Problem 37 - Algebra and Trigonometry

Question

33–40 These problems involve distinguishable permutations. Distributions A man bought three vanilla ice-cream cones, two chocolate cones, four strawberry cones, and five butterscotch cones for his 14 children. In how many ways can he distribute the cones among his children?

Step-by-Step Solution

Answer

There are 2,522,520 ways to distribute the cones among the children.

Step 1: Understand the Problem

We need to find the number of ways to distribute different types of ice-cream cones among 14 children. The total number of cones is the sum of cones of each flavor. Let's count: 3 vanilla, 2 chocolate, 4 strawberry, and 5 butterscotch cones, giving us a total of 14 cones.

Step 2: Use Permutation Formula for Indistinguishable Objects

The problem involves permutations of items where some items are indistinguishable (cones of the same type are identical and indistinguishable from each other). When dealing with indistinguishable items, we use the formula: \[ \frac{n!}{n_1! \times n_2! \times ... \times n_k!} \] where \(n\) is the total number of items, and \(n_1, n_2, ...\) are the counts of each indistinguishable type.

Step 3: Apply the Formula

We substitute into the formula, using \(n = 14\) (total cones), and the counts for each type of cone: 3 for vanilla, 2 for chocolate, 4 for strawberry, and 5 for butterscotch. Thus, the number of ways is: \[ \frac{14!}{3! \times 2! \times 4! \times 5!} \]

Step 4: Calculate Factorials

Calculate each factorial: - \(14! = 87,178,291,200\)- \(3! = 6\)- \(2! = 2\)- \(4! = 24\)- \(5! = 120\)

Step 5: Compute the Number of Permutations

Substitute the factorial values back into the formula: \[ \frac{87,178,291,200}{6 \times 2 \times 24 \times 120} \]Simplify the denominator: \(6 \times 2 \times 24 \times 120 = 34,560\)Finally, divide: \( \frac{87,178,291,200}{34,560} = 2,522,520 \) ways.

Key Concepts

Indistinguishable Objects

When we deal with problems involving distributing items among people or arranging items in a line, a common question arises: are the items distinguishable or indistinguishable? Indistinguishable objects mean that items of a similar type are identical and cannot be differentiated. For example, in the original exercise, the ice-cream cones of the same flavor are indistinguishable. This means that swapping two vanilla cones does not create a new arrangement since they are identical.

This is crucial because if all objects were distinguishable, each unique arrangement would count separately. However, with indistinguishable objects, similar items can be grouped without concern of internal swapping order. This characteristic impacts how we calculate permutations because the formula needs to account for these similarities. As a result, calculating the possible arrangements of indistinguishable objects requires special permutation techniques.

Permutation Formula

The permutation formula is essential when arranging groups of objects, particularly when some items are indistinguishable. The general formula for calculating distinguishable permutations is:

• \(\frac{n!}{n_1! \times n_2! \times ... \times n_k!}\)

where:

• \(n\) is the total number of objects.
• \(n_1, n_2, ..., n_k\) are the numbers of indistinguishable objects in their respective categories.

This formula works by first calculating the total number of permutations (\(n!\)) and then dividing by the factorials of the indistinguishable item counts. This division removes the overcounting caused by indistinguishable swaps, ensuring each unique permutation is counted only once.

For instance, in the exercise, the sequence of ice-cream cones is calculated using the permutation formula to account for the indistinguishability of cones of the same flavor.

Factorials

Factorials are mathematical functions essential in permutations and combinations. Represented by an exclamation mark (\(!\)), they express the product of an integer and all positive integers below it. For example, the factorial of 5 is:

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

Factorials grow rapidly with increasing numbers and are key to calculating permutations because they count all possible arrangements of a set number of items.

In permutation problems, factorials calculate both total permutations and reduce overcounting by considering indistinguishable groups. For example, \(14!\) calculates all possible arrangements of 14 cones, and division by \(3!\), \(2!\), \(4!\), \(5!\) accounts for similar cones. Understanding factorials allows you to break down and solve complex permutation challenges.

Permutations and Combinations

Permutations and combinations are foundational concepts in probability and stats, focusing on the arrangement and selection of items.

Permutations consider order, asking how many ways we can arrange a set of objects. For instance, arranging flavors of ice cream cones among children is a permutation. Combinations, however, ignore order and focus solely on selection—such as choosing a subset without regard to arrangement.

Understanding their distinction is vital for solving related mathematical problems.

• Permutations are calculated when arrangement matters.
• Combinations are used when only selection matters, with formula \(\binom{n}{r}\) ("n choose r").

In our context, the exercise focuses on permutations, as we distribute and arrange cones considering indistinguishability. Recognizing whether a permutation or combination problem helps apply the correct formula and logic.