My Math Forum NEED HELP: A bit Confusion on Permutations

 Probability and Statistics Basic Probability and Statistics Math Forum

 March 12th, 2017, 01:11 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, 05:28 PM #2 Newbie   Joined: Jan 2017 From: Malaysia Posts: 20 Thanks: 3 Math Focus: calculus Anyone can help me???
 March 14th, 2017, 06:59 PM #3 Newbie   Joined: Jan 2017 From: Malaysia Posts: 20 Thanks: 3 Math Focus: calculus Anyone? (3rd Attempts)
 March 15th, 2017, 07:32 PM #4 Math Team   Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 12,930 Thanks: 884 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, 07:52 PM   #5
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Quote:
 Originally Posted by palongze Anyone can help me???
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 08:00 PM.

 March 16th, 2017, 09:03 AM #6 Math Team   Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 12,930 Thanks: 884 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, 09:07 AM   #7
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Quote:
 Originally Posted by Denis 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...
There are 5 non A letters.

There are $\binom{5}{2} = 10$ ways 2 non-A chars can be distributed within the 5 slots.

There are $5 \cdot 4 = 20$ possible ordered pairs of non-A chars.

so $200$ is correct

Last edited by romsek; March 16th, 2017 at 09:14 AM.

 March 16th, 2017, 10:32 AM #8 Math Team   Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 12,930 Thanks: 884 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, 03:41 PM   #9
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Quote:
 Originally Posted by Denis 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...
OP's original post is a mystery to me.

 Tags bit, confusion, permutations

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