If different permutations of the letters of the word 'MATHEMATICS' are listed as in a dictionary, how many words (with or without meaning) are there in the list before the first word that starts with C ?
907200
Understanding the Permutation Problem
The question asks us to find the number of different arrangements (permutations) of the letters in the word 'MATHEMATICS' that would appear before the first word starting with the letter 'C' when these arrangements are listed in alphabetical or dictionary order.
Analyzing the Word 'MATHEMATICS'
First, let's identify the letters in the word 'MATHEMATICS' and count the frequency of each distinct letter. The word has 11 letters:
- M: 2 times
- A: 2 times
- T: 2 times
- H: 1 time
- E: 1 time
- I: 1 time
- C: 1 time
- S: 1 time
Total number of letters is $2+2+2+1+1+1+1+1 = 11$.
The distinct letters in alphabetical order are A, C, E, H, I, M, S, T.
Listing Words in Dictionary Order
When words are listed in dictionary order, they are arranged alphabetically. Words starting with 'A' come first, followed by words starting with 'B', then 'C', and so on. In this case, the distinct letters available to start a word are A, C, E, H, I, M, S, T.
The first word starting with 'C' will appear after all words that start with a letter alphabetically preceding 'C'. Looking at the distinct letters available (A, C, E, H, I, M, S, T), the only letter that comes before 'C' is 'A'.
Therefore, the words that appear in the list before the first word starting with 'C' are exactly those words that start with 'A'.
Counting Permutations Starting with 'A'
To count the number of permutations of 'MATHEMATICS' that start with 'A', we fix the first letter as 'A'.
This means we take one 'A' and place it at the beginning. The remaining 10 letters are the original letters minus one 'A'.
The original letters were: M(2), A(2), T(2), H(1), E(1), I(1), C(1), S(1).
After fixing one 'A' at the start, the remaining 10 letters are:
- M: 2 times
- A: 1 time (one was used for the first position)
- T: 2 times
- H: 1 time
- E: 1 time
- I: 1 time
- C: 1 time
- S: 1 time
So, we need to find the number of permutations of the 10 letters: A, C, E, H, I, M, M, S, T, T.
The formula for the number of permutations of $n$ objects where there are \(n_1\) identical objects of type 1, \(n_2\) identical objects of type 2, ..., \(n_k\) identical objects of type k is given by:
\( \text{Number of permutations} = \frac{n!}{n_1! n_2! \cdots n_k!} \)
In this case, $n = 10$ (the remaining letters). The counts of repeated letters among these 10 are M (2 times) and T (2 times). Other letters (A, C, E, H, I, S) appear 1 time each.
So, the number of permutations of the remaining 10 letters is:
\( \frac{10!}{2! \times 2! \times 1! \times 1! \times 1! \times 1! \times 1! \times 1!} = \frac{10!}{2! \times 2!} \)
Let's calculate this value:
- \( 10! = 10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 3,628,800 \)
- \( 2! = 2 \times 1 = 2 \)
So, the number of permutations is:
\( \frac{3,628,800}{2 \times 2} = \frac{3,628,800}{4} \)
\( \frac{3,628,800}{4} = 907,200 \)
Thus, there are 907,200 permutations of 'MATHEMATICS' that start with 'A'. These are precisely the words that appear before the first word starting with 'C' in the dictionary listing.
Conclusion
The number of words in the list before the first word that starts with 'C' is the number of words starting with 'A'. This number is calculated as the permutations of the remaining 10 letters after fixing 'A' at the beginning, considering the repeated letters M (2) and T (2).
The calculated value is 907,200.
| Step | Description | Calculation/Result |
|---|---|---|
| 1 | Identify letters and frequencies in 'MATHEMATICS' | M(2), A(2), T(2), H(1), E(1), I(1), C(1), S(1) - Total 11 |
| 2 | Identify distinct letters in alphabetical order | A, C, E, H, I, M, S, T |
| 3 | Determine words before first 'C' word | Words starting with 'A' |
| 4 | Fix 'A' at the start, list remaining letters and frequencies | Remaining 10 letters: M(2), A(1), T(2), H(1), E(1), I(1), C(1), S(1) |
| 5 | Calculate permutations of remaining letters | \( \frac{10!}{2! \times 2!} \) |
| 6 | Final Calculation | \( \frac{3,628,800}{4} = 907,200 \) |
Revision Table: Permutations of MATHEMATICS
Understanding how to calculate permutations, especially with repeated letters, is key to solving problems like this. Remember the formula \( \frac{n!}{n_1! n_2! \cdots n_k!} \) for permutations with repetitions.
Additional Information: Permutations and Dictionary Order
Permutations deal with arrangements of objects. When listing permutations in dictionary order, we arrange them alphabetically, just like words in a dictionary. The first letter determines the primary sorting order, followed by the second letter, and so on. To find words before a certain starting letter, we count all words beginning with letters that come earlier in the alphabet. For 'MATHEMATICS', with distinct letters A, C, E, H, I, M, S, T, words starting with 'A' come before words starting with 'C'.
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