Welcome to the world of permutations. We have a pool of 5 numbers, and we are going to pull 3 of them at a time to make a number. Oh, and we put the numbers back after each use because it was specified as a replacement problem. How do we solve this? Let's build a tree to answer it. We are going to be building sets of 3 numbers. Start with 1 and build. 111, 112, 113, 114, 115. 121, 122, 123, 124, 125. 131, 132, 133, 134, 135. 141, 142, 143, 144, 145. 151, 152, 153, 154, 155. That's our first tree. See how it works? Starting with 1, we built looking at each possibility in turn from the right. We made our changes starting at the right and moving to the left, back to the place where our 1 was. See that? We now have 5 + 5 + 5 + 5 + 5 combinations beginning with 1. That's 25 combinations of 3 digits from a replacement set of 5 digits beginning with 1. If we can get 25 combinations of 3 digits from a set of 5 numbers beginning with 1, then how many combinations of 3 digits can we get from the 5 numbers beginning with 2? Beginning with 3? With 4? With 5? We'd get 25 combinations from each starting number. If we add the possibilities from each number, we'd get 25 + 25 + 25 + 25 + 25 = 125. But don't bounce just yet. Look at it this way. We're building a set of 3 numbers from a base of 5 numbers with replacement. We have 5 different choices for our first number. See that? We can pick any of the 5 numbers to begin the number we're making. We also have 5 different choices for our second. And 5 for our third. We have 5 x 5 x 5 possibilities for building 3 numbers from a base of 5 numbers with replacement. And 5 x 5 x 5 = 125. Our answer to the question asked is 125. One last thing. If we were to make a formula for finding the number of combinations (call that P) from a base set of numbers (call that n) using replacement and taken in groups of a certain number (call that r), our formula would be: P = n to the power of r, or P = nr Are we good?
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