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[Statistics] I'm so incredibly confused by combinations and permutations

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Lots of feedback on the 56P2.  I always want students to circle back to probability tree diagrams when in doubt.  Like you said, the first number has 5 choices, and for each of them, there is 4 choices for the last number.  

Neat thing we can learn from probability trees is, the order of the branches don’t affect the final answer, so we can “move” the last digit up to the second stage.  Now we have the twenty branches.  And off of each of them, we have 8 more branches, then 7 more.  How many branches will we have at the end?
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You were right about having 8 and 7 options for the middle two spots. Why are you doing 56P2? 56 represents that number of ways to pick those two digits that meet the criteria. There are 56 options for the middle two digits, which is less than 100.

56P2 would give us the number of ways to pick 2 two-digit numbers to be the middle digits in separate cases (ie 74 and 38 as one of 3080 possibilities).
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You're also not really answering the question.  You just want the probability that the first digit is 1.  The other digits don't matter.  The way you're counting, you're picking the first digit first, and the other numbers you're multiplying by don't depend on *which* digit you picked, so you have the same number of numbers starting with 1, with 3, etc.
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> 56 options right?

Right.

> If I calculate 56P2 I get 3080 which seems way wrong to me.

Yup, why calculate 56P2? 56 is the number you wanted, it accounts for the multiple orderings.

WLOG assume 1 and 3 are our first and last digits chosen, leaving only 0, 2, 4, 5, 6, 7, 8, and 9 to choose the last two from. We have 8 choices of digit, and if repetition were allowed we'd have 8^(2) = 64 possible middle two digits. However we aren't allowing repetition, so of course we instead have 8 choices for the first of the middle two digits, then 7 for the second of the two, making 8 * 7 = 56 possible middle two digits without repetition.

Happy to see you checking this with some common sense upper bounds to notice this cannot be the case.
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20 percent or 0.2

Use Probability concept. Nothing special about the digit "1", all other odd digits have the same permutations.
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Since they allow all digits from 0-9 even in the first position, then the total number of unique numbers is, as you mentioned: 5×4×8×7 where 5,4 are the permutation choices for the first and last digits, and 8,7 for the middle two. We multiply because the order of digits matter.

To find the probability: we need the number of numbers with a leading 1 digit. If you think about it that's pretty much the same expression above just holding the first digit constant: 1×4×8×7.

Dividing this by the above, you get the probability of observing a number with a leading 1 = 20%.
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The naive answer is 1/5. Given all possible combinations that fit the requirements, you may expect that since the first digit must be odd then there is a 1 in 5 chance that it is 1. The question you have to ask yourself is if there is anything about the requirments where the first digit is not evenly distributed across all odd numbers. I don't see anything that would lead me to believe that so I'd answer 1/5.
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Does this help you see an answer?

An employee is accused of stealing, because someone gained entry to a warehouse by typing a four-digit entry code and unlocking an exterior door.  The suspicion is that the theft must have been an inside job, because otherwise how would the thieves found a working entry code?  An interview with the software engineer who created the security system revealed the following:

* All 10 digits (from 0 to 9) could be entered, except that the system would only accept odd numbers for the first and last digits of the code.  In those positions of the code, if someone typed an even number, the system would wait for them to type an odd number.
* The system would not accept any digit that had already been typed earlier when entering the code.  If someone retyped a digit, that retyping was ignored and the system waited for them to type a new digit.
* Any code that started with a 1 was accepted and unlocked the door.

With that software, what was the probability that someone trying random combinations of digits would gain entry on their first try?  (Assume they were truly random in their selection of digits.)  Does the fact that the thieves got past this security system justify concluding that this was an inside job?

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