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Chapter 6: Problem 9

How many strings can be formed by ordering the letters SCHOOL using some or all of the letters?

Short Answer

Expert verified
To solve this problem it is required both the general formula of permutations and the special case for duplicate elements.

Step by step solution

01

Count the total number of letters

First, count the total number of unique letters in the word SCHOOL. The word SCHOOL contains 6 letters, which are S, C, H, O, O.L.
02

Compute the permutations

For each number n from 1 to 6, compute the number of permutations. The formula to calculate permutations is P(n, k) = n! / (n-k)!, where '!' denotes the factorial. However, the word 'SCHOOL' has two repeating letters 'O'. Hence, the formula for permutation with repetition is used, where P(n, k) = n! / [p1! * p2! * ... * pk!], p - number of times an item is repeated.
03

Sum of permutations

Now, the total number of strings that can be formed by ordering the letters of 'SCHOOL' using some or all of the letters is the sum of the permutations obtained from the previous step, remember to exclude the calculations for when no letters are used.
04

Consider the special case

Take into account that the formula will get skewed because of the repeating letter 'O'. So for each sums calculated from Step 3, you need to divide by 2! whenever you're using both O's.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Factorial calculation
In mathematics, the factorial of a non-negative integer is a fundamental concept. It helps us understand how to arrange items. The factorial of a number \( n \), denoted as \( n! \), is the product of all positive whole numbers up to \( n \). For example:
  • 5! = 5 \times 4 \times 3 \times 2 \times 1 = 120
  • 3! = 3 \times 2 \times 1 = 6
Factorials are key in permutations and combinations, aiding in computing how many ways items can be arranged. Factorial calculations are straightforward when dealing with small numbers, but the values grow fast, reflecting the increase in possible arrangements when more items are considered.
Counting unique letters
In forming permutations with repetition, a crucial step is counting the unique letters in a word. For the word 'SCHOOL', we first list all individual letters: S, C, H, O, O, L. Notice that the letter 'O' appears twice, so when counting unique letters, we consider the distinct ones only.

The unique letters in 'SCHOOL' are: S, C, H, O, and L totaling five unique letters. Identifying this is critical for subsequent steps in solving permutations where repetition exists.
Forming strings with repetitions
When forming strings using letters, repetitions can change the number of unique permutations. Let's use the word 'SCHOOL' as our example.
  • The formula for permutations with repetition is \( \frac{n!}{p_1! \times p_2! \times \ldots } \).
  • Here, \( n \) is the total number of items (letters), and \( p_1 \), \( p_2 \), etc., are the counts of each repeating element.
For 'SCHOOL', the calculation is adjusted for the two 'O's. If using all six letters,
  • calculate \( \,\,\frac{6!}{2!} \,\, \).
  • This equal to \( \,\,\frac{720}{2} = 360 \,\,\) unique combinations.
Repetition modifies how many distinct strings we can create since identical letters reduce the variation.
Special case handling in combinatorics
In combinatorics, special cases arise when some elements repeat. Handling these involves adjusting general formulas to account for repeat elements.

In our example with 'SCHOOL', the repeat 'O's mean we divide by the factorial of the repetition count
  • Since there are two 'O's, we divide by \( 2! \).
When calculating permutations of fewer than six letters, inclusion of both 'O's needs similar adjustments.
Each distinct arrangement must consider this to avoid overcounting. This ensures that our calculations reflect true unique permutations.

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