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Combinations with permutations Options
Jessica
#1 Posted : Tuesday, December 13, 2011 8:49:43 AM
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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.

pateman
#2 Posted : Thursday, December 15, 2011 4:25:14 PM
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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. :)

itsmyflow
#3 Posted : Thursday, December 15, 2011 9:28:38 PM
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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 

vinay
#4 Posted : Tuesday, December 20, 2011 3:51:22 AM
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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.

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error
#5 Posted : Tuesday, December 20, 2011 2:59:20 PM
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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.

mathgenie99
#6 Posted : Sunday, January 15, 2012 3:15:01 PM
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Sorry frnd,

But I don't understand with your problem.

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