Ok. I think I'm beginning to understand.

When you say MATHEMATICS can be arranged into 11!/2!2!2! or 11!/8, you mean that is the number of total possible four letter words without repeating? That is to say that by dividing 2!x2!x2!x1 is "removing" potential duplicates caused by the repeating letters? I'm sorry to keep asking questions. I'm just trying to fully understand the concept behind this problem. It could be the difference between a letter grade for me. What has thrown me for a loop is that in the hints, there's a suggestion of dividing possible four letter words into subsets which is why I did it in three sets (all letter combinations non repeating, double letters with double letters, and double letters with single letters). What did I not account for when I did that? Everything below is what was handed out...

**Original Problem**

The letters of the word MATHEMATICS are written on cards and dropped into a hat. 4 cards are drawn without replacement. What is the probability the 4 cards spell MATH?

**HInts given with the problem are...**

2. It is known that

P r[4 cards spell MATH] = n(ways to spell MATH)/n(all arrangements of 4 cards)

3. The number in the numerator of this fraction is easy to calculate.

4. The number in the denominator is not easy to calculate. Every ”easy”

solution that can be initially guessed relies upon an assumption that is unwar-

ranted. For example, one might guess that the number in the denominator is

equal to P (11, 4), but this guess can only work if each letter that occurs in the

word MATHEMATICS occurs only one time. The same problem arises in

the solution to problem 35 on the homework.

5. We should focus on calculating the number that should go in the denominator

of the fraction, since that is the really hard part of the problem. Here is an

outline of how that can be done.

Pr(4 cards spell MATH)= n(ways to spell MATH)/n(all arrangements of 4 cards)

i. We have to come up with a partition of the set that we are considering. In this

example, we want to compute the cardinal number of the set of 4 letter words

that can be made using the letters in the word MATHEMATICS. Point 4

suggests that difficulties are going to arise because this word has 3 letter types

which occur more than once. In fact, the letters A, M, and T occur exactly

twice. Each other letter type (and there are five of them) occurs exactly once.

So, we can partition this set into two subsets. One subset will be the set of

all four letter words which contain at least one M, A, or T. The other subset

will be the complement of the first subset. Explicitly, it is the subset of four

letter words which contain no A, M , or T. So the second subset contains only

four letter words composed of letter types that occur only once in the word

MATHEMATICS. It is easy to count the size of this second subset.

ii. However, it is still not easy to count the size of the first subset. We’ll need

to partition this subset even further. Since this subset contains at least one M,

A, or T, we may break it into the following disjoint subsets (what follows is a

description of the subsets)

1.Four letter words containing A but not T or M

2.Four letter words containing T but not A or M

3.Four letter words containing M but not A or T

4.Four letter words containing A and M but not T

5.Four letter words containing A and T but not M

6.Four letter words containing T and M but not A

7.Four letter words containing A, T, and M

You can check that these sets are all disjoint. For example, the word MAHE is

contained in the set desribed in 4. It cannot be in any of the other sets. There

is a similarity with this problem to problems in which Venn diagrams are used.

Here is a description of how to calculate the size of the set described in 3:

We want to calculate the number of four letter words that contain the letter M,

but not the letter A nor the letter T. There are two M’s to choose from, so a

four letter word can contain either one or two M’s. Because of this, it makes

sense to calculate the number of four letter words which contain exactly one M

and then calculate the number of four letter words which contain exactly two

M’s (but no A and T, of course) We will then add these two numbers together

to calculate the number of four letter words containing an M but not the letter

A nor the letter T. The number of elements in the sets described in 1 and 2 are

similarly calculated.

Here is a description of how to calculate the size of the set described in 4:

The four letter word now needs to contain an A and an M, but no T. There are

various ways this can occur. Here is a list of those ways:

A. The four letter word contains exactly one A and exactly one M

B. The four letter word contains exactly two A’s and exactly one M

C. The four letter word contains exactly one A and exactly two M’s

D. The four letter word contains exactly two A’s and exactly two M’s

Once the we calculate the number of words in each of these four cases, the

numbers can be added. This will give us the number of four letter words that

contain an A and an M, but no T. The calculation of the size of the sets described

in 5 and 6 is done similarly.

Here is a desciption of how the size of the set described in 7 may be calculated:

The word must have four letters, and there are three letter types that must be

chosen in 7. There are various ways this can occur. Here is a list of the possible

ways:

A. The four letter word contains exactly 2 A’s, exactly one M, and exactly one T

B. The four letter word contains exactly one A, exactly two M’s, and exactly one T

C. The four letter word contains exactly one A, exactly one M, and exactly two T’s

D. The four letter word contains exactly one A, exactly one M, and exactly one T

In the sets described in A., B., and C. all four letters in the word are used up.

This is not the case for the set decribed in D. The number of words in each of

these sets should be calculated than added.

Once again, I'm sorry for continuing. I greatly appreciate your help and patience.