)!/((q1!.q2!...qt!)*(no of groups of size q1)!*(no of groups of size q2)!*...*(no of groups of size qt)!)

Also, does the formula change if any of the group size is 0?

Thank you
@habibkhan when bags are identical why do we divide by 2!?  I am not getting this.

And how do we know that bags are identical? or the partition is unordered in general?

1 vote
1
514 views
2
356 views
3
1,051 views