Math Is Fun Forum

juantheron

Member

Registered: 2011-10-19

Posts: 312

permutation

The Total no. of ways in which

balls of Different Color can be Distributed among

persons

so that each person gets at least one balls is

bobbym

bumpkin

From: Bumpkinland

Registered: 2009-04-12

Posts: 109,606

Re: permutation

Hi juantheron;

---

In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

Bob

Administrator

Registered: 2010-06-20

Posts: 10,623

Re: permutation

hi juantheron

EDIT:  Hi bobbym, I did all that typing and posted;  then I see you've jumped in with an answer.  How did you get it, please?

MY ANSWER BEFORE I READ bobbym's:

I think I'd start by choosing three balls and giving one each so we know that condition is met.

Then there are lots of cases for the other two balls.

(i)  Don't give them out at all.

(ii) Give out one only.  (the case of the person who gets two will contain repeats eg.  A, B and C are initially given out then someone (A) gets D is the same as B, C,  D are initially given out then someone (D) gets A as well.

(iii) Give out two balls to two different people.  Check for repeats.

(iv) Give out two balls to the same person.  Check for repeats.

Oh boy this is getting complicated.  Let's say they stand in line; person one, then person two, then person three.

Let the colours be denoted by A, B, C, D, E.

Start with allocating:

P1 = A   P2 = B  P3 = C

then the other balls coukld golike this:

P1 = A + D        P2 = B    P3 = C

etc etc

or

P1 = A + D + E etc etc.

I admit I don't know the answer, but that's how I'd start.

bobbym: haven't you got a formula for this?  eeeeekkkk!

Here's another thought.

You can give out just 3 balls.

You can stick two balls together and treat it as one allocation, and allocate the three objects.

You can stick three balls together and treat it as one allocation, and allocate the three objects.

You can stick two together, and another two together, so you still have three objects to allocate.

I think I like this better as you don't have to worry about repeats.

So the problem is then in two parts:

(1)  How many ways can you create different allocations of three objects?

(2)  How many ways can you then hand these out to the three people?

Bob

---

Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob

bobbym

bumpkin

From: Bumpkinland

Registered: 2009-04-12

Posts: 109,606

Re: permutation

Hi Bob;

All distribution problems to both distinguishable and indistinguishable boxes and balls have been tallied into what is called the 12 fold way by the combinatoricist, Rota.

This one just reduces to a formula and a generating function.

Formula:

where the strange symbol on the left is the Stirling numbers of the second kind.

GF

We take the coefficient of x^5 and times it by 120 to get the answer of 150.

---

In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

Bob

Administrator

Registered: 2010-06-20

Posts: 10,623

Re: permutation

Thanks bobbym.

Now I've got to get that answer by my method.

I may be quiet for a while!

Bob

---

Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob

bobbym

bumpkin

From: Bumpkinland

Registered: 2009-04-12

Posts: 109,606

Re: permutation

Hi Bob;

Work through them carefully and make sure you do not overcount. Happy Easter!

---

In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

Bob

Administrator

Registered: 2010-06-20

Posts: 10,623

Re: permutation

Happy Easter to you too.

Bob

---

Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob

Bob

Administrator

Registered: 2010-06-20

Posts: 10,623

Re: permutation

I'm getting 150 if all thye balls are used,

and another 60 + 180 if some are not used.

Does that sound right?

Bob

---

Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob

bobbym

bumpkin

From: Bumpkinland

Registered: 2009-04-12

Posts: 109,606

Re: permutation

Hi;

If some people do not get a ball you should get 243.

---

In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

juantheron

Member

Registered: 2011-10-19

Posts: 312

Re: permutation

Thanks Bobbym and Bobbundy .

ans = 150

bobbym

bumpkin

From: Bumpkinland

Registered: 2009-04-12

Posts: 109,606

Re: permutation

Your welcome.

---

In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

juantheron

Member

Registered: 2011-10-19

Posts: 312

Re: permutation

but bobbym I didnot understand how we multiply Coeff. of x^5 by 120

bobbym

bumpkin

From: Bumpkinland

Registered: 2009-04-12

Posts: 109,606

Re: permutation

Standard move in what is called an exponential generating function. The 5! = 120 counts the permutations.

---

In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

Bob

Administrator

Registered: 2010-06-20

Posts: 10,623

Re: permutation

If some people do not get a ball you should get 243.

I didn't mean that.

Each person has to get at least one ball.  But it doesn't say you have to use all five.

So I have got

(i)  Use only three = 60 ways.  = 5 x 4 x 5 ( x 1 )

(ii) Pair up two balls, treat as one object, don't use one ball = 180 ways = (5 x 4)/2 x 3 x 2  (x3)

(iii) Group three balls, treat as one object,  using all five balls = 60 ways = (5 x 4 x 3)/6 x 2 x 1 (x3)

(iv) Make one pair, make another pair, treat each as one object, using all five balls = 90 ways = (5 x 4)/2 x (3 x 2)/2 x 1 (x3)

The numbers in brackets are to cover each of the three people being the 'special one' who gets extra in their object.

So I have (iii) + (iv) = 150 ways.

Bob

---

Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob