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**phanthanhtom****Member**- Registered: 2012-06-22
- Posts: 215

I'm having trouble fitting words into the following definitions.

Given a set of n non-recurring objects, S = {a1, a2, a3, ..., an}, and a number k such that 0 <= k <= n. Find words that fit in the following definition.

a, A way of arranging all n objects (order matters, formula n!).

b, A way of choosing AND arranging k out of the n objects (order matters, formula n!/k!)

c, A way of choosing k out of the n objects (order does NOT matters, formula n!/k!(n-k)!)

d, A way of choosing AND arranging k objects from S, repetition allowed, order matters.

e, A way of choosing k objects from S, repetition allowed, order does NOT matters.

Note: If not mention, repetition is not allowed.

P.S. I have no experience using maths formula on this forum, so I prefer not to do so. But I hope I can learn how to do so.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 82,980

Hi;

You can use this site to do your latexing for you.

http://latex.codecogs.com/editor.php

I would think that for the rest of those you would consult your textbook. Is it the Rosen book?

**In mathematics, you don't understand things. You just get used to them.I have the result, but I do not yet know how to get it.All physicists, and a good many quite respectable mathematicians are contemptuous about proof.**

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**phanthanhtom****Member**- Registered: 2012-06-22
- Posts: 215

Given a set of n non-recurring objects,

, and a number k such that . Find words that fit in the following definition.a, A way of arranging all n objects (order matters, formula ).

b, A way of choosing AND arranging k out of the n objects (order matters, formula )

c, A way of choosing k out of the n objects (order does NOT matters, formula )

d, A way of choosing AND arranging k objects from S, repetition allowed, order matters.

e, A way of choosing k objects from S, repetition allowed, order does NOT matters.

Note: If not mentioned, repetition is not allowed.

P.S. Sorry for the older post, it's just an experiment. I forgot to put the formula tag in. Again, sorry.

*Last edited by phanthanhtom (2013-06-13 03:26:15)*

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 82,980

For d) that is all you have?

**In mathematics, you don't understand things. You just get used to them.I have the result, but I do not yet know how to get it.All physicists, and a good many quite respectable mathematicians are contemptuous about proof.**

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**phanthanhtom****Member**- Registered: 2012-06-22
- Posts: 215

What do you mean? I don't know the formula!

*Last edited by phanthanhtom (2013-06-13 03:30:39)*

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 82,980

How many are being picked from S? n objects? 3 objects k - n objects?

**In mathematics, you don't understand things. You just get used to them.I have the result, but I do not yet know how to get it.All physicists, and a good many quite respectable mathematicians are contemptuous about proof.**

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**phanthanhtom****Member**- Registered: 2012-06-22
- Posts: 215

d. For example if k=5, then (a1, a2, a2, a3, a3) would be a solution (repetition allowed), and it would be different from, say, (a1, a2, a3, a2, a3).

e. The above 2 would be the same solution, as order doesn't matter.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 82,980

Hi;

d, A way of choosing AND arranging k objects from S, repetition allowed, order matters.

Choosing k objects from n distinct objects with repetition and order matters is n ^ k

I have the result, but I do not yet know how to get it.

All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

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**phanthanhtom****Member**- Registered: 2012-06-22
- Posts: 215

For k=3, n=5:

a. (a1, a2, a3, a4, a5) and (a1, a3, a4, a2, a5) are different solutions. All solutions contains all 5 objects.

b. (a1, a2, a3), (a1, a2, a4), (a1, a3, a2) are different solutions. (a1, a1, a4) is not, for repetition not allowed.

c. (a1, a2, a3) and (a2, a3, a5) are different solutions. (a1, a2, a3) and (a2, a3, a1) are the same. (a1, a3, a3) is not a solution.

I'm afraid I would be away now because it's late. I'll be back GMT 8:00 am. So I look for the answers then.

And I only need the words. I'm sure I know the formulae, just I don't have enough time.

*Last edited by phanthanhtom (2013-06-13 03:50:35)*

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**phanthanhtom****Member**- Registered: 2012-06-22
- Posts: 215

And the formula for e is

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 82,980

Hi;

You might want to take a look here when you have time.

I have the result, but I do not yet know how to get it.

All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

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**phanthanhtom****Member**- Registered: 2012-06-22
- Posts: 215

Thanks, but I stress again that I need the official NAME, not the formulae.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 82,980

Hi;

They are usually defined or named in terms of balls or objects in boxes or urns. I have never heard of anything else. That page is from Rota, it covers every type of combinatorics problem there is.

I have the result, but I do not yet know how to get it.

All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

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**phanthanhtom****Member**- Registered: 2012-06-22
- Posts: 215

Thank you. The Vietnamese names for them roughly translates to:

A. permutation

B. partial permutation

C. combination

D. partial permutation with repetition

E. combination with repetition

I am looking for formal and better alternatives, for I translate by my knowledge all these names, and it would be really important for BIMC.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 82,980

A partial permutation? Is that when you do not use all the elements of the set?

I have the result, but I do not yet know how to get it.

All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

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**phanthanhtom****Member**- Registered: 2012-06-22
- Posts: 215

Yes you're right.

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**phanthanhtom****Member**- Registered: 2012-06-22
- Posts: 215

In other words, partial permutation <=> k < n

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 82,980

They are best done with exponential generating functions.

I have the result, but I do not yet know how to get it.

All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

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**phanthanhtom****Member**- Registered: 2012-06-22
- Posts: 215

Again, I am looking for alternative names. I finished reading that twelvefold Wikipedia article. Formulae enough.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 82,980

Okay, if I see anything different I will add it here.

I have the result, but I do not yet know how to get it.

All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

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**phanthanhtom****Member**- Registered: 2012-06-22
- Posts: 215

Thanks

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