Okay, so I need to figure out how many different ways the letters of the word 'MATHEMATICS' can be arranged such that all the vowels are together. Hmm, let me start by recalling some basics of permutations with constraints.
First, the word is 'MATHEMATICS'. Let me write down the letters to see them clearly: M, A, T, H, E, M, A, T, I, C, S. Wait, let me count them to make sure. M (1), A (2), T (3), H (4), E (5), M (6), A (7), T (8), I (9), C (10), S (11). So there are 11 letters in total.
Now, the problem wants the vowels to always come together. Let me first identify the vowels in the word. The vowels are A, E, A, I. So vowels are A, E, A, I. Wait, that's four vowels, but note that there are two A's. So in total, the vowels are A, E, A, I. So we have 4 vowels with a repetition of A twice. The consonants would then be the remaining letters: M, T, H, M, T, C, S. Let's list them: M, T, H, M, T, C, S. That's 7 consonants.
Now, the strategy for arranging the letters with all vowels together is to treat the group of vowels as a single entity or "block". So instead of considering the vowels as separate letters, we consider them as one block. Then, we have to find the number of ways to arrange this block along with the consonants. But also, we need to consider the number of ways the vowels can be arranged within the block.
Wait, but first, let's confirm the number of consonants and vowels. Total letters: 11. Vowels: A, E, A, I (4 letters). Consonants: 11 - 4 = 7 letters. The consonants are M, T, H, M, T, C, S. So consonants include M (twice), T (twice), H, C, S. So consonant repetitions: M and T each appear twice, others are unique.
Similarly, vowels: A appears twice, E and I once each.
So, back to the approach. If we group all vowels together as a single block, then the total number of "items" to arrange is the block plus the consonants. Since there are 7 consonants, the total items would be 7 consonants + 1 vowel block = 8 items. However, we need to remember that within the consonants, there are duplicates (M and T each appearing twice), and within the vowel block, there are duplicates (A appearing twice).
Therefore, the total number of arrangements would be calculated as follows:
First, compute the number of ways to arrange the 8 items (7 consonants + 1 vowel block). But since the consonants themselves have repeated letters, we need to divide by the factorial of the number of duplicates. Similarly, the vowel block, when considered internally, will have permutations with duplicates.
So step by step:
Treat the vowels as a single block. So we have 7 consonants and 1 block, totaling 8 elements.
The consonants have duplicates: two Ms and two Ts. The rest (H, C, S) are unique. So the number of ways to arrange the consonants and the vowel block is 8! divided by (2! * 2!) to account for the duplicates in the consonants.
Then, for the vowel block itself, since it contains 4 vowels with two As, the number of arrangements within the block is 4! divided by 2! (for the duplicate As).
Multiply these two results together to get the total number of arrangements.
Let me verify that. So total arrangements = (number of arrangements of consonants and vowel block) * (number of arrangements within the vowel block).
Yes, that makes sense. Because for each arrangement of the consonants and the vowel block, the vowels themselves can be arranged in different ways within the block.
So let me calculate each part.
First, the number of arrangements of the 8 items (7 consonants + 1 block). The consonants have two Ms and two Ts, so duplicates. So permutations would be 8! / (2! * 2!). Let me compute that.
8! = 40320
2! * 2! = 2 * 2 = 4
So 40320 / 4 = 10080
Then, the number of arrangements within the vowel block. The vowels are A, E, A, I. So 4 letters with two As. So permutations would be 4! / 2! = 24 / 2 = 12.
Therefore, total arrangements would be 10080 * 12. Let me compute that.
10080 * 12: 10080 * 10 = 100800, 10080 * 2 = 20160, so total is 100800 + 20160 = 120960.
Wait, so is 120,960 the total number of arrangements?
But hold on, let me make sure I didn't miss any other duplicates. The consonants: M, T, H, M, T, C, S. So two Ms and two Ts. So 2! for Ms and 2! for Ts. Correct.
Vowels: A, E, A, I. Two As. So 2! division. Correct.
Yes, that seems right.
But let me cross-check. Alternatively, another way to think is: when we consider all letters, the total number of arrangements without any restrictions is 11! divided by (2! for M, 2! for T, 2! for A). Wait, in the original word 'MATHEMATICS', the duplicates are:
Letters: M, A, T, H, E, M, A, T, I, C, S.
So duplicates:
M: 2
A: 2
T: 2
Others: H, E, I, C, S: 1 each.
So total permutations without restrictions: 11! / (2! * 2! * 2!) = 11! / (8).
But that's without restrictions. However, the problem here is with the restriction that vowels are together. So the answer we found earlier was 120,960. Let's compute 11! / (2! * 2! * 2!) to see what it is. 11! is 39916800. Divided by 8, gives 4989600. So the total permutations without restrictions is 4,989,600. The answer with restrictions is 120,960, which is significantly smaller, which makes sense because we're imposing a restriction. So 120,960 is roughly 2.4% of the total, which seems plausible.
