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Permutation?

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Permutation? [#permalink] New post 29 Mar 2006, 19:46
Hello guys,

Can someone explain how to get the answer for the following problem?

Each of the integers from 0 to 9, inclusive, is written on a separate slip of blank paper and the ten slips are dropped into a hat. If the slips are then drawn one at a time without replacement, how many must be drawn to ensure that the numbers on two of the slips drawn will have a sum of 10?

A) 3
B) 4
C) 5
D) 6
E) 7
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Re: Permutation? [#permalink] New post 29 Mar 2006, 20:19
john2005 wrote:
Hello guys,

Can someone explain how to get the answer for the following problem?

Each of the integers from 0 to 9, inclusive, is written on a separate slip of blank paper and the ten slips are dropped into a hat. If the slips are then drawn one at a time without replacement, how many must be drawn to ensure that the numbers on two of the slips drawn will have a sum of 10?

A) 3
B) 4
C) 5
D) 6
E) 7


let's assume the worst case scenerio

1st drawn = 0
2nd drawn = 1
3rd drawn = 2
4th drawn =3
5th drawn = 4
6th drawn= 5

or any combination that has those numbers in the first 6 picks, from there any number you get would add to 10 when added to the proper previously slip drawn.

so the answer is E
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Re: Permutation? [#permalink] New post 29 Mar 2006, 20:21
conocieur wrote:
john2005 wrote:
Hello guys,

Can someone explain how to get the answer for the following problem?

Each of the integers from 0 to 9, inclusive, is written on a separate slip of blank paper and the ten slips are dropped into a hat. If the slips are then drawn one at a time without replacement, how many must be drawn to ensure that the numbers on two of the slips drawn will have a sum of 10?

A) 3
B) 4
C) 5
D) 6
E) 7


let's assume the worst case scenerio

1st drawn = 0
2nd drawn = 1
3rd drawn = 2
4th drawn =3
5th drawn = 4
6th drawn= 5

or any combination that has those numbers in the first 6 picks, from there any number you get would add to 10 when added to the proper previously slip drawn.

so the answer is E


more than enough. E.
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 [#permalink] New post 29 Mar 2006, 20:44
Can be (1,9) (2,8) (3,7) (4,6)

Worst case scenario:

Draw in this way: 1,2,3,4,5,0. Left over number are 6,7,8 and 9 all of which will result in a sum of 10.

The number of draws to ensure two of the slips add up to 10 = 7. (6 numbers 1,2,3,4,5,0 plus any one of the remaining numbers 6,7,8,9)
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 [#permalink] New post 30 Mar 2006, 09:48
E.

in worse case, it goes like this... 0, 1,2,3,4,5, so far six have been drawn and none of the two make to 10,
and the seventh one can be any (6,7,8,9) which added to the prvious one can make to 10.

so 7 must be drawn to get to 10.
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 [#permalink] New post 30 Mar 2006, 19:41
Where did the problem come from?
  [#permalink] 30 Mar 2006, 19:41
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