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**phanthanhtom****Member**- Registered: 2012-06-22
- Posts: 290

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: 107,147

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.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

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**phanthanhtom****Member**- Registered: 2012-06-22
- Posts: 290

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: 107,147

For d) that is all you have?

**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.**

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**phanthanhtom****Member**- Registered: 2012-06-22
- Posts: 290

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: 107,147

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.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

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**phanthanhtom****Member**- Registered: 2012-06-22
- Posts: 290

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: 107,147

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

**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.**

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**phanthanhtom****Member**- Registered: 2012-06-22
- Posts: 290

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: 290

And the formula for e is

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 107,147

Hi;

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

**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.**

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**phanthanhtom****Member**- Registered: 2012-06-22
- Posts: 290

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: 107,147

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.

**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.**

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**phanthanhtom****Member**- Registered: 2012-06-22
- Posts: 290

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: 107,147

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

**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.**

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**phanthanhtom****Member**- Registered: 2012-06-22
- Posts: 290

Yes you're right.

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**phanthanhtom****Member**- Registered: 2012-06-22
- Posts: 290

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

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 107,147

They are best done with exponential generating functions.

**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.**

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**phanthanhtom****Member**- Registered: 2012-06-22
- Posts: 290

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: 107,147

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

**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.**

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**phanthanhtom****Member**- Registered: 2012-06-22
- Posts: 290

Thanks

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