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(a) How many distinct permutations can be made from the letters of the word columns? (b) How many of these permutations start with the letter \(m ?\)

Short Answer

Expert verified
a) The total number of distinct permutations for the word 'columns' is 5040. b) The number of these permutations starting with the letter 'm' is 720.

Step by step solution

01

Understanding Permutations

Permutations refer to the arrangement of objects in a specific order. For a word, a permutation would imply an arrangement of its letters. The number of permutations of a word is given by the factorial of the number of letters. The factorial function (denoted by a !) is defined for an integer n, and denoted by n!. It is the product of all positive integers less than or equal to n.
02

Calculate Total Permutations for 'Columns'

The word 'columns' has 7 letters. So, the total number of ways we can arrange these 7 letters is 7! (7 factorial), which is \(7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 5040 \) possibilities.
03

Calculate Permutations starting with 'm'

For the permutations starting with 'm', 'm' occupies the first spot fixed, leaving 6 spots that can be filled by the remaining 6 letters. Therefore, the number of ways we can arrange the remaining 6 letters is 6!, which is \(6 \times 5 \times 4 \times 3 \times 2 \times 1 = 720\) possibilities.

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Key Concepts

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

Factorial
Understanding the factorial concept is essential when dealing with permutations. A **factorial**, represented by an exclamation point (!), is a mathematical function that multiplies a series of descending natural numbers. For any non-negative integer \( n \), \( n! \) equals the product \( n \times (n-1) \times (n-2) \dots \times 2 \times 1 \). This concept simplifies the process of calculating permutations, as it automatically accounts for all possible ways to arrange objects.

For example, if we have a word with 7 unique letters, such as "columns," the number of different permutations is given by \( 7! \), calculated as \( 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 5040 \). This calculation gives all possible arrangements of the letters, taking into account each possible position they can occupy.
Arrangements
Arrangements, or permutations, describe how objects can be ordered or arranged distinctively. This concept is crucial in various contexts, particularly when dealing with a set of items where order significantly matters.

For example, arranging the letters of the word "columns" involves permutations. To compute all possible distinct arrangements of its 7 letters, we use the factorial of 7, which is \( 7! \). This calculation captures each unique sequence that the letters can form.

When focusing on permutations that start with a specific letter, say "m," the problem simplifies slightly. Here, "m" is fixed in the first position, leaving six other letters to be arranged. The number of permutations starting with "m" is determined by arranging these 6 remaining letters, or \( 6! \). Thus, the number of such unique arrangements is \( 720 \).

These concepts help us understand how to tackle problems that involve arranging items in different sequences or orders.
Probability and Statistics
Though permutations primarily deal with arrangements, they connect deeply with probability and statistics. In probability, permutations help calculate the likelihood of different outcomes based on arrangements. Knowing how to calculate permutations becomes a foundational skill in understanding more complex probabilistic models.

An exercise like determining how many words can start with 'm' from "columns" incorporates basic probability principles. Such calculations assume each permutation is equally likely. Therefore, if you understand how many distinct permutations exist, it's straightforward to determine the probability of a specific arrangement occurring. If there are 5040 total permutations and 720 start with 'm,' the probability that a randomly chosen permutation starts with 'm' is \( \frac{720}{5040} \), simplifying to \( \frac{1}{7} \).

Introducing these concepts at an early stage can develop intuition for solving similar statistical problems which require an understanding of likelihood and frequency of specific outcomes.

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Most popular questions from this chapter

Pollution of the rivers in the United States has been a problem for many years. Consider the following events: \(A=(\) The river is polluted. \(\\}\) \(B=\\{A\) sample of water tested detects pollution. \(\\}\) \(C=\\{\) Fishing permitted. \(\\}\) Assume \(P(A)=0.3 . P(B \mid A)=0.75, P\left(B \mid A^{\prime}\right)=0.20,\) \(P(C \mid A \cap B)=0.20, P\left(C \mid A^{\prime} \cap B\right)=0.15, P\left(C \mid A \cap B^{\prime}\right)=\) \(0.80,\) and \(\mathrm{P}\left(\mathrm{C} \mid \mathrm{A}^{\prime} \mathrm{n} B^{\prime}\right)=0.90 .\) (a) Find \(P(A \cap B \cap C)\). (b) Find \(P\left(B^{\prime} \cap C\right)\). (c) Find \(P(C)\). (d) Find the probability that the river is polluted, given that fishing is permitted and the sample tested did not detect pollution.

The probability that Tom will be alive in 20 years is \(0.7,\) and the probability that Nancy will be alive in 20 years is 0.9 . If we assume independence for both. what is the probability that neither will be alive in 20 years?

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