Question

A shelf has room for 10 books. (a) Given an inventory of 25 books, how many years will it take to display all combinations of 10 books if the display is changed once a week? (b) How many years will it take if the display is changed five times a week?

Short answer

(a) 62862 years, (b) 12572 years.

Step-by-step solution

Determine Combinations

To find the number of different combinations of 10 books from a collection of 25, use the combination formula \( \binom{n}{r} \), where \( n \) is the total number of books and \( r \) is the number of books to choose. Calculate \( \binom{25}{10} \).

Calculate Total Combinations

Calculate \( \binom{25}{10} = \frac{25!}{10!(25-10)!} \). This simplifies to \( \frac{25 \times 24 \times 23 \times 22 \times 21 \times 20 \times 19 \times 18 \times 17 \times 16}{10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1} \).

Simplify the Expression

Simplify the expression to find \( \binom{25}{10} = 3268760 \). This means there are 3,268,760 different combinations.

Determine Weekly Changes for Part (a)

If the display is changed once a week, find the number of weeks needed to display all combinations: Divide 3,268,760 by 52 weeks per year. \[ \frac{3268760}{52} \approx 62861.92 \text{ years} \].

Determine Weekly Changes for Part (b)

If the display is changed five times a week, calculate the weeks needed by dividing 3,268,760 by (52 weeks multiplied by 5 times a week). \[ \frac{3268760}{260} \approx 12572.15 \text{ years} \].

Round to Whole Years

Finally, round the years to the nearest whole number: 62862 years for part (a) and 12572 years for part (b).

Key concepts

Understanding the Combination Formula

When working with problems in combinatorics, especially when you're tasked with choosing or arranging items, the **combination formula** becomes an invaluable tool. This formula helps in determining how many ways you can choose a subset of items from a larger pool, where the order of selection does not matter. The combination formula is expressed as \( \binom{n}{r} \), which is read as \( n \) choose \( r \).

The formula itself is:
\[\binom{n}{r} = \frac{n!}{r!(n - r)!}\]
Where:

• \( n! \) (n factorial) is the product of all positive integers up to \( n \).
• \( r! \) is the factorial of the number of items you want to choose.
• \((n - r)!\) is the factorial of the difference between the total number of items and the number of items being chosen.

When using this formula, you only consider different groupings, regardless of how they're ordered. This is what distinguishes combinations from permutations, the latter considering the arrangement as significant.
For the exercise above, we used \( \binom{25}{10} \) to find out how many ways we can select 10 books out of 25, which provided the extensive number of possible book displays: 3,268,760.

Factorial Calculation Made Simple

Factorial calculation plays a crucial role in many combinatorial formulas, including combinations. The factorial of a number \( n \), denoted as \( n! \), means multiplying all integers from \( 1 \) up to \( n \). This gives an idea of how many ways you can order \( n \) items.

For instance, 5! is calculated as:
\[ 5 \times 4 \times 3 \times 2 \times 1 = 120 \]
Consequently, that means you can arrange 5 items in 120 different ways. Now, as numbers increase, factorials grow really fast, which is why knowing how to simplify calculations by cancelling terms is essential.

In the original exercise, calculating \( \binom{25}{10} \) involves finding:
\[ \frac{25!}{10! \times (25-10)!} \]
This breaks down to dividing the product of 15 successive numbers (from 25 down to 16) by the factorial of 10 (10!). Understanding these steps allows you to perform large factorial calculations without overwhelming yourself by doing each calculation from scratch.

Exploring Discrete Mathematics

**Discrete mathematics** is the field of study that deals with mathematical structures which are fundamentally discrete instead of continuous. This includes areas like graph theory, logic, and of course, combinatorics.

One of the central topics in discrete mathematics is solving problems related to combinations and permutations, just like the one in our exercise. These problems help us understand different ways to organize or select a group of items, illustrating how principles from this branch of math apply in practical scenarios.
Combinatorial problems often require using both the combination formula and factorial calculations, as we saw.

In discrete math:

• We deal with countable, opposed to infinite, scenarios.
• Problems often involve determining efficient strategies for organization or arrangement.
• Understanding these concepts can be handy in computer science, cryptography, and statistical analysis.

So, mastering discrete mathematics has real-world usefulness in areas that require rigorous planning or optimization, beyond just solving textbook problems.