PERMUTATION

Saumyojit said:
are u saying that : 5,0,1,1 & 0,5,1,1
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I would say that 5,0,1,1 & 0,5,1,1 are two different ways of buying 7 cakes (from 4 available types).

The fact that both options have c3=1 and c4=1 does not create any conflict or replication in that statement.
 
Dr.Peterson said:
A couple of us have mentioned multisets. That is what this case (5,1,1,0) is; it is not a set for which the standard permutation method works.

If you have seen problems that ask for the number of words you can make from a word with duplicate letters, such as GOOD, that is what you need to do here. The formula is [MATH]\frac{n!}{n_1!n_2!...n_k!}[/MATH], where [MATH]n[/MATH] is the total number of letters, and [MATH]n_i[/MATH] is the number of copies of the same ith letter out of k distinct letters. This was demonstrated in post #14.
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THen why dr peterson said due to repetiton we need to divide. Where is the repetion then?
 
Saumyojit said:
But in 5,1,1,0 two 1's are of two different cakes . like 5 of c1 1 of c2 1 of c3

and in 5,0,1,1 5 are of first cake,zero of cake 2 ,one of cake 3 ,one of cake 4 .
They are of different cakes .

are u saying that : 5,0,1,1 & 0,5,1,1
these two unique arrangements are including replication of c3 and c4 in both the cases?
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These are different because the first has 5 of the first kind of pastry, and the second has 5 of the second kind. They happen to both have one each of the last two kinds, but that is not significant.

The important thing is that in counting the number of ways to assign the numbers 5, 1, 1, 0 to the four kinds, you can't just permute as if the two 1's were distinct. An alternative way to count would be to say that there are 4 ways to place the 5, and 3 ways to place the 0, leaving 1 in the other two places, for a total of 4*3 = 12 arrangements. The formula gives 4!/(2!1!1!) = 24/2 = 12, which agrees.

In any case, since you mentioned stars and bars in the OP, you really need to learn that method! This longer method is useful to learn for other problems, but is wasteful for this one.
 
Saumyojit said:
THen why dr peterson said due to repetiton we need to divide. Where is the repetion then?
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At this point, we are only counting ways to arrange the numbers 5, 1, 1, 0. We are not talking about cakes/pastries at all. The repetition is of the number 1, only.

This is an important point in combinatorics: we often move between different models, and have to leave behind thoughts of how the meaning of one phase of our work relates to others. They are just abstract counts.
 
This post is to illustrate why there is a division in the multiset permutation formula.

If the divide was missing then there would be repetition. For example the word SEEK would have 4! permutations...

EEKS (eEKS)
EEKS (EeKS)
EESK (eESK)
EESK (EeSK)
EKES (eKES)
EKES (EKeS) 		EKSE (eKSE)
EKSE (EKSe)
ESEK (eSEK)
ESEK (ESeK)
ESKE (eSKE)
ESKE (ESKe) 		KEES (KeES)
KEES (KEeS)
KESE (KeSE)
KESE (KESe)
KSEE (KSeE)
KSEE (KSEe)		SEEK (SeEK)
SEEK (SEeK)
SEKE (SeKE)
SEKE (SEKe)
SKEE (SKeE)
SKEE (SKEe)

(I show uppercase and lowercase eE in brackets to illustrate what is happening.)

So the true number of unique permutations is actually 4!/2! = 12
 
Dr.Peterson said:
At this point, we are only counting ways to arrange the numbers 5, 1, 1, 0. We are not talking about cakes/pastries at all. The repetition is of the number 1, only.

This is an important point in combinatorics: we often move between different models, and have to leave behind thoughts of how the meaning of one phase of our work relates to others. They are just abstract counts.
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In SEEK i know there 2E so dividing by 2 fact
but in 5,1,1,0 or 0,1,5,1 repetion is happening due to present of one cake type 2 in both the cases thats why ?
 
At this point, we are only counting ways to arrange the numbers 5, 1, 1, 0. We are not talking about cakes/pastries at all.
dr peterson said :

But arranging the numbers would give me a new set everytime how can i omit that We are not talking about cakes/pastries at all
ordering is deciding qunatity of cakes of each type
 
You have a very questioning mind @Saumyojit . That is a good trait for a mathematician.

Combinations can be very confusing, especially when beginning.

Now I recommend that you move on from this. Perhaps do some different questions in the subject for a while. Keep the questions simple at first and build up. If you are still confused then try to occasionally flip viewpoints to convince yourself about things. Consider what happens when you put different numbers into a question. Good luck!
 
Cubist said:
You have a very questioning mind @Saumyojit . That is a good trait for a mathematician.

Combinations can be very confusing, especially when beginning.

Now I recommend that you move on from this. Perhaps do some different questions in the subject for a while. Keep the questions simple at first and build up. If you are still confused then try to occasionally flip viewpoints to convince yourself about things. Consider what happens when you put different numbers into a question. Good luck!
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I have done all the easy to medium sums . U answer to the point.

We are only counting ways to arrange the numbers 5, 1, 1, 0. We are not talking about cakes/pastries at all.


how can i omit that We are not talking about cakes/pastries at all
ordering is deciding qunatity of cakes of each type

IN 5110 and 5101 choosing 5 cakes of type 1 and each one of 2 and 3 && second arrangement means choosing 5 cakes of type 1 and each one of 2 and type 4 .
So due to presence of 5 cakes of type 1 in both the sets we will get repetiton or for the individual set due to presence of more than 1, type1 cake we are telling there is repetion involved?
 
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I'm not sure what you are saying you understand, and what you are just quoting from others.

Do you see that we have changed perspective on the problem, turning a question about choosing types of cakes (pastries -- not really quite the same thing!) into an abstract question about arranging numbers (or, with the other approach, stars and bars)? Yes, an arrangement of these numbers corresponds to a choice of cakes, but it is easier to think about it in terms of the former.

But the choice 5, 1, 1, 0, and the choice 5, 1, 1, 0 are the same choice, whether you think of them as numbers or as types of cakes. That is why you divide the number of permutations by 2.

I do think you need to move on in some sense. Perhaps try this: Go back and write out your complete solution to this problem, not omitting details. In doing that, you may be able to see more easily what is happening, and how to state it clearly. It may also help us see more clearly where you are not thinking right.
 
I got it at last.
5011 combo if u arrange it as no's only then there are 24 arrangements out of 12 are distinct.
But it seems strange though that 5,1,1 denotes quantity of each type of cakes
 
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