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How many ways can we select three letters from the letters

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How many ways can we select three letters from the letters [#permalink] New post 15 Nov 2005, 18:35
How many ways can we select three letters from the letters of RSTUV?
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 [#permalink] New post 15 Nov 2005, 20:35
I say order matters so it's a permutation (rearranging the letters makes a different way)

So it's 5!/3!

5*4= 20

any good?
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Re: Combination: Letters [#permalink] New post 15 Nov 2005, 20:43
TeHCM wrote:
How many ways can we select three letters from the letters of RSTUV?


Answer is 5P3 = 5!/2! = 5.4.3 = 60
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 [#permalink] New post 15 Nov 2005, 23:24
Order does matter so this is a permutation problem. Since we have no repeating elements, then use the standard formula for selecting K elements from a pool of N elements:

N!/(N-K)!

5!/(5-3)!

5!/2!---> 5*4*3---> Answer is 60 ways
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 [#permalink] New post 15 Nov 2005, 23:37
i think it`s a combination problem and so it`s 5c3. combination means selection and permutation means arrangement !
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Re: Combination: Letters [#permalink] New post 16 Nov 2005, 05:29
TeHCM wrote:
How many ways can we select three letters from the letters of RSTUV?


I feel it is a combination since we have not been asked about any particular order.

5C3.
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 [#permalink] New post 16 Nov 2005, 06:29
oops, I made an error. n-k should be my denominator so it should be 2! in the denominator instead of 3!. I get 60 as well.
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 [#permalink] New post 16 Nov 2005, 09:03
Hmmm, the OA is actually 10.

Order matters so the formula is 5!/(2!)(3!) = 10

I got this question off a website
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 [#permalink] New post 16 Nov 2005, 09:11
Christof was right. This is a combination question and not a permutation question because of the term "select." Agreed the answer should be 10.
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 [#permalink] New post 16 Nov 2005, 09:11
TeHCM wrote:
Hmmm, the OA is actually 10.

Order matters so the formula is 5!/(2!)(3!) = 10

I got this question off a website


The OE talks about 5C3..does it say order mattars?
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 [#permalink] New post 16 Nov 2005, 10:02
duttsit wrote:
TeHCM wrote:
Hmmm, the OA is actually 10.

Order matters so the formula is 5!/(2!)(3!) = 10

I got this question off a website


The OE talks about 5C3..does it say order mattars?


Sorry....I meant to type order doesn't matter. Its a combination problem.

Last edited by TeHCM on 16 Nov 2005, 11:03, edited 1 time in total.
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 [#permalink] New post 16 Nov 2005, 10:44
Yes, since you are only picking numbers, picking ABC is supposedly to be the same with picking CBA, so it is a combination problem. We have C(5,3)=5!/2!3!. For easier calculation I always do this: C(5,3)=C(5,2)=5*4/2!=10

If the question asks how many different three letter words can you form, then order matters since ABC is a different word from CBA. In that case the answer would be P(5,3)=5*4*3=60.
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 [#permalink] New post 16 Nov 2005, 11:41
we are selecting 3 letters ....so order should not matter
hence
5C3 = 10
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 [#permalink] New post 25 Nov 2005, 20:03
can each letter can picked only once?
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Last edited by rlevochkin on 25 Nov 2005, 20:10, edited 2 times in total.
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 [#permalink] New post 25 Nov 2005, 20:07
It does not matter in what order letters are chosen, so it's a combination problem. if each letter can be chosen only once, then C(5,3)=5!/3!(5-3)!=10
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  [#permalink] 25 Nov 2005, 20:07
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