0.10 Sets and counting  (Page 4/9)

Since there are 26 letters and 10 digits, we have the following choices for each.

Therefore, the number of possible license plates is $26\cdot 10\cdot 10\cdot 10\cdot 10=\mathrm{260,000}$ .

Since there are two choices for each question, we have

Applying the multiplication axiom, we get $2\cdot 2\cdot 2=8$ different ways.

We list all eight possibilities below.

TTT, TTF, TFT, TFF, FTT, FTF, FFT, FFF

The reader should note that the first letter in each possibility is the answer corresponding to the first question, the second letter corresponds to the answer to the second question and so on. For example, TFF, says that the answer to the first question is given as true, and the answers to the second and third questions false.

Suppose we put four chairs in a row, and proceed to put four people in these seats.

There are four choices for the first chair we choose. Once a person sits down in that chair, there are only three choices for the second chair, and so on. We list as shown below.

So there are altogether $4\cdot 3\cdot 2\cdot 1=\text{24}$ different ways.

The problem is very similar to [link] .

Imagine a child having three building blocks labeled $A$ , $B$ , and $C$ . Suppose he puts these blocks on top of each other to make word sequences. For the first letter he has three choices, namely $A$ , $B$ , or $C$ . Let us suppose he chooses the first letter to be a $B$ , then for the second block which must go on top of the first, he has only two choices: $A$ or $C$ . And for the last letter he has only one choice. We list the choices below.

Therefore, 6 different word sequences can be formed.

Finally, we'd like to illustrate this with a tree diagram.

All six possibilities are displayed in the tree diagram.

There are four choices for the first letter of our word, three choices for the second letter, and two choices for the third.

Applying the multiplication axiom, we get $4\cdot 3\cdot 2=\text{24}$ different arrangements.