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?
