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How many different 3 digit possible integers can be formed

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How many different 3 digit possible integers can be formed [#permalink] New post 22 Aug 2004, 19:08
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How many different 3 digit possible integers can be formed using only the digits 1,2,3,4,5 and 6 if not digit is repeated?

pls explain your answer
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 [#permalink] New post 22 Aug 2004, 19:30
6C3*3! = 120
# of ways of selecting 3 out of the given 6 numbers: 6C3
The 3 selected numbers can be arranged 3! ways
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 [#permalink] New post 23 Aug 2004, 04:39
good job guys, the answer is indeed 120

when I did this problem, I did 6C3, which turned out to be 20
subsequently, I realized that there're some fundamental problems with my approach to permutation.

How come 6C3 does not stand for number of ways to select 3 numbers out of 6? I also wrongly believed that it takes care of number of ways of picking these 3 numbers.
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 [#permalink] New post 23 Aug 2004, 06:06
That's a really important question. The notation 6C3 refers to combinations, when the order doesn't matter. Permutations refers to arrangments, when the order does matter.

I don't ever use the "C" notation. I think it's confusing. I just talk about permutations and combinations. Permutations are easier. You simply ask yourself how many spaces you have to fill up, and how many options go into each space, and multiply. In this problem, there are 3 spaces, and 6 options for the first, 5 for the second, and 4 for the third. So the answer's 120.

For combinations, all you really need to know is that you take the permutations answer and divide by (the number of spaces)!. Remember, combinations is when the order DOES NOT matter. Dividing eliminates the repeats. In this case, IF this were a combos problem, we'd divide by 3!, getting 20. The "C" notation and its resulting formula does that automatically, but it also takes away some amount of thinking from the process, which could hurt you down the road. You've got to keep your head in it.

What Paul did, by the way, in his solution, is perfectly fine, but it's redundant. He used combinations to solve, dividing the permuations by 3!, and then multiplied by 3! again using a different logic. It worked, of course, because it was right, but it added two or three extra steps, as well as too many extra logical conclusions that could have been potentially wrong.

I hope that makes sense! Let me know if you'd like me to ellaborate further.
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 [#permalink] New post 23 Aug 2004, 13:53
Yes Ian, you are perfectly right :) I have the bad habit of working with combinations instead of permutations. It's like saying that there are 3 spots available for the 3 digit numbers. You have 6 available numbers to pick from. The first spot could be filled by one of the 6 numbers. The second spot could be filled by 1 of the remaining 5 numbers. And the last spot could be filled out by 1 of the remaining 4 numbers. Hence, your solution is the best and fastest: 6P3 or 6*5*4 = 120
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  [#permalink] 23 Aug 2004, 13:53
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