Combinations! How many letter combinations can you make with 9 letters?

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The discussion revolves around calculating the number of letter combinations from the letters M-A-T-E-M-A-T-I-K, specifically focusing on the arrangement of 9 letters with repeated characters. Participants explore the use of combinations and factorials to derive the correct formula, debating the accuracy of their math book's answer. The consensus suggests that the correct approach is to use the multinomial coefficient or the formula 9! / (2! * 2! * 2!) to account for the repeated letters. There is confusion regarding the counting of doubled letters, leading to a clarification that the initial solution may have miscounted. The final agreement emphasizes the importance of correctly accounting for repetitions in combinatorial problems.
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Combinations! How many...

Homework Statement


How many letter combinations with 9 letters are you able to make with following letters : M-A-T-E-M-A-T-I-K?


Homework Equations


Well its pretty obvious you need to use Combinations.

Please explain how you solve this problem, don't write use combinations :smile:.
I need to know how you think and from which angle you "attack" the problem.
 
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You have 9 letter positions to fill. First let's place the M's. There are two of them, so I have C(9,2) ways. Now let's 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?
 
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.
 
Elruso said:
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.

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 I've misunderstood the question, it seems that you may be right.
 
Elruso said:
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,
\binom{9}{2 \ 2 \ 2 \ 1 \ 1 \ 1} .​
 
Last edited:
Hurkyl said:
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,
\binom{9}{2 \ 2 \ 2 \ 1 \ 1 \ 1} .​

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.
 
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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