Krdytkyu

I am a student in middle school, and I would really appreciate if anyone could explain this question in simpler terms.

On the refrigerator, MATHEMATICS is spelled out with $11$ magnets, one letter per magnet. Two vowels and four consonants fall off and are put away in a bag. If the T's, M's, and A's are indistinguishable, how many distinct possible collections of letters could be put in the bag?

I think that if we use complementary counting, we could subtract something from $C(11, 6)$, but I do not know what that something would be…

If the number of collections of two vowels is $A$, and that of four consonants is $B$, then the answer is $A·B$.

First to calculate $A$. There are four vowels: A, A, E, I. If two A's are selected, then there is only one collection, i.e. (A, A). If only one A is selected, then there are two collections: (A, E), (A, I). If no A is selected, then there is one collection: (E, I). Thus $A = 4$.

Next to calculate $B$. There are seven consonants: M, M, T, T, H, C, S.

Case 1: Three consonants are selected from H, C, S. To complete the collection, one more consonant has to be selected from M, M, T, T, so it is either an M or a T. Thus there are$$
C(3, 3) × 2 = 2
$$
collections.

Case 2: Two consonants are selected from H, C, S. To complete the collection, two more consonants have to be selected from M, M, T, T, so it is can be (M, M), (M, T) or (T, T). Thus there are$$
C(3, 2) × 3 = 9
$$
collections.

Case 3: One consonant is selected from H, C, S. To complete the collection, three more consonants have to be selected from M, M, T, T, so it is can be (M, M, T) or (M, T, T). Thus there are$$
C(3, 1) × 2 = 6
$$
collections.

Case 4: No consonant is selected from H, C, S. Then there is only one collection, i.e. (M, M, T, T).

Therefore, $B = 2 + 9 + 6 + 1 = 18$.

The final answer id $4 × 18 = 72$.

It's easier to count the number of ways to leave two vowels and three consonants on the refrigerator in a random order.

Let's start with the consonants. We can construct a sum equal to 3 as just 3 or as 2+1 or as 1+1+1. The first partition doesn't work, because we don't have three identical letters. It has to be either one pair and one single or three singles. Then
$$N_{text{consonants}} =
binom {2 color{gray}{smalltext{ pairs}}}
{1 color{gray}{smalltext{ pair}}} times
binom {4 color{gray}{smalltext{ singles}}}
{1 color{gray}{smalltext{ single}}} +
binom {5 color{gray}{smalltext{ singles}}}
{3 color{gray}{smalltext{ singles}}} = 18.$$
Now we have to do the same for the vowels and multiply.

We can use casework for this.
For the vowels, there are two identical letters and two distinct letters. So, the number of ways for the vowels is $1+$3C2$=4$.
The $1$ is because there is one way to choose two identical vowels. The 3C2 is because there are that many ways to choose $2$ objects out of $3$ distinct vowels. Note that it is $3$ and not $4$ because it is distinct objects.
Similarly, the number of ways for the consonants is $1+2*$4C2$+$5C4$=1+12+5=18$.
So, the total number of ways is $4*18=boxed{72}$.

We count the number of ways to choose the vowels and the consonants separately. There are four vowels, of which two are As. If there are no As, then we must choose both the remaining vowels, so there is $1$ choice; if there is one A, then we can choose the remaining vowel in $2$ ways; and if there are two As, then there are no vowels left to choose, so there is $1$ choice. This makes $1 + 2 + 1 = 4$ distinct pairs of vowels.

There are seven consonants, of which two are Ts and of which two are Ms. Since we must choose four consonants, we must use at least one of the Ts and Ms.
If we use one T and no Ms, we have only $1$ choice (use the three remaining consonants); the same is true if we use one M and no Ts.
If we use both Ts and no Ms, there are $tbinom{3}{2} = 3$ choices for the two remaining consonants; the same is true if we use both Ms and no Ts, or if we use one T and one M.
If we use both Ts and one M, there are $tbinom{3}{1} = 3$ choices for the single remaining consonant; the same is true if we use both Ms and one T.
Finally, if we use both Ts and both Ms, there are no more letters left to choose, so we get $1$ more choice.
In total, we have $2(1) + 5(3) + 1 = 18$ distinct collections of consonants.

Therefore, the number of distinct collections of letters is $4 cdot 18 = boxed{72}.$