Question

In how many ways can seven books be arranged on a shelf if a) the books are all different? b) two of the books are identical? c) the books are different and the mathematics book must be on an end? d) the books are different and four particular books must be together?

Short answer

a) 5040 ways, b) 2520 ways, c) 1440 ways, d) 576 ways

Step-by-step solution

Understanding the Problem

We need to find the number of ways to arrange seven books on a shelf under four different scenarios: all books different, two books identical, including a specific book on an end, and grouping specific books together.

All Books Different

If all seven books are different, the number of ways to arrange them is the factorial of seven, denoted as 7!. This is calculated as follows: \( 7! = 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 5040 \)

Two Books Identical

If two of the seven books are identical, the number of distinct arrangements is the factorial of seven divided by the factorial of two (to account for the indistinguishable pair). This can be expressed as \( \frac{7!}{2!} \). Calculating this, we get: \( \frac{7!}{2!} = \frac{5040}{2} = 2520 \)

Mathematics Book on an End

If a specific mathematics book must be on an end, there are two possible end positions for this book. For each of these positions, the remaining six books can be arranged in 6! ways. This calculation is: \( 2 \times 6! = 2 \times (6 \times 5 \times 4 \times 3 \times 2 \times 1) = 2 \times 720 = 1440 \)

Four Particular Books Together

Treat the four particular books as one 'superbook', making it 4 books in total now. These can be arranged as 4!, and within the 'superbook', the 4 books can be arranged internally in 4! ways. Multiply these two results together: \( 4! \times 4! = (4 \times 3 \times 2 \times 1) \times (4 \times 3 \times 2 \times 1) = 24 \times 24 = 576 \)

Key concepts

Permutations

Understanding permutations is key to solving problems involving the arrangement of objects. Permutations refer to different ways of arranging a set of items where the order matters. For instance, when you have seven different books and you want to know how many ways you can arrange them on a shelf, you are looking for the number of permutations.

When arranging all seven different books, we use the formula for permutations of n items, which is denoted as n!. Here, n equals 7, so the number of permutations is calculated as follows:

\(7! = 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 5040\)

If any two objects are fixed or must be in specific positions, we adjust our calculations to account for these constraints. Understanding permutations helps us tackle various arrangement problems with or without additional conditions.

Factorials

Factorials are fundamental in combinatorics and permutations. The factorial of a number n, denoted as n!, is the product of all positive integers up to n. Factorials grow very quickly with larger numbers. For example:

\(5! = 5 \times 4 \times 3 \times 2 \times 1 = 120\)

For the problem of arranging seven books:

\(7! = 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 5040\)

Factorials also help when dealing with scenarios like identical objects. When two books are identical, you reduce the number of unique permutations by dividing by the factorial of the number of identical objects. For two identical books out of seven:

\(\frac{7!}{2!} = \frac{5040}{2} = 2520\)

Understanding factorials enables you to compute complex arrangement problems efficiently.

Identical Objects

Identical objects pose unique challenges in permutations since they look the same and thus create fewer unique arrangements. For example, if you have seven books and two of them are identical, you need to adjust your permutations count.

With seven books, the arrangements would be:

\(7! = 5040\)

However, since two books are identical, you divide by the factorial of the number of identical books (2! in this case):

\(\frac{7!}{2!} = \frac{5040}{2} = 2520\)

If more objects were identical, say three out of seven, you would adjust further:

\(\frac{7!}{3!} = \frac{5040}{6} = 840\)

This logic is vital for accurately determining permutations where some items are indistinguishable from each other.

Specific Positions

Placing objects in specific positions changes how you calculate permutations. For example, if a specific book must be at the end of the arrangement, you have limited positions for that book. For a mathematics book that must be at either end of the shelf, you have 2 choices (start or end).

Once the position for the mathematics book is fixed, the remaining six books can be arranged in the remaining slots:

\(6! = 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 720\)

Since there are two positions for the math book, multiply by 2:

\(2 \times 6! = 2 \times 720 = 1440\)

Similarly, if certain books must be together, treat them as a single 'superbook'. For instance, if four books must stay together, treat them as one object initially:

\(4! \times 4! = 24 \times 24 = 576\)

These methods ensure accurate permutations by accounting for specific constraints or positional requirements.