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# From a list of 1-8, how many ways can you pick 3 numbers so

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Senior Manager
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From a list of 1-8, how many ways can you pick 3 numbers so [#permalink]

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20 Aug 2007, 13:25
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From a list of 1-8, how many ways can you pick 3 numbers so that the sum of the numbers is 12?

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Manager
Joined: 18 Jun 2007
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20 Aug 2007, 13:33
Not sure if I am totally off..

I just tried using combinations.. since 1 + 3 + 5 , is the same as 3 + 1 + 5 or any other combo..I left the repetitions out.

I come up 11 as the answer? is this correct?

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Senior Manager
Joined: 03 Jun 2007
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20 Aug 2007, 13:37
44 ? I dont have a shortcut to do this. Can someone suggest a shorter method. This problem took abt 4 mins. Or Am I not thinking ?

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VP
Joined: 10 Jun 2007
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20 Aug 2007, 17:17
misterJJ2u wrote:
From a list of 1-8, how many ways can you pick 3 numbers so that the sum of the numbers is 12?

I don't know if this is right, but here we go....

Start off with 2 numbers and eliminate the same number combinations.

12 = 8+4 => Look at how many two numbers can make up 4, you have: (1+3) only. Look at how many numbers can make up 8, you have (1+7), (2+6), (3+5). So total of 4 combinations here.

12 = 7+5 => Same way:
For 5, (1+4), (2+3)
For 7, (6+1), (5+2), (4+3)
Can't use (5+2) here cuz it is the same number. Total of 4.

12 = 6+6 => Same way:
For 6, (1+5), (2+4)
Total of 2.

Total combinations = 4+4+2 = 10

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Manager
Joined: 18 Jun 2007
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20 Aug 2007, 17:53
Bkk145- I used the same method and came up with 11.. wonder what the correct answer is.

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Manager
Joined: 03 Sep 2006
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20 Aug 2007, 20:21
My 5 cents:

1) 12 = 8 + 3 + 1 --> 6 combinations ('cause it could be 8 + 1 + 3 and so on)
2) 12 = 7 + 4 + 1 --> 6 combinations (ALREADY answer is definitely >10 or 11)

SO
1) (** + **) + 1 = 12.
** + ** should = 11, and we have 10 combinations for this.
2) (** + **) + 2 = 12
** + ** should = 10, and we have 9 combinations for this.
... and so on

I would look into the unswer choices and use POI - it'll be faster

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Manager
Joined: 14 May 2006
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21 Aug 2007, 02:17
misterJJ2u wrote:
From a list of 1-8, how many ways can you pick 3 numbers so that the sum of the numbers is 12?

x + y + z = 12

x=1 then y+z = 11. So 3 combinations here.
x=2 then y+z = 10. So 2 combinations here.
x=3 then y+z = 9. So 3 combinations here. (exclude 6+3)
x=4 then y+z = 8. So 3 combinations here.
x=5 then y+z = 7. So 2 combinations here. (exclude 5+2)
x=6 then y+z = 6. So 2 combinations here.
x=7 then y+z = 5. So 2 combinations here.
x=8 then y+z = 4. So 1 combination here.

So total combinations = 18.

I see no patterns in this problem.

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CEO
Joined: 29 Mar 2007
Posts: 2553

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21 Aug 2007, 11:49
Whatever wrote:
My 5 cents:

1) 12 = 8 + 3 + 1 --> 6 combinations ('cause it could be 8 + 1 + 3 and so on)
2) 12 = 7 + 4 + 1 --> 6 combinations (ALREADY answer is definitely >10 or 11)

SO
1) (** + **) + 1 = 12.
** + ** should = 11, and we have 10 combinations for this.
2) (** + **) + 2 = 12
** + ** should = 10, and we have 9 combinations for this.
... and so on

I would look into the unswer choices and use POI - it'll be faster

I did the same thing as you. Seriously we need some answer choices.

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21 Aug 2007, 11:49
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