Alternatively, let me check the method again. When we group the vowels together, the total number of elements becomes 8: 7 consonants and 1 vowel block. The consonants have duplicates (2 Ms and 2 Ts), so we divide by 2! for Ms and 2! for Ts. Then, the vowel block has duplicates (2 As), so when we permute the vowels inside the block, we divide by 2!.
Thus, total number of arrangements is 8! / (2! * 2!) * (4! / 2!) = (40320 / 4) * (24 / 2) = 10080 * 12 = 120,960. That seems correct.
Wait, but let me also think of another approach to verify. Suppose we consider the vowels as a single block. Then, how many positions can this block occupy?
In the original word, there are 11 letters. If vowels are together, they form a block of 4 letters. So the block takes up 4 positions, and the remaining 7 consonants take up 7 positions. So effectively, the number of "slots" is 8: the vowel block can be in positions 1-4, 2-5, ..., 8-11. But the way we usually handle this is by considering the vowel block as a single entity, so we have 8 entities total: 7 consonants and 1 vowel block. Then, we arrange these 8 entities, considering the duplicates in consonants. Then multiply by the arrangements within the vowel block.
Yes, that's the same as before. So 8 entities, consonants have duplicates, so 8! / (2! * 2!), and vowels have duplicates, so 4! / 2!. Multiply together. So 8! / (2! * 2!) * 4! / 2! = (40320 / 4) * (24 / 2) = 10080 * 12 = 120,960. Same result.
Alternatively, if I were to use the formula for permutations with some elements grouped together, it's similar to arranging n items where k items are grouped, which becomes (n - k + 1)! * k! (if all k items are distinct). But in this case, the k items (vowels) are not all distinct, and the surrounding consonants also have duplicates. So the formula adjusts by dividing by the factorial of the duplicates in both the consonants and the vowels.
Another way: think of the problem as two steps.
Step 1: Arrange the consonants among themselves, considering duplicates.
Wait, but no, because when we treat the vowels as a block, we have to interleave the consonants and the vowel block. So perhaps the better approach is as initially: total arrangements = (number of ways to arrange consonants and vowel block) * (number of ways to arrange vowels within the block). Considering duplicates in both steps.
So consonants include two Ms and two Ts. So arranging them with the vowel block: 8 objects, with duplicates, so 8! / (2! * 2!). Then vowels have duplicates, so arranging them is 4! / 2!.
Multiply the two. So yes, 8! / (2! * 2!) * 4! / 2! = 120,960.
Therefore, the answer should be 120,960.
Wait, but let me check with a smaller example to ensure that this approach works.
Suppose we have a word like 'APPLE'. Let's say we want to arrange the letters so that the vowels (A, E) are together. The word 'APPLE' has letters A, P, P, L, E. So vowels are A and E. Consonants are P, P, L.
Total letters: 5. Vowels: 2 (A, E). Consonants: 3 (P, P, L).
If we group vowels together as a block, we have the block + 3 consonants. But consonants have two Ps. So number of arrangements would be: number of arrangements of the block and consonants: 4 items, with two Ps. So 4! / 2! = 12. Then, number of arrangements within the vowel block: 2! = 2. So total arrangements: 12 * 2 = 24.
Alternatively, total permutations without restrictions: 5! / 2! = 120 / 2 = 60. If we calculate the number of arrangements where vowels are together: 24. Let's check by another method.
Vowels A and E can be together as AE or EA. So treat them as a single entity. Then, we have four entities: [AE], P, P, L. The number of distinct arrangements is 4! / 2! = 12, multiplied by 2 (for AE and EA), giving 24. So same result.
So this shows that the method works for the smaller example. Therefore, applying the same method to 'MATHEMATICS' should be valid.
Thus, returning to the original problem, I think the answer is 120,960. So the number of different ways is 120,960.
But just to be thorough, let me verify once again:
Total letters: 11. Vowels: 4 (A, E, A, I). Consonants: 7 (M, T, H, M, T, C, S).
Group vowels as a block. So total items to arrange: 7 consonants + 1 block = 8 items.
But consonants have two Ms and two Ts, so the number of distinct arrangements of these 8 items is 8! divided by (2! for Ms and 2! for Ts). So 8! / (2! * 2!) = 40320 / 4 = 10080.
Within the vowel block: 4 vowels with two As, so arrangements are 4! / 2! = 24 / 2 = 12.
Multiply them: 10080 * 12 = 120,960.
Yes, that seems correct.
Therefore, the answer should be 120,960.
Final Answer The number of different ways is \boxed{120960}.
Prompt: In how many different ways can the letters of the word 'MATHEMATICS' be arranged so that the vowels always come together?