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# Multinomial/combinatorics question

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1. Question: Consider the word COMBINATORICS. What is the probability on the arrangements containing the word COMIC without interruption?

I think.....we have 2 options for C (appearing at front or back), 2 options for O and 2 options for I. The word COMIC can then be placed in 13-5+1 = 9 places, and the remaining words can be arranged in 8! ways. So this gives 2 x 2 x 2 x 9 x 8!

We then divide this by (13!/3.*2!), which is the total number of combinations.

But then the next question is:

How many different ways can the word COMIC be made from the letters of COMBINATORICS if the position the letters came from in COMBINATORICS distinguishes alike letters?

So I'm confused...should my answer to this part be my answer to the previous part, and for the previous part should the number of ways be just 9 * 8!

Thanks for any help, really stressing out
2. Does anyone have any advice on this ?
3. (Original post by combinatorix)
Question: Consider the word COMBINATORICS. What is the probability on the arrangements containing the word COMIC without interruption?

I think.....we have 2 options for C (appearing at front or back), 2 options for O and 2 options for I.
I think you have 1 option for C, beacuse you have to select 2 C letters from 2, to form COMIC and you can not distinguish them (which one is at the start and which is at the end position).
So you have 1 option for O and I, too because you can not distinguish them.
So you have 8 different letter, it means 8! options, and the COMIC can be placed in 9 places so 9*8!=9!
The word COMIC can then be placed in 13-5+1 = 9 places, and the remaining words can be arranged in 8! ways. So this gives 2 x 2 x 2 x 9 x 8!

We then divide this by (13!/3.*2!), which is the total number of combinations.
We divide by 13!/(2!*2!*2!)

But then the next question is:

How many different ways can the word COMIC be made from the letters of COMBINATORICS if the position the letters came from in COMBINATORICS distinguishes alike letters?

So I'm confused...should my answer to this part be my answer to the previous part, and for the previous part should the number of ways be just 9 * 8!

Thanks for any help, really stressing out
Here the answer is that you worked out for Q1

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