Distribution of Distinct objects

[Saumyojit]

How many ways 5 different ice creams can be distributed to three person without any restrictions

So, here I will assume that there are infinite no of ice cream of each type or One of each type .

If former then , person 1 can receive one out of 5 ice creams then 2nd person can recieve one out of 4 then 3 . 5p3
or
Latter then,
(in each event )person 1 can recieve any one out of 5 ice creams then next person can recieve the same ice cream or
any one out of other 4 ice cream so 5 then last person is 5 = 5^3
That means there are numerous no of ice cream of each type
or

Can this happen that suppose in one event : one person gets all the five ice cream and other two gets nothing ? then also 5^3 . and there are only 5 ice cream all total .

without any restrictions what does it implies ?

Note: I am thinking from the shopkeeper side as if i am shopkeeper and i am distributing . (so not 3^5)

 

[pka]

How many ways 5 different ice creams can be distributed to three person without any restrictions
So, here I will assume that there are infinite no of ice cream of each type or One of each type .
If former then , person 1 can receive one out of 5 ice creams then 2nd person can recieve one out of 4 then 3 . 5p3 or
Latter then, (in each event )person 1 can recieve any one out of 5 ice creams then next person can recieve the same ice cream or
any one out of other 4 ice cream so 5 then last person is 5 = 5^3
That means there are numerous no of ice cream of each type or
Can this happen that suppose in one event : one person gets all the five ice cream and other two gets nothing ? then also 5^3 . and there are only 5 ice cream all total . without any restrictions what does it implies ?

There are \(3^5\) functions from a set of five to a set of three. Yes it is possible that one person gets all five of those ice creams.
The problem does say without restrictions. If it were stated that each person gets at least one then we would count the number of surjections (onto functions).

 

[Saumyojit]

person 1 can receive one out of 5 ice creams then 2nd person can recieve one out of 4 then 3 . 5p3

what if there was restiriciton then above would work right?

There are \(3^5\) functions from a set of five to a set of three. Yes it is possible that one person gets all five of those ice creams.
The problem does say without restrictions. If it were stated that each person gets at least one then we would count the number of surjections (onto functions).

okay yes , no of distributions will be = 3^5.
There are all total 5 ice creams ( no case of infinite) and three persons.
Now the question suggest to look from the point of Distributor so i can distribute one icecream to all the three persons.
and the second ice cream to three persons and so on .


Discuss about the below case:

Now one could think from the point of person

MAIN POINT: In one event: So, if one person recieves ice cream1 suppose , then in that event person 2 can recive that ice cream also or any other 4 and and same goes person3 in that event.
In that logic 5^3 can happen . But then it means there are infinite no of ice creams then? which is not mentioned in question.


But the question suggest to look from the point of Distributor.

 

[lex]

Once you understand what the question means, then it doesn't matter whose perspective you look at it from - the answer will be the same.

If the distribution is completely without restriction, no limit on supply and people can be given no ice-cream or all five flavours, then it appears there are [MATH]2^{15}[/MATH] results.
There are 15 combinations of ice-cream flavour and person. For each of these fifteen combinations there is a (binary) question to be answered:
"to give or not to give" (that is the question).
So there are [MATH]2^{15}[/MATH] different answers.

If you want to look at it from the perspective of the ice-cream. An ice-cream flavour has to decide to be given or not, to each of the three people:
[MATH]2^3[/MATH] possibilities. There are 5 ice-cream flavours, therefore [MATH](2^3)^5[/MATH] possibilities.
If you want to look from the perspective of the receivers. A person can receive or not receive for each of the 5 ice-creams: [MATH]2^5[/MATH] possibilities.
As there are 3 people, a total of [MATH](2^5)^3[/MATH] possibilities.

(A caveat - I'm not very experienced in 'discrete mathematics' and can often fool myself into the wrong answer. So, it is advisable to be critical of any proposed 'solution').

 

[pka]

what if there was restiriciton then above would work right?

If for example we had m distinct items to give to n people with the restriction that each person was the receive at least one of the items then of course \(m\ge n\), then we count the number of onto function from a set of \(m\) to a set of \(n\).
To get the number of onto functions: \(Surj(m,n)=\sum\limits_{k = 0}^n {{{( - 1)}^k}\dbinom{n}{k}{{(n - k)}^m}} \)
In your case \(Surj(5,3)\) see here

 

[Saumyojit]

@pka @JeffM
Out of 6 apples, 10 mangoes and 4 bananas, the number of ways of distributing fruit to three persons such that each person gets at least one mango is


Please tell where am i wrong?
See my approach :

M1 M2 M3 M4 M5 M6 M7 M8 M9 M10 A1 A2 A3 A4 A5 A6 B1 B2 B3 B4

Now bcoz three person will have three mangoes so I exclude M8 M9 M10 (considering it's not there). I have already distributed those 3 to 3 person each.
As it stands,
M1 M2 M3 M4 M5 M6 M7 A1 A2 A3 A4 A5 A6 B1 B2 B3 B4

( fruits are identical of Same type)


Now for partiton between three people , there will be two partiton points
P1 P2

Suppose one of the arrangements is this
M1 M2 M3 M4 M5 P1 M6 M7 A1 A2 A3 A4 A5 A6 P2 B1 B2 B3 B4

It means person1 has 6 mangoes where as person2 has 3 mangoes and six apples whereas person3 has one mango and 4 banannas.


P1 P2 M1 M2 M3 M4 M5 M6 M7 A1 A2 A3 A4 A5 A6 B1 B2 B3 B4

It means person1 has 1 mango where as person2 has 1 mango and person3 has all the remaining fruits.

So there are all total 19 objects including P1 and P2. Rearranging them will give 19! arrangemnts but we need to consider only unique so -> 19! / ( 7! * 4!* 6! * 2!) . Is this right?

2! is for two partiton points which are the same like objects although i have named p1 and p2 .

 

[Saumyojit]

Okay I think I found out where my approach is going wrong.

M1 M2 M3 M4 A1 B1 P1 M5 A2 A3 A4 A5 A6 P2 B2 B3 B4 M6 M7
Here person 1 gets 5 mangoes one apple one ?.

As I am rearranging the position of fruits , so I will get another arrangement
M1 M2 A1 B1 M3 M4 P1 M5 A2 A3 A4 A5 A6 P2 B2 B3 B4 M6 M7

But I am counting the same combination of fruits got by person 1 twice through my approach..and so on.



Here also, person 1 gets 5 mangoes one apple one ?.

 

[pka]

We can go-ahead and give each one mango. Now we have seventeen pieces of fruit to three people with no restrictions.
The number of ways of ways to place \(K\) identical objects into \(N\) distinct cells is: \(\dfrac{(N-1+K)!}{(N-1)!\cdot K!}\)
Do you understand why we treat all seventeen remaining pieces of fruit as identical objects?
There are three distinct persons (cells) into which to place the fruit.
Here is a hint: we could five all the remaining fruit to the youngest child.