Problem 30
Question
In \(23-30,\) find the number of different arrangements that are possible for the letters of each of the following words. MATHEMATICS
Step-by-Step Solution
Verified Answer
There are 4,989,600 different arrangements of 'MATHEMATICS'.
1Step 1: Count the Total Number of Letters
The word 'MATHEMATICS' has a total of 11 letters.
2Step 2: Identify and Count Repeated Letters
In 'MATHEMATICS', the letters M, A, and T each appear twice. The remaining letters (H, E, I, C, S) each appear once.
3Step 3: Calculate the Factorial of Total Letters
Calculate the factorial of the total number of letters: \[11! = 11 \times 10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 39,916,800\]
4Step 4: Calculate the Product of Factorials of Repeated Letters
For the repeated letters M, A, and T (each repeated twice), calculate the product of their factorials: \[2! \times 2! \times 2! = 2 \times 2 \times 2 = 8\]
5Step 5: Calculate the Number of Different Arrangements
Use the formula for permutations of multiset to find the number of different arrangements:\[\frac{11!}{2! \times 2! \times 2!} = \frac{39,916,800}{8} = 4,989,600\]
6Step 6: State the Final Result
The total number of different arrangements of the letters in the word 'MATHEMATICS' is 4,989,600.
Key Concepts
Factorial CalculationRepeated LettersArrangements of Letters
Factorial Calculation
Factorial calculation is an essential concept when determining permutations, especially when dealing with multisets. A factorial, denoted with an exclamation mark (!), represents the product of all positive integers up to a given number. For example, to compute the factorial of 11, we multiply all integers from 1 through 11 together:\[11! = 11 \times 10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 39,916,800\]The factorial helps us understand the total number of ways in which a set of items can be arranged. This calculation is crucial in step 3 of the solution above, as it gives us the number of possible arrangements if all letters were unique. Understanding how to calculate factorials sets the groundwork for tackling more complex permutation problems, especially in combinatorial mathematics.
Repeated Letters
In many words, especially longer ones like 'MATHEMATICS', some letters can repeat. Repeated letters affect the overall number of unique arrangements since swapping these identical letters doesn’t create a new permutation. Therefore, we must adjust our calculations to account for these repetitions.For the word 'MATHEMATICS', the letters 'M', 'A', and 'T' are each repeated twice. To account for these repeated letters in our calculation, we find the factorial for each repeated letter count. This results in:
- \(2!\) because 'M' repeats twice
- \(2!\) because 'A' repeats twice
- \(2!\) because 'T' repeats twice
Arrangements of Letters
Arranging letters involves calculating the number of ways to organize them to form different words or sequences. With permutations of multisets, especially with repeated elements, there's a specific formula to apply. In the case of 'MATHEMATICS', we've identified the presence of repeated letters.To find the number of unique arrangements, we use the permutation formula for multisets:\[\frac{n!}{k_1! \times k_2! \times k_3! \ldots}\]Where \(n!\) is the factorial of the total number of letters, and each \(k_i!\) refers to the factorial of each group's repeated letters. For our example:\[\frac{11!}{2! \times 2! \times 2!} = \frac{39,916,800}{8} = 4,989,600\]This formula ensures only unique permutations are counted by dividing the total permutations by those that are repeated due to identical letters. The calculation yields 4,989,600 unique arrangements of 'MATHEMATICS', beautifully encapsulating how systematic accounting of each letter's repetition shapes the total possibilities.
Other exercises in this chapter
Problem 29
In \(23-30,\) find the number of different arrangements that are possible for the letters of each of the following words. STATISTICS
View solution Problem 29
In how many ways can 3 different job openings be filled if there are 8 applicants?
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In how many ways can 5 roles in the school play be filled if there are 9 possible people trying out? (Each role is to be played by a different person.)
View solution Problem 31
A box contains 9 red, 4 blue, and 6 vellow chips. In how many ways can 6 chips be chosen if: a. all 6 chips are red? c. 2 chips are blue? e. 4 chips are yellow?
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