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1 vote

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.

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.

2 votes

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.

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.

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..

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

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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