Math Is Fun Forum

bbcnyc

Member

Registered: 2008-10-13

Posts: 1

Combinations and Permutations

I keep doubting myself that I am calculating the following problem incorrectly:

How many sequences of three letters can be made from the letters {A, B, C, D, E} if

(a) The letters must be all different.

I would assume that I would use the permutation equation to solve this.  Pn,r = n!/(n-r)!

(b)  Repetitions are allowed.

I would assume I would use the combination equation to solve this.  Cn,r = n!/r!(n-r)!

If anyone can verify whether that is correct or incorrect I would greatly appreciate it.  Thanks!

John E. Franklin

Member

Registered: 2005-08-29

Posts: 3,588

Re: Combinations and Permutations

b.) > a.), so ofcourse you are wrong with at least 1 equation...

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igloo myrtilles fourmis

mathsyperson

Moderator

Registered: 2005-06-22

Posts: 4,900

Re: Combinations and Permutations

It might be easier just to think about it terms of choices.

a)

For the first letter, you have 5 options: A, B, C, D or E.
For the second letter, you have 4 options - every letter except the one you had first.
Similarly, you can pick any of the 3 unchosen letters as the final one.

The number of sequences is therefore 5x4x3 = 60.

b)
This time, you have 5 options for each pick, since the later choices aren't restricted by what you've already picked.

The number of sequences here is 5x5x5 = 125.

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Why did the vector cross the road?
It wanted to be normal.

bbcnyc

Member

Registered: 2008-10-13

Posts: 1

Re: Combinations and Permutations

Thank you so much!  Thinking in terms of order and no order was completely throwing me off.