Permutation and Combination Help Kinda Desperate (I have my test at 1) (1 Viewer)

cristal

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So I've been doing some practice questions but this particular question got me stuck from my practice (ai won't co-operate) and I've tried my best but I still haven't gotten the answer right
Question: Bob is about to hang his 8 shirts in the wardrobe. He has four different styles of shirts and 2 shirts for each style. How many different arrangements in can Bob make (to hang them in his wardrobe) if no two identical shirts are together/ next to each other?
 

cheesynooby

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@cheesynooby am i tripping or is this got to do with derangement


also do u have the answer so i dont look like noob when i get it wrong :) @cristal
i dont think its a derangement but u should just be able to find the complement
8!/2!2!2!2! total combinations ignoring restriction
satisfying restriction: 8*1*6!*7/2!2!2!2! (8 possible shirts, 1 choice for its matching pair, other shirts do not matter, 7 possible spots for the matching pairs) but overcounting for if there are other pairs (so uhhh yeah its incomplete)
ok this method is probably not very good because it seems like a recursive thing or something with cases idk
 

alphxreturns

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i dont think its a derangement but u should just be able to find the complement
8!/2!2!2!2! total combinations ignoring restriction
satisfying restriction: 8*1*6!*7/2!2!2!2! (8 possible shirts, 1 choice for its matching pair, other shirts do not matter, 7 possible spots for the matching pairs) but overcounting for if there are other pairs (so uhhh yeah its incomplete)
ok this method is probably not very good because it seems like a recursive thing or something with cases idk
yeah take a look at this @cristal @cheesynooby

i think the approach with A,B,_,C,_,D,_ where "_" can be 0 and any letter(s) but the restricted ones works the most logically (but theres 7 cases)

doubt this will be asked, though

I tried to do this with Ai and I get this:
View attachment 47125

is this ai fking stupid
i tried AI too, it did the exact same thing. cannot trust
 

C2H6O

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Ik im too late here but would you do total - combinations with shirts next to each other? So 8!/2!2!2!2! - 4! (all 4 sets of shirts together) - 4C1*5!/2! (choose 1 set to separate, 3 sets of shirts together while 1 set separate, cancelling that the separate set can be swapped) - 4C2*6!/2!2! - 4C3*7!/2!2!2!

idk this gives the wrong answer guys any ideas?
 

jane1820

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Are deragment in the syllabus? My teacher refuses to teach them n im scared shes gonna put them in the test (math facult are bitches they made everyone in yr 12 ext 1 drop class of 25 - this is the first time my school doesnt have a ext 1/2 class)
 

cheesynooby

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Are deragment in the syllabus? My teacher refuses to teach them n im scared shes gonna put them in the test (math facult are bitches they made everyone in yr 12 ext 1 drop class of 25 - this is the first time my school doesnt have a ext 1/2 class)
i dont think so but they were in some of my school's past papers a couple years back so either they were in the old syllabus or 💀
 

pl4smaa21

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Are deragment in the syllabus? My teacher refuses to teach them n im scared shes gonna put them in the test (math facult are bitches they made everyone in yr 12 ext 1 drop class of 25 - this is the first time my school doesnt have a ext 1/2 class)
No I don't think so it doesn't appear here
1743500821839.png
 

pl4smaa21

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i dont think so but they were in some of my school's past papers a couple years back so either they were in the old syllabus or 💀
but they do appear in enrichment questions in the math ext 1 Cambridge textbooks soooo maybe schools try to "over-prepare" students by making them do stuff beyond what's expected so they have greater ease in what is expected --> pretty gud marks

I actually saw this somewhere idk if this was legit but it was supposedly from a Dr Du 4 unit trial. I looked near the end and I saw a question related to epsilon-delta proofs, WHICH IS INSANE. Like srsly epsilon delta proofs are apart of real analysis in which you literally PROVE calculus lmao
here's an image:
1743501015928.png

it doesn't even have numbers 😭
 

pl4smaa21

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Ik im too late here but would you do total - combinations with shirts next to each other? So 8!/2!2!2!2! - 4! (all 4 sets of shirts together) - 4C1*5!/2! (choose 1 set to separate, 3 sets of shirts together while 1 set separate, cancelling that the separate set can be swapped) - 4C2*6!/2!2! - 4C3*7!/2!2!2!

idk this gives the wrong answer guys any ideas?
I tried to do it but kinda gave up on it and this was the idea I had:
Calculate number of unrestricted selections = 8!/((2!)^4)
2. Determine number of invalid cases, by cases
CASE 1: 4 violations
- By 4 violations I mean that all four shirt styles are adjacent
- Symbolically imagine if the 4 styles are A , B , C and D
- We'd have A A B B C C D D shirts
- One permutation with 4 violations would be AACCDDBB, another would be BBAADDCC
- The way to calculate for four violations would be to treat AA,BB,CC,DD as individual grouped objects and permutate = 4! = 24
- So we have 24 ways that 4 violations would lead to inapplicability

