|
||||||||||||
|
||||||||||||
| ||||||||||||
Hi, You have the four digits 4, 2, 3 and 0 and you want to arrange them in all possible orders. These are called the permutations of the digits 4, 2, 3 and 0. There are not very many possibilities so you can write them down. There are four possibilities for the first digit, 4, 2, 3 and 0. No matter which digit you chose for the first digit there are three remaining digits for the second digit. Thus your choices so far are
There are $4 \times 3 = 12$ of them. For each of the twelve, two digit numbers above how many choices are there for the third digit. Write them all down. How many are there? What about the last digit? I hope this helps, | ||||||||||||
|
||||||||||||
Math Central is supported by the University of Regina and The Pacific Institute for the Mathematical Sciences. |