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Rank: Newbie
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Joined: 11/8/2011 Posts: 1
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I need help on this one problem because i keep trying and can't seem to get it right. It says, Using the letters in the word CHARISMA, find the number of combinations that can be formed using 2 letters at a time. I'm confused on how to do this.
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Rank: Newbie
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Joined: 12/15/2011 Posts: 1
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I am not an expert at this, but let's take it logically.
CHARISMA has 8 letters, so in order to form 2-letter chunks, you first need to select one letter, let's take "C" for example. Then you only have 7 letters left, because you already took the "C". This gives us 56 combinations. Now, you have 8 letters, so you need to multiply the result by 8, which gives us 448 combinations.
If my logic is correct, 448 is the answer. :)
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Rank: Newbie
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Joined: 12/15/2011 Posts: 3
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"I need help on this one problem because i keep trying and can't seem to get it right. It says, Using the letters in the word CHARISMA, find the number of combinations that can be formed using 2 letters at a time. I'm confused on how to do this."
Charisma indeed has 8 letters but there are 2 a's
assuming that one "a" is the same as another "a"
technically there are only 7 letters usable to form 2 letters combinations. since only 2 letters are used. you need to choose one out of the 7.. then one of out 6 letters left..making the number possible of combinations->
7 * 6 = 42 numbers of combinations
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Rank: Newbie
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Joined: 12/20/2011 Posts: 2
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For the second place, you can use the a again.
(there being two of them). Hence for the second place also there can be 7 choices so the answer will be 7*7 = 49.
<a href="http://www.chemtopper.com/math.php">Online Math Tutor </a>
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Rank: Newbie
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Joined: 12/20/2011 Posts: 2
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Lets look at 4 letters: ABCC (the C appears twice) Which combinations can you get? AB AC BA BC CA CB CC
7 different ones. This is 3*2 + 1
In your case: 7*6 + 1=43
Why not 42? Well, it misses the AA solution Why not 49? Well, it's just wrong. If the first letter is C then you don't have 7 letter to choose from...
This of course assumes that you don't think CA is different from CA just because you use different A's.
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Rank: Newbie
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Joined: 1/15/2012 Posts: 1
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Sorry frnd,
But I don't understand with your problem.
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