In how many permutations do the letters H, E, C and N appear in that order, but not necessarily consecutively?

Help! I don't even know where to start with this question, can someone give me a leadway??

First - consider HEC as the first three letters, in positions 1, 2, and 3. Then N can be in position 4, or 5, or 6, or... or 12 or 13. That's a total of 10 choices.
Now consider HE as the first two letters, skip position 3 (it gets one of the other letters M, A, T or I) and put C in position 4. Then N can be in position 5, or 6, or... or 12 , or 13, for a total of 9 choices.

Continue this line of thinking, and you see that for HE in positions 1 and 2, the total number of possibilities is:


Now consider H as the first letter, skip position 2, and put E in position 3. Using the above logic, the total number of choices for C and N is . See the pattern? Continuing on: for H in the first position the total possibilities are:




Now repeat with H in position 2. By the above reasoning the posibilities are:


Can you take it from here?

But I'm in a class that's working on permuations, and that's not the way you approach it, that's working with sums.

Here's another way to do it - figure out the total number of ways that the 13 letters can be arranged without duplicates (hint - take care with the fact that there are duplictate letters: 3 AS's, 2 T's, 2 I's, and 2 M's), then divide by the number of ways that H,E,C, and N can be arranged (since only one of those arangments is allowed). Post back and tell us how you're doing with this.

okay, so figuring out the number of ways that the 13 letters can be arranged would be:

13!/3!2!2!2!. then with that value, divide by the number of ways that H, E, C and N can be arranged would be 4!

so the answer is 5405400?

Quote:

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Originally Posted by chemistrynerd12

so the answer is 5405400?

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Yes - I believe that is correct.

I see now that I made an error in my first posting. The process I used shows how many ways the letters H, E, C, and N can be positioned in a 13-letter word, but I forgot that for each of those arrangements there are a bunch of ways that the other letters (M,A, T and I) can be arranged. For example, one way to arrange them is:

HECNMMAAATTII,

And another is:

HECNMMAAATITI

Note that both have H,E,C and N in the same positions, but the other letters are arranged differently. So to account for this we have to multiply the summations that I talked about by the number of ways that M, A, I and T and I can be arranged, which is 9!/(3!2!2!2!). If you do this you get the same answer as you found. It's nice to know that through two different methods we arrive at the same answer!

Okay, yes that makes sense, wow what a confusing question.
I wouldn't had gotten and understood the answer without your help.

Thank you!

You're welcome!