[chemistrynerd12]

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??

[ebaines]

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?

[chemistrynerd12]

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.

[ebaines]

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.

[chemistrynerd12]

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?

[ebaines]

so the answer is 5405400?

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!

[chemistrynerd12]

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

Thank you!

[ebaines]

You're welcome!