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Jack has eight cans of different heights. [#permalink]
Expert Reply
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
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Re: Jack has eight cans of different heights. [#permalink]
Row A - _ _ _ 8
Row B- 1 _ _ _ _

Now let's select 6c3 for row A=20

But we cannot select 2,3,4 in row a so -1

We cannot select 2,3 combine in row A with 4 ways
Similary 3,4 with 4 ways

I.e 20-4-4-1=11 ways

Posted from my mobile device
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Re: Jack has eight cans of different heights. [#permalink]
Dear EMPOWERgmatRichC,
This is a similar question to the 6 people of different heights ( and you had given the solution there)

Please share how to go about 8 cans here.
I gave it a shot and got the follows:

c1 and c8 will be fixed

c1 _ _ _
_ _ _ c8

Now if we do a little trial and error we see that c2 and c7 can take 2 places each only

1 2 3 4
5 6 7 8

1 2 4 5
3 6 7 8

1 3 4 5
2 6 7 8

Now remaining 3,4,5,6 ---> can take any of the 4 places

So ways = 4! (abv) X 2(for 2) X 2(for 7) = 96

Where am I going wrong?
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Jack has eight cans of different heights. [#permalink]
Expert Reply
NCC wrote:
Dear EMPOWERgmatRichC,
This is a similar question to the 6 people of different heights ( and you had given the solution there)

Please share how to go about 8 cans here.
I gave it a shot and got the follows:

c1 and c8 will be fixed

c1 _ _ _
_ _ _ c8

Now if we do a little trial and error we see that c2 and c7 can take 2 places each only

1 2 3 4
5 6 7 8

1 2 4 5
3 6 7 8

1 3 4 5
2 6 7 8

Now remaining 3,4,5,6 ---> can take any of the 4 places

So ways = 4! (abv) X 2(for 2) X 2(for 7) = 96

Where am I going wrong?


Can you have a taller person in first row. NO

But 3,4,5,6 arranged in 4! will have all those possibilities.

1 3 4 5
2 6 7 8

cannot be written as

1 6 4 5
2 3 7 8

because 6 is taller to 3.
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Re: Jack has eight cans of different heights. [#permalink]
So then how do we manually count all arrangements?
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Re: Jack has eight cans of different heights. [#permalink]
Expert Reply
Hi NCC,

Yes, this question is essentially the same concept as the prompt listed here: https://gmatclub.com/forum/a-photograph ... 86046.html), BUT by adding two additional items, many more arrangements become possible. I'd work through the brute-force in much the same way (although the answers are written in such a way that you can potentially avoid writing out all of the possible options).

I would start in the same way that you do - by "locking in" the smallest and largest cans.

_ _ _ 8
1 _ _ _

Next, if we put all of the remaining cans in order of size, then we have our 1st OPTION...

5 6 7 8
1 2 3 4

From here, both the '2' and the '7' have the greatest limitations (the only other option for the '2' is to go behind the '1' and the only other option for the '7 is to go in front of the '8'), so we have 4 frameworks to consider:

_ _ 7 8
1 2 _ _

2 _ 7 8
1 _ _ _

_ _ _ 8
1 2 _ 7

2 _ _ 8
1 _ _ 7

Once you have worked through the first 3 frameworks (again, with some basic brute-force), you'll have enough information to choose the correct answer.

Final Answer:

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Contact Rich at: Rich.C@empowergmat.com
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Re: Jack has eight cans of different heights. [#permalink]
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Re: Jack has eight cans of different heights. [#permalink]
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