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Chapter 5: Problem 54

A family has eight children. If this family has exactly three boys, how many different birth and gender orders are possible?

Short Answer

Expert verified
There are 56 different birth and gender orders possible.

Step by step solution

01

- Understand the Problem

We need to determine the number of different ways to arrange eight children where exactly three of them are boys and the other five are girls.
02

- Formulate the Combination Problem

To solve this problem, we use combinations. We need to choose 3 positions out of 8 for the boys. The rest will automatically be girls. The number of ways to choose 3 positions from 8 is given by the combination formula: \[ \binom{8}{3} = \frac{8!}{3!(8-3)!} \]
03

- Calculate the Combinations

Calculate \( \binom{8}{3} \): \[ \binom{8}{3} = \frac{8!}{3!5!} = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = 56 \]
04

- Interpret the Combination Result

The number 56 represents the number of different birth and gender orders where there are three boys and five girls.

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Key Concepts

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

Combinations
Combinations are a way to find the number of ways to choose a set number of items from a larger group, where the order does not matter.
In our exercise, we need to choose 3 boys out of a total of 8 children.
This is a classic example of a combination problem where we are not concerned with the order in which the boys are chosen, only that they are chosen.
The combination formula is: \( \binom{n}{k} = \frac{n!}{k!(n-k)!} \)
Here, \(n\) is the total number of items, and \(k\) is the number of items to choose.
In our problem, \(n = 8\) and \(k = 3\).
Factorials
Factorials are a key part of combinations and permutations.
They are denoted by an exclamation mark (!), and represent the product of all positive integers up to a given number.
For example, \( 5! = 5 \times 4 \times 3 \times 2 \times 1 \).
In our combination formula, we used factorials for calculating the number of ways to arrange 8 children where 3 are boys.
Specifically, we needed \( 8! \), \( 3! \), and \( 5! \).
This allowed us to plug into the formula and simplify to get our answer.
Binomial Coefficient
The binomial coefficient, often denoted as \( \binom{n}{k} \), represents the number of ways to choose \(k\) items from \(n\) items without regard to order.
It is used in various fields such as combinatorial analysis and probability.
In our problem, we used the binomial coefficient to determine how many ways we can select 3 boys from a group of 8 children.
The calculation looked like this: \[ \binom{8}{3} = \frac{8!}{3!(8-3)!} = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = 56 \]
The result, 56, tells us that there are 56 unique ways to arrange the birth and gender order of the 8 children.
Probability
Probability is the measure of the likelihood of an event to occur.
In the context of our exercise, while we didn't directly calculate a probability, understanding combinations helps in understanding probabilities.
For instance, if you wanted to find the probability of having exactly 3 boys in 8 children, you would need to consider the number of ways (combinations) to get 3 boys over the total possible arrangements.
Specifically,
if there are 56 favorable ways to have 3 boys, knowing the total number of arrangements can help in finding the probability of this specific arrangement.
Probability leverages combinatorial methods to count favorable outcomes and total outcomes.

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Most popular questions from this chapter

\(\text{True or False:}\) In a combination problem, order is not important.

If a person rolls a six-sided die and then flips a coin, describe the sample space of possible outcomes using 1,2,3 4,5,6 for the die outcomes and \(\mathrm{H}, \mathrm{T}\) for the coin outcomes.

Packaging Error Through a manufacturing error, three cans marked "regular soda" were accidentally filled with diet soda and placed into a 12 -pack. Suppose that three cans are randomly selected from the 12 -pack. (a) Determine the probability that exactly two contain diet soda. (b) Determine the probability that exactly one contains diet soda. (c) Determine the probability that all three contain diet soda.

Suppose you are dealt 5 cards from a standard 52 -card deck. Determine the probability of being dealt three of a kind (such as three aces or three kings) by answering the following questions: (a) How many ways can 5 cards be selected from a 52 card deck? (b) Each deck contains 4 twos, 4 threes, and so on. How many ways can three of the same card be selected from the deck? (c) The remaining 2 cards must be different from the 3 chosen and different from each other. For example, if we drew three kings, the 4 th card cannot be a king. After selecting the three of a kind, there are 12 different ranks of card remaining in the deck that can be chosen. If we have three kings, then we can choose twos, threes, and so on. Of the 12 ranks remaining, we choose 2 of them and then select one of the 4 cards in each of the two chosen ranks. How many ways can we select the remaining 2 cards? (d) Use the General Multiplication Rule to compute the probability of obtaining three of a kind. That is, what is the probability of selecting three of a kind and two cards that are not like?

(a) Roll a single die 50 times, recording the result of each roll of the die. Use the results to approximate the probability of rolling a three. (b) Roll a single die 100 times, recording the result of each roll of the die. Use the results to approximate the probability of rolling a three. (c) Compare the results of (a) and (b) to the classical probability of rolling a three.
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