# Permutations - Problem 3

An applied case of permutations is looking at an ATM pin. But the problem we're looking at is basically an ATM pin dealing with four digits and we're trying to figure out the number of pin numbers if repetition is allowed or if repetition is not allowed.

So whenever I'm dealing with a problem like this I try to draw out my spots to give me a visual on what's going on and before we started this we actually had a debate as to if our pin number can start with 0, nobody actually knows which tells you that most people don't start it with 0, but we're going to assume that we can to give use the idea that this is a little bit more uniform than otherwise maybe.

So basically first spot, we said we can throw in 0, we're not sure if we can so we can throw in 0, 1, 2, 3, 4, 5 all the way up to 9 which gives us 10 digits so we can have 10 things going in our spot. Because repetition is allowed, there is still 10 things that can go in second 10 things, 10 in the third, 10 in the fourth giving us 10 to the fourth which is the same thing as 10,000.

What if repetition is not allowed? I still like to draw out my little slots and then basically fill in the blacks. So our first term could be anything 0 through 9 giving us 10 things. Our second, we already chose one number and when can't reuse that number again so this is just going to be 9 times 8 times 7 which given permutations is just going to be 10 permute 4. 10 permute there's 10 things available and the permutation is basically saying the 4 numbers that we're looking for.

So difference between repetition and non-repetition when we are dealing with repetition there in no really permutation involved in terms of our formula, there is just something to a power. When repetition is not allowed, then we're dealing with a permutation just a number of things that are available to us the number of digits 0 to 9 permute a number that we actually want in this case 4. Just a slight difference between our repetition and our non-repetition.

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