Question

How many six-letter arrangements can you make using all of the letters \(\mathrm{A}, \mathrm{B}, \mathrm{C}, \mathrm{D}, \mathrm{E}\) and \(\mathrm{F},\) without repetition? Of these, how many begin and end with a consonant?

Short answer

Total arrangements: 720. Arrangements beginning and ending with a consonant: 288.

Step-by-step solution

Calculate total arrangements

To find the total number of six-letter arrangements using all the letters \(A, B, C, D, E, F\), we use the formula for permutations since there is no repetition. The formula for permutations of n distinct objects is n!. Here, we have 6 letters. So, the total number of arrangements is \(\frac{6!}{(6-6)!} = 6! = 720\)

Identify consonants and vowels

The letters provided are \(A, B, C, D, E, F\). Identify which of these are consonants and which are vowels. Consonants: \(B, C, D, F\). Vowels: \(A, E\)

Calculate arrangements that begin and end with a consonant

First, calculate the number of ways to choose the first and last letters from the consonants. There are 4 choices for the first letter and 3 choices for the last letter after choosing the first. This gives us 4 * 3 = 12 combinations. The remaining 4 positions (middle letters) can be filled with the remaining 4 letters (2 vowels and 2 consonants). The number of ways to arrange 4 letters is 4!. Therefore, the total number of arrangements is 12 * 4! = 12 * 24 = 288.

Key concepts

Factorials

To understand permutations, we need to first grasp the idea of factorials. Factorials are mathematical expressions generally denoted by an exclamation mark (!). The factorial of a positive integer n is the product of all positive integers less than or equal to n. For example, \(5! = 5 \times 4 \times 3 \times 2 \times 1 = 120\). Factorials help in determining the number of ways we can arrange a set of objects. In permutations, factorials are used to calculate the total number of arrangements possible for a given set.

Combinatorics

Combinatorics is a branch of mathematics focused on counting, arranging, and combining objects. It includes topics like permutations and combinations. In combinatorics, we often deal with questions like 'How many ways can we arrange these objects?' or 'How many ways can we select objects from a group?'. When solving our problem, we use combinatorics to calculate the different ways letters can be arranged in a specific order. This involves understanding the formulas and principles behind permutations and combinations.

Permutations Without Repetition

In our exercise, we are specifically dealing with permutations without repetition. This means we arrange letters where none of them repeats. To formulate this, if we have n distinct objects, the number of permutations is given by \(n!\). For example, with 6 unique letters (A, B, C, D, E, F), we first compute factorial of 6 as \(6! = 720\), which provides the total arrangements of the letters. In permutations without repetition, each arrangement is unique because each letter or object is used only once.

Arrangement of Letters

Let's break down how we arrange letters using permutations. Given our letters (A, B, C, D, E, F), we computed the total permutations as \(6! = 720\). Next, for arrangements with specific conditions, like beginning and ending with consonants, we narrow our choices. Identify that consonants are B, C, D, and F. To start and end with a consonant, we have 4 options for the first letter and 3 for the last. Therefore, that's 4 * 3 = 12 initial combinations. The middle positions are filled with the 4 remaining letters, calculated as \(4! = 24\). Thus, the specific arrangements are 12 * 24 = 288, giving the requested conditions.