Harder Counting -- can someone do this? (1 Viewer)

[Deven-7]

Here's the problem in Cambridge 3u 10F extension:

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

A: 864


But how do I work it out?

 

[underthesun]

http://www.boredofstudies.org/community/showthread.php?s=&threadid=11435

 

[Fosweb]

Yeah - i posted this exact question (but with AABBCCDD) in another thread.

 

[underthesun]

I still haven't really seen the solution, meibi Ai shult start werking en stop bladjing..

de eich es si iz kaming klouser en klouser..

 

[Deven-7]

what the... yeh! i remember reading that thread already too! how dumb of me

thanx guys

 

[Richard Lee]

Counting

Hi, there.
Try this calculation:
4P3(5P5/2-4P4)=864

 

[ND]

Re: Counting

Originally posted by Richard Lee
Hi, there.
Try this calculation:
4P3(5P5/2-4P4)=864

Would you mind explaining it?

 

[Deven-7]

whao.. how did u get that?

 

[Richard Lee]

Go to the thread: Probability-q1, then Probability-q2. Follow the questions, you will find the answer by yourself!

Good luck!