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1 vote
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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.
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2 Answers

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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
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2 votes
2 votes

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..

2 Comments

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
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@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?
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