Question

In how many different ways can you arrange all of the letters of each word? a) hoodie b) decided c) aqilluqqaaq d) deeded e) puppy f) baguette

Short answer

a) 360, b) 420, c) 138600, d) 20, e) 60, f) 10080.

Step-by-step solution

- Understand the arrangement formula

The number of ways to arrange a word where some letters may repeat is given by the formula: \[ \frac{n!}{k_1! k_2! \, ... \, k_r!} \]where \( n \) is the total number of letters and \( k_i \) are the frequencies of the repeated letters.

- Calculate for 'hoodie'

For 'hoodie':Total letters \( n = 6 \). Repeated letters: \( o \) appears 2 times.Using the formula: \[ \frac{6!}{2!} = \frac{720}{2} = 360 \]There are 360 ways.

- Calculate for 'decided'

For 'decided':Total letters \( n = 7 \). Repeated letters: \( d \) appears 3 times, \( e \) appears 2 times. Using the formula: \[ \frac{7!}{3! \cdot 2!} = \frac{5040}{6 \cdot 2} = \frac{5040}{12} = 420 \]There are 420 ways.

- Calculate for 'aqilluqqaaq'

For 'aqilluqqaaq':Total letters \( n = 11 \). Repeated letters: \( q \) appears 4 times, \( a \) appears 3 times, \( l \) appears 2 times.Using the formula: \[ \frac{11!}{4! \cdot 3! \cdot 2!} = \frac{39916800}{24 \cdot 6 \cdot 2}=\frac{39916800}{288} = 138600 \]There are 138600 ways.

- Calculate for 'deeded'

For 'deeded':Total letters \( n = 6 \). Repeated letters: \( d \) appears 3 times, \( e \) appears 3 times.Using the formula: \[ \frac{6!}{3! \cdot 3!} = \frac{720}{6 \cdot 6} = \frac{720}{36} = 20 \]There are 20 ways.

- Calculate for 'puppy'

For 'puppy':Total letters \( n = 5 \). Repeated letters: \( p \) appears 2 times.Using the formula: \[ \frac{5!}{2!} = \frac{120}{2} = 60 \]There are 60 ways.

- Calculate for 'baguette'

For 'baguette':Total letters \( n = 8 \). Repeated letters: \( t \) appears 2 times, \( e \) appears 2 times.Using the formula: \[ \frac{8!}{2! \cdot 2!} = \frac{40320}{2 \cdot 2} = \frac{40320}{4} = 10080 \]There are 10080 ways.

Key concepts

factorials

In mathematics, a factorial is a function that multiplies a number by every whole number less than it. Represented with an exclamation mark, for example, 5 factorial is written as 5!. The calculation is as follows: 5! = 5 × 4 × 3 × 2 × 1 = 120.
Factorials play a crucial role in combinatorics, particularly when calculating permutations (arrangements) of objects. They help find out how many ways we can arrange a set of elements.

combinatorics

Combinatorics is a branch of mathematics focusing on counting, arrangement, and combination of objects. It's important for solving problems related to permutations, like those with repeated elements. Combinatorics provides tools to calculate the number of distinct ways to arrange letters or objects.
For example, arranging the letters in the word 'hoodie' uses combinatorial principles. It involves calculating the total number of permutations and adjusting for repeated elements using a formula.

repeated elements

When dealing with permutations, repeated elements affect the total number of distinct arrangements. The presence of repeated letters reduces the number of unique permutations.
We use the formula \[ \frac{n!}{k_1! k_2! \, ... \, k_r!} \] where \( n \) is the total number of elements, and \( k_1, k_2, ... , k_r \) are the frequencies of the repeated elements. This adjusts for the indistinguishable arrangements caused by repeated elements.
For instance, in 'deeded', both 'd' and 'e' are repeated, which modifies the total arrangements compared to a word with all unique letters.

arrangements of letters

The arrangements of letters in a word refer to the different ways you can order the letters. Each unique permutation counts as a different arrangement. The challenge comes when some letters are repeated.
To find the number of arrangements, we first consider the total number of all possible arrangements of the letters as if they were unique. Then, we divide by the factorials of the frequency of each repeated letter to adjust for overcounting.
For example, in the word 'puppy', the letter 'p' is repeated twice, which impacts the number of unique ways the letters can be arranged. The correct calculation uses the permutation formula considering the repetition of 'p'.