7.(a) How many permutations are there of the letters of the word "abracadabra"?

(b) How many if the first letter must be "b"?

### (a) Number of Permutations of "abracadabra"

To determine the number of permutations of the word "abracadabra," we need to account for the fact that some of the letters are repeated. The word "abracadabra" has 11 letters in total, with the following frequency of each letter:

- **a**: 5 times
- **b**: 2 times
- **r**: 2 times
- **c**: 1 time
- **d**: 1 time

The formula for the number of permutations of a multiset (where some items are indistinguishable) is given by:

\[
\text{Number of permutations} = \frac{n!}{n_1! \times n_2! \times \cdots \times n_k!}
\]

Where:
- \( n \) is the total number of items,
- \( n_1, n_2, \ldots, n_k \) are the frequencies of the distinct items.

For "abracadabra":

\[
\text{Number of permutations} = \frac{11!}{5! \times 2! \times 2! \times 1! \times 1!} = \frac{39916800}{120 \times 2 \times 2 \times 1 \times 1} = \frac{39916800}{480} = 83160
\]

### (b) Number of Permutations if the First Letter is "b"

If the first letter must be "b", then we are left with the problem of finding the permutations of the remaining 10 letters: "a", "a", "a", "a", "a", "b", "r", "r", "c", "d".

The frequency distribution of the remaining letters is:
- **a**: 5 times
- **b**: 1 time
- **r**: 2 times
- **c**: 1 time
- **d**: 1 time

The number of permutations is:

\[
\text{Number of permutations} = \frac{10!}{5! \times 2! \times 1! \times 1! \times 1!} = \frac{3628800}{120 \times 2 \times 1 \times 1 \times 1} = \frac{3628800}{240} = 15120
\]

### Summary:
- **(a)** The number of permutations of the letters in "abracadabra" is **83,160**.
- **(b)** The number of permutations where the first letter is "b" is **15,120**.

Would you like any details on this solution, or do you have any further questions?

#### Related Questions:
1. How would the permutation count change if the word "abracadabra" had one less "a"?
2. What if the first and last letters must both be "b"?
3. How do the calculations change if the word has all unique letters?
4. What is the number of permutations if the first three letters must be "abr"?
5. How does the presence of identical letters affect the permutation count?

#### Tip:
When dealing with permutations of words with repeated letters, always remember to divide by the factorial of the frequency of each repeated letter to avoid overcounting.

7.(a) How many permutations are there of the letters of the word 'abracadabra'?

(b) How many if the first letter must be 'b'?

Explore how to calculate the permutations of the word 'abracadabra', accounting for repeated letters, and determine the permutations when the first letter is 'b'. This detailed explanation covers combinatorial concepts, including the multiset permutation formula and factorials, ideal for Grades 10-12.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation