How many five-digit numbers can be formed from the digits,2,3,4,5,6,7,8, and 9, if no digit may be repeated?

Question

Discret Math and its Application

Answer 1

So you have 8 digits and only 8 spaces to fit them in.  Since you cannot repeat any digit, once you choose a digit, you can't choose it again.  Think of the 8 numbers as the numbers on a basketball players jersey (do they wear jerseys? ... doesn't matter).  Now, you can only take 5 with you to the theater, but the order they sit in the rows makes a different.  So you have 8 options for the person right next to you.  Once he's chosen, for the next seat there are only 7 more choices left. And so on until the fifth seat, for which you will have only 4 possible choices. 

 

If you draw a tree diagram, you will see the first seat having 8 branches, then each of those will have seven branches, then each of the next one will have 6 branches and so on until the 5th seat having 4 branches each.  If you were to count all these final possibilities, you get the answer.  But that takes too long, so we use the concept of permutations.

 

Let's say you had a simpler problem.  There are only 5 numbers (or b-ball player) and you can choose to arrange only 3 of them in any order.  For the first number, you have 5 choices, the next one 4 and the last number 3 choices.  The total ways to arrange them without repeating a number would be 5 x 4 x 3 = 60 ways.   This would be the same as counting all the last set of branches for this smaller problem.  (This is also P(5,3) permutation of 5 things, 3 at a time.)

 

So what you need to do is ...