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In how many ways can you select three integers x1, x2 &

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Senior Manager
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In how many ways can you select three integers x1, x2 & [#permalink] New post 08 Sep 2003, 03:11
In how many ways can you select three integers x1, x2 & x3 from 1 to 12 such that x1<x2<x3. (Hint: Two ways to solve this. One lengthy...uhh and one very shorte... ahh)
-Vicks
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 [#permalink] New post 08 Sep 2003, 08:06
javropu wrote:
id say 220


can u explain javs?
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 [#permalink] New post 08 Sep 2003, 08:52
id do the following:
12*11*10/(3!)=220
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 [#permalink] New post 08 Sep 2003, 09:45
javropu wrote:
id do the following:
12*11*10/(3!)=220



Yeah..My answer cant be right.
But ...did you use 12C3 ??
Isnt 12C3 the total # of ways to draw 3 from 1-12 numbers without any restrictions?

Please explain

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 [#permalink] New post 08 Sep 2003, 20:38
right javs...
12C3 is the number of ways to select three numbers, and there is only one
way to arrange these numbers such that x1<x2<x3 :!:
i hope u got it praet...
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 [#permalink] New post 11 Sep 2003, 03:04
Vicky wrote:
right javs...
12C3 is the number of ways to select three numbers, and there is only one
way to arrange these numbers such that x1<x2<x3 :!:
i hope u got it praet...
-Vikcs



Clever way of phrasing the problem..same as the # of ways to pick 3 #'s from 1-12..

Thanks!
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still confused [#permalink] New post 11 Sep 2003, 12:19
if order does matter than why are we using a combination formula?


shouldn't we use the permutation formula?
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Re: still confused [#permalink] New post 11 Sep 2003, 17:27
lakrana wrote:
if order does matter than why are we using a combination formula?


shouldn't we use the permutation formula?


If order matters ,we use permutations..
Permutations say the every position is unique..so 123 is different from 321...

But in our case...we dont need the positions to be unique..

hth
praetorian
Re: still confused   [#permalink] 11 Sep 2003, 17:27
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