I was solving this attached problem from one of the GRE mini exams. I understand the solution completely. However, i’m curious: if the problem stated that 3 of the digits were 7 instead of 2 (for example 777321), would we divide by 3 or 3 factorial throughout the process in order to get rid of the repeats?
**to be clear, not when solving for the total number of cases, but when trying to figure out how many cases with the restriction applied fit the problem.
I understand that the permutations with repeats formula is n!/r!. However, when encountering restrictions, the math does not always use factorials (for example, in the picture of my attached work in the 2nd row, we do 3 times 2 times1 times1 times 2, then divide by 2, but this is not a factorial)
My doubt comes from the fact that 2! = 2, so in terms of the hypothetical 3 7s, would we be dividing by 3 or 3! to rid of counting identical permutations more than once.
Thanks.
It’s essentially a counting problem and those “formulas” just work as tools in the grand scheme of things.
You want 1 to be to the right of both the 7s, so you just consider cases like:
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771XX
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XXX1X where both the 7s have to appear in the three X’s preceding 1
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XXXX1 where both the 7s have to appear in the four X’s preceding 1
Now doing the counting:
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2!
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you have 2 ways to fill out the last digit leaving you with 3 ways to make the numbers preceding 1.
For example, if you select 2 as your last digit then you can have 773, 377, and 737. Symmetrically, you have the same thing if we swap the 2s and 3s, thus leaving you with 6 possible ways.
With factorials, you would have counted it like 2 \cdot \frac{3!}{2!} to account for the permutation w repetition (like u mentioned)
- Same logic as 2), which then leads you to \frac{4!}{2!} = 12 ways.
The total number of ways to permute 5 objects with a repeat would be \frac{5!}{2!} = 60 ways.
Hence, you have \frac{2 + 6 + 12}{60} = \frac 13 as your final answer.
Essentially, your formulas are just tools to help you count. If you understand this, then you should now be able to extend this to your new problem.
As a bonus, there’s a super easy way to do this problem in which u can just ignore the presence of 3 and 2 and compute the probability that both the 7s are before the 1 in \{7,7,1\}. You then easily arrive at \frac 13.
