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?
1 set
125
120 when you want to know something like that you multiply the digits together Ex. 12345-1 times 2 times 3 times 4 times 5=120
The first digit can have 5 possible numbers, the second digit can have 4, the third 3, the fourth 2. 5
Since the order of the digits does not matter there are only five combinations: 1234, 1235, 1245, 1345 and 2345.
1 set
It is 120 if the digits cannot be repeated.
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If repetition of digits isn't allowed, then no13-digit sequencescan be formed from only 5 digits.
Since there are only five different digits, a 6-digit number can only be generated if a digit can be repeated. If digits can be repeated, the smallest 6-digit number is 111111.
Their order is switched. For example, in the two numbers 12345 and 13245, the second and third digits are transposed.
125
There are 625 of them - too many to list.
If the two must be different digits, then there 20 possibilities. If they can be the same one, then there are 25 possibilities.
It has five digits each of them representing numerical quantities
120 when you want to know something like that you multiply the digits together Ex. 12345-1 times 2 times 3 times 4 times 5=120
They are: 12343+2 = 12345