How many different permutations can be made out of the letters of the word ‘PERMUTATION’?
Understanding Permutations of Words
The question asks us to find the number of different permutations that can be made from the letters of the word 'PERMUTATION'. A permutation is an arrangement of objects in a specific order. When dealing with words, the objects are the letters.
First, let's analyze the word 'PERMUTATION':
- We need to count the total number of letters in the word.
- We also need to identify if any letters are repeated and how many times they are repeated.
Counting Letters and Identifying Repetitions
Let's list the letters in 'PERMUTATION' and count their occurrences:
- P: 1
- E: 1
- R: 1
- M: 1
- U: 1
- T: 2
- A: 1
- I: 1
- O: 1
- N: 1
The total number of letters in the word 'PERMUTATION' is 11.
The letter 'T' is repeated 2 times. All other letters appear only once.
Applying the Permutation Formula with Repetitions
When we have a set of \(n\) objects where some objects are identical, the number of distinct permutations is calculated using the formula:
$$ \text{Number of Permutations} = \frac{n!}{n_1! n_2! \dots n_k!} $$
Where:
- \(n\) is the total number of objects.
- \(n_1, n_2, \dots, n_k\) are the frequencies of each distinct repeated object.
In the case of the word 'PERMUTATION':
- \(n = 11\) (total letters)
- The letter 'T' is repeated \(n_1 = 2\) times.
- All other letters appear once, so their frequencies are 1. The factorials of these frequencies are \(1! = 1\), which do not affect the denominator's value other than multiplying by 1.
Using the formula, the number of distinct permutations of the letters in 'PERMUTATION' is:
$$ \frac{11!}{2!} $$
Comparing with Given Options
Let's compare our calculated result with the given options:
- Option 1: \(\frac{{11!}}{{2!}}\)
- Option 2: \(\frac{{10!}}{{2!}}\)
- Option 3: \(\frac{{11!}}{{3!}}\)
- Option 4: \(\frac{{10!}}{{3!}}\)
Our calculated number of permutations is \(\frac{{11!}}{{2!}}\), which matches Option 1.
Conclusion
The number of different permutations that can be made out of the letters of the word ‘PERMUTATION’ is \(\frac{{11!}}{{2!}}\).
| Word | PERMUTATION |
|---|---|
| Total Letters (n) | 11 |
| Repeated Letter | T |
| Frequency of Repeated Letter | 2 |
| Formula Used | \(\frac{n!}{n_1! n_2! \dots}\) |
| Number of Permutations | \(\frac{11!}{2!}\) |
Revision Table: Permutations and Factorials
| Concept | Definition | Formula/Example |
|---|---|---|
| Permutation | An arrangement of objects in a specific order. Order matters. | Permutations of ABC taken 2 at a time: AB, BA, AC, CA, BC, CB (6 permutations) |
| Factorial | The product of an integer and all the integers below it down to 1. Denoted by \(n!\). | \(5! = 5 \times 4 \times 3 \times 2 \times 1 = 120\) \(0! = 1\) (by definition) |
| Permutations with Repetition | Arrangements of objects where some objects are identical. | Number of permutations of n objects with \(n_i\) repetitions of type i: \(\frac{n!}{n_1! n_2! \dots}\) |
Additional Information: Combinations vs. Permutations
It's important to distinguish between permutations and combinations.
- Permutations: Order matters. (e.g., arranging letters in a word, arranging people in a line).
- Combinations: Order does not matter. (e.g., selecting a committee from a group, choosing items from a menu).
The formula for combinations of \(n\) objects taken \(r\) at a time (without repetition) is \(\binom{n}{r} = \frac{n!}{r!(n-r)!}\). For permutations of \(n\) objects taken \(r\) at a time (without repetition), it is \(P(n, r) = \frac{n!}{(n-r)!}\).
In this problem, since we are arranging the letters to form different 'words' (even if they aren't real words), the order of the letters matters. Therefore, we use the permutation formula, specifically the one for arrangements with repeated items because the letter 'T' is repeated in 'PERMUTATION'.
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Important Questions from Permutations and Combinations
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The largest coefficient of ( x + 1)20 is:
