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March 12th, 2017, 03:11 AM  #1 
Newbie Joined: Jan 2017 From: Malaysia Posts: 20 Thanks: 3 Math Focus: calculus  NEED HELP: A bit Confusion on Permutations
Here the questions and my working steps. I'm a bit confuse because not all the letter will used and there are repetition on letter A. So I separate into 4 case where letter A not include, than A, double A and last all A included. Are this step correct? 
March 13th, 2017, 06:28 PM  #2 
Newbie Joined: Jan 2017 From: Malaysia Posts: 20 Thanks: 3 Math Focus: calculus 
Anyone can help me???

March 14th, 2017, 07:59 PM  #3 
Newbie Joined: Jan 2017 From: Malaysia Posts: 20 Thanks: 3 Math Focus: calculus 
Anyone? (3rd Attempts)

March 15th, 2017, 08:32 PM  #4 
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 10,906 Thanks: 716 
Your problem is somewhat unclear. Are you looking for combinations of 5 letters only? Take combination AAAKN: are you counting it as 1 or as 6? Take JYAKN : 1 or 3? May be clearer if you rewrite the problem using digits instead of letters; like 1 2 3 4 5 6 6 6 
March 15th, 2017, 08:52 PM  #5 
Senior Member Joined: Sep 2015 From: Southern California, USA Posts: 1,604 Thanks: 817  We helpers look for unanswered problems by seeing which ones have no replies. When you reply to your own post you defeat that and thus we skip over your problem. To solve this just note that A is the only letter that repeats so once you've placed all the other letters the A's will just fit into the remaining spaces. The first letter has 8 slots, the second 7, etc. for a total of $\dfrac{8!}{3!} = 6720$ This assumes that each arrangement uses the full 8 letters. If that's incorrect you need to be clearer about what the problem actually is. Last edited by romsek; March 15th, 2017 at 09:00 PM. 
March 16th, 2017, 10:03 AM  #6 
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 10,906 Thanks: 716 
Romsek, can you confirm this: IF 5letter combos involved, and all 3 A's are included, and A's are undistinguishable (yikes, whatta long word!), then number of combos = 200. That's what I get "loopingly"! Merci a lot... 
March 16th, 2017, 10:07 AM  #7  
Senior Member Joined: Sep 2015 From: Southern California, USA Posts: 1,604 Thanks: 817  Quote:
There are $\binom{5}{2} = 10$ ways 2 nonA chars can be distributed within the 5 slots. There are $5 \cdot 4 = 20$ possible ordered pairs of nonA chars. so $200$ is correct Last edited by romsek; March 16th, 2017 at 10:14 AM.  
March 16th, 2017, 11:32 AM  #8 
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 10,906 Thanks: 716 
The OP shows 1120. Must have light red, normal red and dark red colors!! If that's the case, then his 1120 should be 1200, right? Thanks et al... 
March 16th, 2017, 04:41 PM  #9 
Senior Member Joined: Sep 2015 From: Southern California, USA Posts: 1,604 Thanks: 817  

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