/wsr/ - Worksafe Requests » Thread #1456981

[5 / 3 / 1]

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Anonymous Fri 19 Apr 2024 20:02:02 No.1456981 View View Reply Original Report

Quoted By:

saw this math problem and couldnt figure it out:(im a fucking retard)

how many combinations of 12345 that make up single positive integers
are there, counting only the combinations where the 1 is left of 2

(like 12'345 and 15'234 and so on)

i guess you find out how many total combinations are there starting with 1, then take out all the combinations starting with 2 and then the hard part is calculating the rest? i cant find anything online desu

pic related im taking a massive shit

Anonymous

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Anonymous Fri 19 Apr 2024 21:06:27 No.1457008 Report

Quoted By: >>1457010

Since 2 can only appear to the right of 1, I simply took all of 1's positions and saw how many places 2 could be to the right of 1. The Xs represent the places that 2 cannot occupy and the checks represent the places it can

Anonymous

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Anonymous Fri 19 Apr 2024 21:18:08 No.1457010 Report

Quoted By: >>1457018

>>1457008
I think this logic is incorrect. For example, on the 3rd row, just because the 2 can be in the 4th position or the 5th position doesn't mean there's only two possibilities. There's more. For example, 54123, 45123, 34125, 43125, etc. There's two possibilities for the position of the 2, but more possibilities for the rearranging of other numbers you're not counting.

My immediate instinct is that the answer is 60. There's 120 total number combinations, and since 1 and 2 appear equally in all positions, 1 will be in front of 2 half the time, and 2 will be in front of 1 half the time. Therefore half of them will be invalid.

Anonymous

Anonymous Fri 19 Apr 2024 21:29:42 No.1457018 Report

Quoted By:

>>1457010
Yes, 5!/2 = 3*4*5

Anonymous

Anonymous Sat 20 Apr 2024 05:59:18 No.1457083 Report

Quoted By:

there are 10 different positions for 1 & 2 based on that rule (you can count them)
each one leaves 3 slots for the remaining 3 numbers in any order, so 3 x 2 x 1 = 6 possible arrangements for those

so 10 x 6 or 60 total arrangements

you can use the following python code to print out all of them and confirm there are 60:
[code:lit]
import itertools;
perms = [x for x in itertools.permutations([1,2,3,4,5]) if x.index(1) < x.index(2)]
print(f"There are {len(perms)} permutations:")
[print(x) for x in perms]
[/code:lit]

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