The Fundamental Counting Principle



~If event A has n ways of occurring and a second event B m ways of occurring, then the product (n)(m) will be the number of ways both could occur in succession. A very basic multiplication rule that is very useful in solving many problems related to probability.

~Example 1: Consider the digits, 1,2,3,4,5,6. How many 3-digit numbers can be made using a digit once and only once?

~Solution: Use a model like, _ _ _, where the far left spot is the hundreds place, then the next spot, the tens, and the far right spot, the ones place. Since there is a restriction that the digits can only be used once, we must keep this in mind when solving this problem. (usually, the phase, “no repetitions allowed” is stated in the wording of the problem).

Now, we must consider the number of possibilities for choosing a digit for each spot. Since there are no restrictions on location, we can start at any spot. Starting from the left, there will be 6 choices, then only 5 for the next spot, and 4 for the remaining spot. By the fundamental principle, our answer will be the product, (6)(5)(4) = 120 numbers can be formed.

~Note: Many problems have either natural or stated restrictions, so, one must go to those positions first and fill them in before the rest. The following is an example.

~Example 2: (stated restriction). Consider the digits, 1,2,3,4,5,6. Let’s ask the same questions as in example 1 above but must have a 5 in the ones place.

~Solution: Since the restriction that a 5 must be in the ones place, we fill that spot first. Since there is only one 5, there is only one choice for that spot, the 5. Now, go to the other spots and find the possibilities. Since no repetition is allowed, there will be 5 choices for our next spot, and 4 for the last spot. Again, using the fundamental principle, we have, (5)(4)(1) = 20 numbers that can be formed.

~Example 3: (two stated restrictions. Take the digits, 1,2,3,4,5,6. How many even 4-digit numbers can be made starting with 2?

~Solution: There are two restrictions (other than no repetition of digits) in this problem. The first is, starting with a 2 and the second would be that the 4-digit number be even. Since a 2 must start and is an even digit, it is eliminated for a choice for the end spot. That would only leave two choices for the end, a 4 or 6, since these numbers must end in an even digit to be an even number. So, go to these spots first and figure the possibilities, before the two spots in the middle. Remember, no repetitions allowed. So, multiplying our choices, we get, (1)(4)(3)(2) = 24 . Keep in mind, these numbers are possible ways not the digits given in the problem.

~Note: Many problems allow you to repeat the digits. However, be aware, there could be restrictions as well.

~Example 4: Let’s do these examples with repetition allowed.

~Solution: For example 1, the 6 digits may be used for each of the 3 spots. We get, (6)(6)(6) = 216 numbers.

For example 2, since our numbers must end in a 5, there is still only one choice for that spot, however, there will be 6 choices for each of the remaining spots. We get, (6)(6)(1) = 36 numbers.

For example 3, there is still only one choice for the first spot, however, there will be 3 choices for the end spot since the 2 can be repeated and is an even digit. There are no restrictions on the middle spots, so there will be 6 choices for each. We get, (1)(6)(6)(3) = 108 numbers.