permutations help! (1 Viewer)

[viraj30]

"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?

can someone please explain it to me in detail. GOD this is taking so much time and it's only first question in extension section of cambridge textbook..i am scared about later questions!

thanks in advance​

 

[viraj30]

bump

 

[Sy123]

Whats the answer in the book say?

 

[viraj30]

Whats the answer in the book say?

864

 

[viraj30]

bump

 

[Shadowdude]

Draw a diagram. Let the first shirt you choose be design 'A'.

This could be done very slowly and tediously with a massive tree diagram.

 

[viraj30]

Draw a diagram. Let the first shirt you choose be design 'A'.

This could be done very slowly and tediously with a massive tree diagram.

tree diagram?? r u serious????

 

[Shadowdude]

On first sight, yeah.

There's probably a better way of doing it - but it eludes me right now.

 

[Sy123]

4! 2! 2! 2! 2! = Arrangements where all 4 types of clothing are next to their respective type (i.e. AABBCCDD)

4C3 (5! 2! 2! 2!) = Arrangements where 3 types of clothing are next to their respective type (i.e. CDDAABBC or DDCAACBB ...) It is multiplied by 4C3 due to the fact that I choose 3 out of the 4 colors to arrange them together.

4C2 (6! 2! 2!)= Arrangements where 2 types of clothing are grouped together. Multiplied by 4C2 to get no. of combinations of this case

4C1 (7! 2!) = Arrangments of 1 type of clothing grouped together. 4C1=No. of combinations.. you get the drift.

Now. 8!- (all of this added together) =negative

Why is it, I dont get what is wrong with my logic.

This is why I dont like probability...

 

[asianese]

FML probability...

 

[RealiseNothing]

I think I got the answer, but it is way too hard to explain through latex etc.

I may write it out on paper and scan it then upload it to this thread tomorrow. That's if no one else has answered it by then.

 

[Rezen]

One way to do this is to use the Inclusion-exclusion principle. If we label the styles by the letters A, B, C and D as Sy123 did we consider the sets of strings such that AA are together, BB are together and so on. Using this we get the formula:

$$
\frac{8!}{2^4}-\binom{4}{1}\frac{7!}{2^3}+\binom{4}{2}\frac{6!}{2^2}-\binom{4}{4}\frac{5!}{2}+4! = 864
$$

 

[Sy123]

One way to do this is to use the Inclusion-exclusion principle. If we label the styles by the letters A, B, C and D as Sy123 did we consider the sets of strings such that AA are together, BB are together and so on. Using this we get the formula:

Fixed