Jack has eight cans of different heights.

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

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

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.

So then how do we manually count all arrangements?

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:
GMAT assassins aren't born, they're made,
Rich

Contact Rich at: Rich.C@empowergmat.com

MartyTargetTestPrep KarishmaB Please help with the valid solution for this one.

­The photograph question itself was hard because it required you to consider the case in which you could have both 4 and 5 ahead. 
Here this question has a lot more added complexity because you have to consider many other cases. I don't have a sequential logic leading to a formula solution here, if you are looking for that. 
I will also explore possibilities and considerations, as done by Chetan and Rich above.