Solution To: I Can't Count That High...I Ran Out Of Fingers And Toes!

Here was the original problem.

The four digits 5, 6, 7, and 8 are placed on separate cards. How many different numbers can be formed using one or more of the cards?



Here is the solution:

The answer is 64.  Now, like I said in the original post, you could have listed out all possible numbers...but that would have taken a lot of time.  Here is the mathematical way to approach the problem:

The problem states that we are to use "one or more" of the cards.  This means we can have 4 digit numbers, 3 digit numbers, 2 digit numbers, and 1 digit numbers.

For the 4 digit numbers:  If you were to make a 4 digit number at random using the given cards, how many choices do you have for the first number in the 4 digit number?  You have 4.  For the second number of your 4 digit number, you have 3 choices (since you can't repeat numbers).  For the third you have 2 choices and for the fourth you have 1 choice.  To find out how many options that gives you for four digit numbers (if you remember some math skills) you would multiply 4x3x2x1 = 24.

(For instance,  this is my blank number _ _ _ _.  For the first spot I have 4 choices.  Let's say I pick 5 for the first spot.  I now have 5 _ _ _.  For the second spot, I now only have 3 choices (6, 7, 8).  Let's say I pick 7.  I now have 5 7 _ _.  Now for the third spot I have 2 choices (and so on...).)

For the 3 digit numbers:  For the first number of your three digit number, you have 4 choices, for the second you have 3 and for the third you have 2.  So you multiply 4x3x2=24

For the 2 digit numbers:  You have 4 choices for the first number and 3 for the second number. So you have 4x3 = 12.

For the 1 digit numbers:  You have 4 choices for the first (and only) number in your 1 digit number.

Since you must take into account all possibilities, you add all you possible outcomes. 

                                                                       24 + 24 +12 +4 = 64

Tada!

Sincerely,
The Math Freak