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

In summary, there are C(9,2)*C(7,2)*5! possible letter combinations with 9 letters using the letters M-A-T-E-M-A-T-I-K. This can also be written as 9! / (2! * 2! * 2!), or as a multinomial coefficient, \binom{9}{2 \ 2 \ 2 \ 1 \ 1 \ 1}.
  • #1
Elruso
5
0
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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  • #2
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?
 
  • #3
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.
 
  • #4
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.
 
  • #5
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,
[tex]\binom{9}{2 \ 2 \ 2 \ 1 \ 1 \ 1} .[/tex]​
 
Last edited:
  • #6
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,
[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.
 
  • #7
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!​
 

1. How do you calculate the number of possible combinations?

To calculate the number of possible combinations, you can use the formula nCr = n! / (r! * (n-r)!), where n represents the total number of items and r represents the number of items in each combination.

2. Can combinations be used in real-world applications?

Yes, combinations are used in many real-world applications such as creating unique passwords, lottery number selections, and creating different combinations of ingredients in recipes.

3. Is there a difference between combinations and permutations?

Yes, there is a difference between combinations and permutations. Combinations focus on the selection of items without considering the order, while permutations consider both the selection and order of items.

4. How can you tell if a problem requires the use of combinations?

A problem may require the use of combinations if it involves selecting a group of items without considering the order or repetition of items. Additionally, if the problem involves choosing a certain number of items from a larger set, combinations may be used.

5. Are there any shortcuts or tricks for calculating combinations?

Yes, there are some shortcuts for calculating combinations. For example, if the problem involves selecting all of the items, the number of combinations will be 1. Additionally, if the problem involves selecting half of the items, the number of combinations can be calculated by dividing the total number of items by 2.

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