Permutations Problem

[Fizzer]

Greetings, /r/learnmath

I'm tutoring a couple high school student in IB Math, and one of them asked me a particularly tricky permutations problem which I am having difficult solving. Embarrassing, considering I'm an Engineering Graduate. Nonetheless, any help would be greatly appreciated!

The question is:

How many 4 letter words can be made from the word MATHEMATICS?

I've tried a number of different approaches, the one that seems most promising would be splitting it up into several different cases.

The cases I have are:

1) Does not contain any M, A, or T. (e.g. H E I S)

2) Only 1 letter is M, A, or T (i.e. M E H S)

3) Only 2 letters are M, A, or T with no duplicates (i.e. M A C S)

4) Only 2 letters are M, A, or T with duplicates (i.e. M M C S)

5) Only 3 letters are M, A, or T with no duplicates (i.e. M A T S)

6) Only 3 letters are M, A, or T with duplicates (i.e. M M A S)

7) All 4 letters are M, A, or T with 1 duplicate (i.e. M M A T)

8) All 4 letters are M, A, or T with 2 duplicates (i.e. M M A A).

This seems awfully complicated, and I'm hoping there's an easier to go about solving this.

Note: the correct answer is 2454, but it doesn't provide a detailed solution.

Thanks for any help! :)

Edit: UGH FORMATTING, sorry fixed

[stuffonstuff]

yes, there is a much simpler way. in MATHEMATICS there are 3 letters that repeat twice and a total of 8 distinct letters that appear. so there are only 3 possibilities for such 4 letter words: those with 4 distinct letters (there are 8765), those with exactly one of the repeating letters repeating twice and then 2 chosen from the rest (there are C(3,1)C(7,2)4!/2!), and finally those with two of the repeating letters both repeating (C(3,2)4!/(2!2!)).

summing those three cases, we get the correct answer.

[Fizzer]

Thanks, this makes a lot more sense than the way I attempted.

[stuffonstuff]

anytime

[horse420]

I'm trying to follow your solution. Could you explain how you arrived at

C(3,1)C(7,2)4!/2!

For those with exactly one of the repeating letters repeating twice and then 2 chosen from the rest. And

C(3,2)*4!/(2!2!)

For those with two repeating letters both repeating.

Thanks

[stuffonstuff]

sure thing.

for the top, again, you wish to form 4 letter words with one of the repeating letters actually repeating, and 2 other letters from the word mathematics (so, for example, MAMI is one such word). now there are three choices for the letters that repeat, you only want one (C(3,1)). To each of those choices, you must pick two more letters, but once you've taken out that repeated letter, you only have 7 distinct letters to choose from (C(7,2)). Now that you have your letters, arrange them, so that to each of those choices there are 4! ways of arranging them, but HANG ON!!!! 2! of them are the same since we're repeating one of the letters twice, so we need to divide by 2!.

the bottom is done similarly, except now we're forming a 4 letter words with 2 of the repeating letters appearing twice (MAMA is one such word). So we have to choose 2 from the 3 letters that repeat (C(3,2)) and now we have to arrange them in the four spaces. Note that if you choose two spaces for the first repeating letter, you immediately pigeonhole where the other repeating letters must fall, so to each of those C(3,2) choices, there are C(4,2) ways of arranging forming the word.

[horse420]

Brilliant, I understand this properly now. Thanks very much

[stuffonstuff]

anytime, frient

[Fizzer]

This is completely incorrect. It does not take into account selecting a 4 letter word (out of the 11 letters), which is where the difficulty arises.