0.10 Sets and counting  (Page 5/9)

In the word ARTICLE, there are 4 consonants.

Since the first letter must be a consonant, we have four choices for the first position, and once we use up a consonant, there are only three consonants left for the last spot. We show as follows:

Since there are no more restrictions, we can go ahead and make the choices for the rest of the positions.

So far we have used up 2 letters, therefore, five remain. So for the next position there are five choices, for the position after that there are four choices, and so on. We get

So the total permutations are $4\cdot 5\cdot 4\cdot 3\cdot 2\cdot 1\cdot 3=\text{1440}$ .

The problem is easily solved by the multiplication axiom, and answers are as follows:

1. a. The number of four-letter word sequences is $5\cdot 4\cdot 3\cdot 2=\text{120}$ .

2. b. The number of three-letter word sequences is $5\cdot 4\cdot 3=\text{60}$ .

3. c. The number of two-letter word sequences is $5\cdot 4=\text{20}$ .

We will identify $n$ and $r$ in each case and solve using the formulas provided.

1. $\mathrm{6P3}=6\cdot 5\cdot 4=\text{120}$ , alternately $\mathrm{6P3}=\frac{6!}{\left(6-3\right)!}=\frac{6!}{3!}=\frac{6\cdot 5\cdot 4\cdot 3\cdot 2\cdot 1}{3\cdot 2\cdot 1}=\text{120}$

2. $\mathrm{7P2}=7\cdot 6=\text{42}$ , or $\mathrm{7P2}=\frac{7!}{5!}=\frac{7\cdot 6\cdot 5\cdot 4\cdot 3\cdot 2\cdot 1}{5\cdot 4\cdot 3\cdot 2\cdot 1}=\text{42}$

Let us suppose we have four people $A$ , $B$ , $C$ , and $D$ . Further suppose that $A$ and $B$ want to sit together. For the sake of argument, we tie $A$ and $B$ together and treat them as one person.

The four people are $\begin{array}{|c|}\hline \text{AB}\\ \hline\end{array}$ $\begin{array}{c}\text{CD}\end{array}$ . Since $\begin{array}{|c|}\hline \text{AB}\\ \hline\end{array}$ is treated as one person, we have the following possible arrangements.

Note that there are six more such permutations because $A$ and $B$ could also be tied in the order $\text{BA}$ . And they are

So altogether there are 12 different permutations.

Let us now do the problem using the multiplication axiom.

After we tie two of the people together and treat them as one person, we can say we have only three people. The multiplication axiom tells us that three people can be seated in $3!$ ways. Since two people can be tied together $2!$ ways, there are $3!2!=\text{12}$ different arrangements.