Case 2 : 3 violations (a bit more tricky)
- Now the groupings have 4C1 ways of occurring as we have to choose one of styles A,B,C or D to be the separated group
- Example case (D separate)
- Our objects would be AA,BB,CC,D , D
- So the number of unrestricted permuations would be 5!
- However, this would include instances where we have D and D together. To get rid of this we look at case 1 of four violations because they coincide with The case that the D's are adjacent in this particular ordering.
- So the number of ways we can have 3 violations with D being the only separate = 5! - 24 = 96
- However, this case has symmetry because we can swap the D's with A's B's or C's in each of the 96 cases which would lead to:
No. of arrangements with exactly three violations = 4 * (5!-24) = 4*96 = 384

Case3 : 2 violations (even more tricky)
- Now our choices on what stays grouped has become 4C2 = 6
- Example Case (A and B violate)
- Objects: AA, BB , C , C , D ,D
- We can arrange these 6 objects in 6! ways without restrictions
- However, in this counting we include cases where:

There are 4 violations
There are 3 Violations including C but not D
There are 3 violations including D but not C
- To figure what to subtract use case 1 and case 2
- For 4 violations we know there are 24 of them
- For 3 violations including C we know there are 5!-24 (due to case 2)
- Similarly for 3 violations including D we know there are 5!-24
- So we subtract these from 6! :

6!-(24+(5!-24)+(5!-24)) = 504

but this is only for if styles A and B are violators in total we have 4C2 = 6 possibilities for what violates
so 6*504 = 3024

Case 4(1violation)
We have 4C1 choices on what violates
Case : A violates
objects: AA,B,B,C,C,D,D
Total permutations without restriction = 7!
subtract:
4 groups together= 24
3 groups together = 384 (case 2)
2 groups together = 3024

7!-(24+384+3024)=1608

total ways with all identicals NEVER adjacent = unrestricted - invalid = 8!/((2!)^4) - (24+384+3024+1608) = -2520 (obviously wrong)

I wrote all this in hopes someone can explain what is wrong in my working because I think I made a mistake somewhere but it feels like this idea could potentially work. Can someone tell me how to do this way properly if possible. please?
 

cheesynooby

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I tried to do it but kinda gave up on it and this was the idea I had:
Calculate number of unrestricted selections = 8!/((2!)^4)
2. Determine number of invalid cases, by cases
CASE 1: 4 violations
- By 4 violations I mean that all four shirt styles are adjacent
- Symbolically imagine if the 4 styles are A , B , C and D
- We'd have A A B B C C D D shirts
- One permutation with 4 violations would be AACCDDBB, another would be BBAADDCC
- The way to calculate for four violations would be to treat AA,BB,CC,DD as individual grouped objects and permutate = 4! = 24
- So we have 24 ways that 4 violations would lead to inapplicability

Case 2 : 3 violations (a bit more tricky)
- Now the groupings have 4C1 ways of occurring as we have to choose one of styles A,B,C or D to be the separated group
- Example case (D separate)
- Our objects would be AA,BB,CC,D , D
- So the number of unrestricted permuations would be 5!
- However, this would include instances where we have D and D together. To get rid of this we look at case 1 of four violations because they coincide with The case that the D's are adjacent in this particular ordering.
- So the number of ways we can have 3 violations with D being the only separate = 5! - 24 = 96
- However, this case has symmetry because we can swap the D's with A's B's or C's in each of the 96 cases which would lead to:
No. of arrangements with exactly three violations = 4 * (5!-24) = 4*96 = 384

Case3 : 2 violations (even more tricky)
- Now our choices on what stays grouped has become 4C2 = 6
- Example Case (A and B violate)
- Objects: AA, BB , C , C , D ,D
- We can arrange these 6 objects in 6! ways without restrictions
- However, in this counting we include cases where:

There are 4 violations
There are 3 Violations including C but not D
There are 3 violations including D but not C
- To figure what to subtract use case 1 and case 2
- For 4 violations we know there are 24 of them
- For 3 violations including C we know there are 5!-24 (due to case 2)
- Similarly for 3 violations including D we know there are 5!-24
- So we subtract these from 6! :

6!-(24+(5!-24)+(5!-24)) = 504

but this is only for if styles A and B are violators in total we have 4C2 = 6 possibilities for what violates
so 6*504 = 3024

Case 4(1violation)
We have 4C1 choices on what violates
Case : A violates
objects: AA,B,B,C,C,D,D
Total permutations without restriction = 7!
subtract:
4 groups together= 24
3 groups together = 384 (case 2)
2 groups together = 3024

7!-(24+384+3024)=1608

total ways with all identicals NEVER adjacent = unrestricted - invalid = 8!/((2!)^4) - (24+384+3024+1608) = -2520 (obviously wrong)

I wrote all this in hopes someone can explain what is wrong in my working because I think I made a mistake somewhere but it feels like this idea could potentially work. Can someone tell me how to do this way properly if possible. please?
lol i also kept getting -2520 when i was doing it (i love breaking the laws of the universe)
 

cristal

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I'm pretty sure I got like -1320 or smthing yesterday/today
thankfully there wasn't anything like this in the test (but I think I lost 5% bc I potentially wrote 5! instead of 4!)
 

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