Combinations! How many.........


by Elruso
Tags: combinations
Elruso
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#1
Jun2-07, 11:25 AM
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1. The problem statement, all variables and given/known data
How many letter combinations with 9 letters are you able to make with following letters : M-A-T-E-M-A-T-I-K?


2. Relevant equations
Well its pretty obvious you need to use Combinations.

Please explain how you solve this problem, don't write use combinations .
I need to know how you think and from which angle you "attack" the problem.
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Dick
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Jun2-07, 04:31 PM
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You have 9 letter positions to fill. First lets place the M's. There are two of them, so I have C(9,2) ways. Now lets do the A's. There 2 of them and 7 places left to fill, so I have C(7,2) ways. So far I've got C(9,2)*C(7,2). Can you finish?
Elruso
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Jun3-07, 03:34 AM
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So iit´s C(9,2)*C(7,2)*C(5,2)*3*2*1?

In my math book the answer is C(9,2)*C(7,2)*5!..... which i find a little strange.

danago
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Jun3-07, 04:04 AM
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Combinations! How many.........


Quote Quote by Elruso View Post
So iitīs C(9,2)*C(7,2)*C(5,2)*3*2*1?

In my math book the answer is C(9,2)*C(7,2)*5!..... which i find a little strange.
When i did it, i got the same answer as you, and then to check, i got mathematica to output every single permutation of those letters into a list. That list contained 45360 elements, so unless ive misunderstood the question, it seems that you may be right.
Hurkyl
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Jun3-07, 04:07 AM
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Quote Quote by Elruso View Post
So iit´s C(9,2)*C(7,2)*C(5,2)*3*2*1?

In my math book the answer is C(9,2)*C(7,2)*5!..... which i find a little strange.
Did you notice those are the same thing?

Incidentally, it seemed most clear to me to write the answer as
9! / (2! * 2! * 2!),
or, as a multinomial coefficient,
[tex]\binom{9}{2 \ 2 \ 2 \ 1 \ 1 \ 1} .[/tex]
Dick
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Jun3-07, 08:47 AM
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Quote Quote by Hurkyl View Post
Did you notice those are the same thing?

Incidentally, it seemed most clear to me to write the answer as
9! / (2! * 2! * 2!),
or, as a multinomial coefficient,
[tex]\binom{9}{2 \ 2 \ 2 \ 1 \ 1 \ 1} .[/tex]
They aren't the same thing. They differ by a factor of two. Whoever wrote the solution seems to have miscounted the number of doubled letters.
Hurkyl
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#7
Jun3-07, 09:32 AM
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Ah, right. This is what was written:
C(9,2)*C(7,2)*C(5,2)*3*2*1
and this is what I thought I read:
C(9,2)*C(7,2)*C(5,2)*3!*2!*1!


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