Permutations and Combinations 3U Cambridge. (2 Viewers)

[trailblazer]

"Bob is about to hang his eight shirts in the wardrobe. He has four different styles of shirt, two identical ones of each particular style. How many different arrangements are possible if no two identical shirts are next to one another?"

 

[lychnobity]

8!/(2! x 4)

 

[gurmies]

Ahh, logic/reasoning please? Sorry, i'm learning here too xD

 

[khorne]

So what the poster is saying is that:

8! ways to make an arrangement

8!/ (4x2!) (as there are 4 sets of 2 repeats)

HOWEVER

this does not solve the problem...it is more complex than that I think. You could try an eliminate all the possibilities of them being together, but this would take too long...

Btw, it is under the topic of cases, so I assume they want us to make some cases up

 

[azureus88]

This question is hard...something got to do with the inclusion-exclusion principle thing, which i dont really get.

 

[gurmies]

Yeah, I thought he was incorrect. Events are mutually exclusive, i.e. if you wish to approach it like 8!/2^4 - combinations with ANY together, you'll have to use cases (e.g. All 4 groups together, only 1, etc...)

 

[khorne]

Well, Cambridge are notorious for difficult problems in their extension section, so don't be suprised...

 

[lychnobity]

8!/(2! x 4)

I realise I can edit, but most people don't read those. But anyway, I realised I did the question wrong, so sorry guys.

So attempt no. 2:

4! x 4C1 x 3 x 3C1 x 4 x 2C1 x 5 x 1C1 x 6 = 207 360

4! - Say the styles were P, Q, R, S. I took one shirt of each style and arranged them, this can be done in 4! ways.

4C1 - out of the remaining shirts, I'm choosing one of them ie there will be one left of P, Q, R, S. Each time I use the nCr, I'm choosing one of the remaining shirts.

3 - There are 3 spots for this shirt to be placed. _P_Q_R_S_ Say the shirt chosen (above) is P. P can't be in 2 of the 5 spots (the _ means a spot). Therefore, P can only be in 3 spots. The logic follows that every time a shirt is placed on the rack, the no. spots available is = (no. shirts on rack + 1) - 2

3C1 - choosing one of the 3 remaining shirts

4 - _P_Q_R_P_S_ . Only Q, R, S are left. Say Q is chosen, it can't be in the spots marked in red. Same applies to R & S, there are 2 spots on the rack that all shirts can't be in.

2C1 - choosing either R or S. Say R was chosen.

5 - _Q_P_Q_R_P_S_ . R can't be in 2 of the spots, marked in red. There are 5 spots left.

1C1 - One shirt left. ie S

6 - _Q_P_R_Q_R_P_S_. S can't be in the red spots. Therefore, 6 spots to be placed in.

Someone check if I'm right. I spent ages typing this up.

Actually, answer seems too big. FML if I'm wrong.

 

[Timothy.Siu]

total arrangments is 2520 with no restrictions anyway so thats wrong

 

[gurmies]

total arrangments is 2520 with no restrictions anyway so thats wrong

Correct, 8!/2^4 is the number of total arrangements...

 

[lychnobity]

FML :mad1:

 

[khorne]

Considering how this hasn't been solved, and no one reads edits: I wanted to give an attempt:

This isn't the full one, because there are some other factors I need to consider, but assuming we had 4 Shirts, 2 are the same A and 2 are the same, B.

We have a total of 4!/2x2 = 6 permutations

But we can see that only two are applicable to the problem.

Thus, we can set up the cases:

Assume both are together (the easiest): Then the total arrangements is 2x1x2x2/2x2 = 2

Now, assume Shirt B is together: it follows that there is 3x2x1/2x2 arrangements, but two of them include shirt A being together also, thus there is 3-2 = 1 arrangements. This is true for A also, thus the answer is

6-2-1-1 = 2

I'm pretty sure a similar tactic could be used for this problem. I've tried, but haven't had too much luck.

 

[steamboat93]

that wont work Mr 100 UAI. It is not a combination.

 

[khorne]

It will work you berk, since I just missused the word combination, because permutation sounds terrible.

 

[grasshopper1]

can i ask the poster what is the actual answer in the textbook?

i came up with an answer of 126 possible arrangements

working was a little dodgy

8!/2x2x2x2=2520

2520/(5x4)

5 being the number of places a shirt can go without being next to another
4 being the number of pairs of identical shirts


but i think im majorly off LOL

 

[khorne]

Finally:

My way: We're trying to find cases which don't fit the restriction.

Total number of shirt ways -> 8!/2^4 = 2520

If they are all together: 4!*2^4/2^4 = 4! = 24 ways

If only 3 are together, 5!/2 = 60 ways - 24 ways (all four together) = 36

If only 2 are together 6!/4 = 180 but...several different combinations of 2 shirts (6) so 180-24-36-36 (because both shirts together can belong to two sets of 3 shirts together)= 84

If only 1 is together, 7!/8 = 630, but repeats again so 630-(3x84 (3 combinations of wrong shirts)) - (3*36 (3 combinations again of wrong set)) - 24 = 246

Now, 2520 - 24 - (4x36) - (6x84) - (4x246) = 864, which is the answer...

If you don't get it, basically I'm finding all the way it can't be done, and each successive shirt left (eg 3 shirts together) I take away the correct arrangements of repeats (eg. 3 shirts together could also include 4th shirt together).

Hope it makes sense, if not, i'll re type it.

 

[jchoi]

I gave up probabilities, it's like too hard... for me. I just don't think the way I'm meant to in it.

 

[gurmies]

I just used a tree-diagram for this...

 

[khorne]

Tree Diagrams are for the weak

 

[wendus]

hahahahahahahaha lol at khorne's signature