Problem 22 - Algebra and Trigonometry with Analytic Geometry

Question

Let the letters \(\mathrm{G}\) and \(\mathrm{B}\) denote a girl birth and a boy birth, respectively. For a family of three boys and three girls, one possible birth order is \(\mathrm{G} \mathrm{G} \mathrm{G} \mathrm{B} \mathrm{B} \mathrm{B.} \mathrm{How} \mathrm{many} \mathrm{}\) birth orders are possible for these six children?

Step-by-Step Solution

Answer

There are 20 possible birth orders for these six children.

Step 1: Understand the Problem

We need to determine the number of different ways to arrange three letters G (representing girls) and three letters B (representing boys) to create different birth orders.

Step 2: Identify the Formula

The problem can be solved by finding the number of unique permutations of the letters GGG and BBB. This is given by combinatorics as the formula for permutations of a multiset: \( \frac{n!}{n_1! \cdot n_2!} \), where \( n \) is the total number of children, and \( n_1 \) and \( n_2 \) are the frequencies of the different items.

Step 3: Apply the Formula

With 3 G's and 3 B's, we have: \( n = 6 \), \( n_1 = 3 \), and \( n_2 = 3 \). Substituting into the formula: \(\frac{6!}{3! \times 3!} = \frac{720}{6 \times 6} = \frac{720}{36} = 20. \)

Step 4: Solve the Equation

Calculate \( 6! = 720 \), \( 3! = 6 \), so the equation is \( \frac{720}{36} = 20 \). There are 20 unique ways to arrange the birth order for 3 girls and 3 boys.

Key Concepts

Combinatorics

Combinatorics is a branch of mathematics that deals with counting, arranging, and analyzing possible configurations in a given set. It is incredibly useful for solving problems related to permutations, combinations, and probabilistic calculations.

In the context of arranging births, combinatorics allows us to understand how different arrangements or sequences can be formed given a set of conditions. For example, when considering the birth order of children represented by letters G and B, combinatorics provides us with the tools to calculate the number of different orders in which these letters can be arranged.

The power of combinatorics lies in its general applicability across various scenarios, from simple problems like arranging letters to more complex situations involving large data sets. It uses principles such as the multiplication rule, permutations, and combinations to assess different possibilities. These concepts form the backbone of combinatorial calculations, enabling structured approaches to problem-solving in everyday applications.

Permutations of a Multiset

A multiset is a collection of items where repetition is allowed, and the order of these items holds significance in permutations. When calculating permutations of a multiset, we aim to discover how many distinct sequences can be generated by arranging these items.

For the exercise involving three boys and three girls, the multiset consists of the letters GGG and BBB. Since both G's and B's repeat, the typical permutation calculation of simply arranging 6 items (6!) overestimates the number of unique permutations. To adjust for this, the permutation formula for a multiset is used:

• Formula: \( \frac{n!}{n_1! \cdot n_2!} \)
• Where \( n \) is the total number of items, \( n_1 \) and \( n_2 \) are the frequencies of the repeated items.

Applying this concept, we can accurately determine distinct birth orders by accounting for the repetitions of each letter type, thus obtaining the correct number of permutations.

Factorial Calculations

Factorial calculations are essential in permutations and combinatorics. The factorial of a number, denoted as \( n! \), is the product of all positive integers up to that number. For instance, \( 6! = 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 720 \).

In the context of our exercise, factorial calculations help determine the total possible arrangements of the letters G and B before accounting for repetitions. We calculated \( 6! \) to find the total number of ways to arrange these six items.

However, since some letters are repeated, we must divide by the factorial of the frequency of each repeated letter, which, in this problem, are \( 3! \) for the girls and \( 3! \) for the boys. This results in dividing \( 720 \) by \( 3! \times 3! = 6 \times 6 = 36 \), yielding the result of 20 different possible birth orders.

Understanding factorials is crucial because they form the base of more complex calculations in permutations and combinations, making them fundamental to studying and applying combinatorial methods.