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Jack has eight cans of different heights.

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Joined: 22 Apr 2017
Posts: 38

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Jack has eight cans of different heights. [#permalink]

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New post 18 Sep 2017, 09:52
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Question Stats:

33% (01:53) correct 67% (01:19) wrong based on 42 sessions

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Jack has eight cans of different heights. He arranges them in two rows such that the height of the cans must increase from left to right in each row. Further, each can in the second row must be taller than the can in front of it (in the first row). In how many different ways can Jack arrange the cans?
A. 1
B. 6
C. 8
D. 10
E. More than 10

Source- Expert's Global
[Reveal] Spoiler: OA

Kudos [?]: 19 [0], given: 40

Expert Post
Math Forum Moderator
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Joined: 02 Aug 2009
Posts: 4986

Kudos [?]: 5506 [0], given: 112

Jack has eight cans of different heights. [#permalink]

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New post 19 Sep 2017, 04:51
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ManishKM1 wrote:
Jack has eight cans of different heights. He arranges them in two rows such that the height of the cans must increase from left to right in each row. Further, each can in the second row must be taller than the can in front of it (in the first row). In how many different ways can Jack arrange the cans?
A. 1
B. 6
C. 8
D. 10
E. More than 10

Source- Expert's Global



Hi..

Hi..

It is more to do with visualisation than combinations.
And not likely to be in actuals..

Let the number of cans be 1 to 8 as per height.
ONLY 1 and 8 will have permt places..
2 and 7 can have Only two places


So four ways as per above conditions are

A) _,_,7,8
1,2,_,_
FOUR places remaining and to choose 2 out of them is 4C2=6

B)_,_,_,8
1,2,_,7
Here again FOUR places BUT the 1st row will have 6 fixed
_,_,6,8
1,2,_,7
AND the second row can take ONLY 1 out of remaining THREE, so 3C1= 3 ways


C) 2,_,7,8
1,_,_,_
SAME logic as B above - 3 ways

D) 2,_,_,8
1,_,_,7
here again FOUR places but in this case 3 and 6 also can have only one place
2,_,6,8
1,3,_,7
so two ways for remaining 2 = 2 ways


TOTAL 6+3+3+2=14 ways

ans more than 10 ..
E



Few examples..
A) 5,6,7,8
....1,2,3,4
B) 4,5,6,8
1,2,3,7
C) 2,6,7,8
1,3,4,5
D) 2,4,6,8
1,3,5,7
E) 2,4,7,8
1,3,5,6
F) 2,5,7,8
1,3,4,6
and so on for all 14
_________________

Absolute modulus :http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html

Kudos [?]: 5506 [0], given: 112

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Joined: 02 Aug 2015
Posts: 70

Kudos [?]: 19 [0], given: 22

Jack has eight cans of different heights. [#permalink]

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New post 19 Sep 2017, 08:17
ManishKM1 wrote:
Jack has eight cans of different heights. He arranges them in two rows such that the height of the cans must increase from left to right in each row. Further, each can in the second row must be taller than the can in front of it (in the first row). In how many different ways can Jack arrange the cans?
A. 1
B. 6
C. 8
D. 10
E. More than 10

Source- Expert's Global


This is how I solved it,

The two rows have to be in the below format as 8 is the biggest number and 1 is the least number.

- - - 1
8 - - -

Now we have 6 cans left. Let's choose 3 cans from the 6 to fill either the 1st or 2nd row. So the no.of combinations is 6c3 = 20.

For each of the 20 ways of selection of 3, there can be only 1 arrangement possible in ascending order. This can go either in row 1 or row 2. If it goes in row 1, there is only 1 way to arrange the nos of row 2, vice versa. So there are a total of 40 ways. Out of 40, half of it will be duplicates. So 20 ways in total.


Hence E.

Cheers!

Kudos [?]: 19 [0], given: 22

Expert Post
Math Forum Moderator
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Joined: 02 Aug 2009
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Kudos [?]: 5506 [0], given: 112

Jack has eight cans of different heights. [#permalink]

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New post 19 Sep 2017, 08:53
Diwakar003 wrote:
ManishKM1 wrote:
Jack has eight cans of different heights. He arranges them in two rows such that the height of the cans must increase from left to right in each row. Further, each can in the second row must be taller than the can in front of it (in the first row). In how many different ways can Jack arrange the cans?
A. 1
B. 6
C. 8
D. 10
E. More than 10

Source- Expert's Global


This is how I solved it,

The two rows have to be in the below format as 8 is the biggest number and 1 is the least number.

- - - 1
8 - - -

Now we have 6 cans left. Let's choose 3 cans from the 6 to fill either the 1st or 2nd row. So the no.of combinations is 6c3 = 20.

For each of the 20 ways of selection of 3, there can be only 1 arrangement possible in ascending order. This can go either in row 1 or row 2. If it goes in row 1, there is only 1 way to arrange the nos of row 2, vice versa. So there are a total of 40 ways. Out of 40, half of it will be duplicates. So 20 ways in total.


Hence E.

Cheers!


hi

ans will not be 20 as it will contain ways where the two conditions of front row less than rear row and left lower than right will NOT meet.
ans will be 14
_________________

Absolute modulus :http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html

Kudos [?]: 5506 [0], given: 112

Jack has eight cans of different heights.   [#permalink] 19 Sep 2017, 08:53
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