Jack has eight cans of different heights.

Question

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

chetan2u · Accepted Answer

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

Diwakar003 · Answer

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!

chetan2u · Answer

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

ankit9291 · Answer

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

NCC · Answer

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?

chetan2u · Answer

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.

NCC · Answer

So then how do we manually count all arrangements?

EMPOWERgmatRichC · Answer

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:

Show Spoiler

E

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

Contact Rich at: [email protected]

jain67 · Answer

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

KarishmaB · Answer

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.

ndwz · Answer

Hi  Bunuel , would you be able to confirm if in this question it doesn't necessarily mean that the cans need to be arranged in rows like 4 in row 1 and 4 in row 2 right? could also be (1,7). (2,6) etc? No where does it say that cans should be equal in both rows? Please let me know where I am going wrong in understanding thanks!

Bunuel · Answer

The question says: "Each can in the second row must be taller than the can in front of it (in the first row)." This clearly implies that each can in the second row must have a can directly in front of it. So you can’t have row distributions like (1,7), (2,6), etc., because then some cans in the second row wouldn’t have any can in front of them to compare with, which would break the rule. So yes, the number of cans in each row must be equal, meaning only a (4,4) split is valid. Also, just to note, this question is a direct copy of another problem posted here: https://gmatclub.com/forum/eight-women- ... 92289.html Same logic, same structure, just reworded with cans instead of people.

srik410 · Answer

So annoying to get this question in the mocks.

If 2n different objects are there and you have to pick n and n in two rows both in ascending order, such that one row always has greater number than the corresponding number in the other row. Then,

Combinations Cn = 2nCn * 1/(n+1)

So, here there are 8C4/5 = 14 combinations.

thekingslay · Answer

is this a gmat style question?

Bunuel · Answer

There is a similar but slightly easier question from GMAT Prep, so I’d say a question like this could definitely be tested on the actual exam. Here is that question: https://gmatclub.com/forum/a-photograph ... 86046.html

shankk · Answer

this doesn't specify whether each row has to have 4 cans each or if the cans can be arranged with different numbers in each row, is this implied? Do we have to assume this in problems like this?

Bunuel · Answer

You can pretty much infer this from the line that says  each can in the second row must be taller than the can in front of it.  If the rows didn’t both have four cans, then this condition wouldn’t make sense, because some cans would not have a corresponding “front” can.

That said, this is not an official question, and the wording is not very precise. In the post above, I shared a link to the official question, which is worded better and makes the requirement explicit by saying that each row contains the same number of items.

Jack has eight cans of different heights.

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Jack has eight cans of different heights.

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