Last visit was: 11 Dec 2024, 03:41 It is currently 11 Dec 2024, 03:41
Close
GMAT Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track
Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Close
Request Expert Reply
Confirm Cancel
User avatar
Konstantin Lynov
Joined: 24 Jun 2003
Last visit: 19 Sep 2004
Posts: 71
Own Kudos:
Location: Moscow
Posts: 71
Kudos: 13
Kudos
Add Kudos
Bookmarks
Bookmark this Post
avatar
stolyar
Joined: 03 Feb 2003
Last visit: 06 May 2014
Posts: 1,012
Own Kudos:
Posts: 1,012
Kudos: 1,751
Kudos
Add Kudos
Bookmarks
Bookmark this Post
User avatar
mciatto
Joined: 10 Jun 2003
Last visit: 13 Jun 2005
Posts: 132
Own Kudos:
Location: Maryland
Posts: 132
Kudos: 38
Kudos
Add Kudos
Bookmarks
Bookmark this Post
User avatar
mciatto
Joined: 10 Jun 2003
Last visit: 13 Jun 2005
Posts: 132
Own Kudos:
Location: Maryland
Posts: 132
Kudos: 38
Kudos
Add Kudos
Bookmarks
Bookmark this Post
stolyar
No info about their heights. I assume that they all have different heights.
Let their heights be 1, 2, 3, 4, 5, 6, 7, 8 units

Possible variants:

5678
1234

2468
1357

I think 2 variants


Stolyar,

What about,

4568
1237

...I think you missed some...
User avatar
Konstantin Lynov
Joined: 24 Jun 2003
Last visit: 19 Sep 2004
Posts: 71
Own Kudos:
Location: Moscow
Posts: 71
Kudos: 13
Kudos
Add Kudos
Bookmarks
Bookmark this Post
the official answer is 14. I can not comment on this. :cry:
Let's think together.
User avatar
rich28
Joined: 02 Jul 2003
Last visit: 26 Sep 2003
Posts: 39
Own Kudos:
Posts: 39
Kudos: 2
Kudos
Add Kudos
Bookmarks
Bookmark this Post
THe only way I can think to get 14 is all four in front are the shortest and must line up in one way- Left to right tallest to shortest. Thus in back we are not given the heights, but the tallest has to be the furtherest left. That leaves the other three spots to be filled. Since heights are not given= 4c3+4c2+4c1=14
User avatar
AkamaiBrah
User avatar
GMAT Instructor
Joined: 07 Jul 2003
Last visit: 24 Jun 2009
Posts: 392
Own Kudos:
Location: New York NY 10024
Concentration: Finance
Schools:Haas, MFE; Anderson, MBA; USC, MSEE
Posts: 392
Kudos: 496
Kudos
Add Kudos
Bookmarks
Bookmark this Post
rich28
THe only way I can think to get 14 is all four in front are the shortest and must line up in one way- Left to right tallest to shortest. Thus in back we are not given the heights, but the tallest has to be the furtherest left. That leaves the other three spots to be filled. Since heights are not given= 4c3+4c2+4c1=14


I think your answer is more coincidental than explanatory. Nice try, though.
User avatar
AkamaiBrah
User avatar
GMAT Instructor
Joined: 07 Jul 2003
Last visit: 24 Jun 2009
Posts: 392
Own Kudos:
Location: New York NY 10024
Concentration: Finance
Schools:Haas, MFE; Anderson, MBA; USC, MSEE
Posts: 392
Kudos: 496
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Konstantin Lynov
Eight people are to make photo, four of them being standing in the first row and four of them in the second. Every man standing in the front row should be lower that that standing exactly behind him, and all people from the same row should be arranged in order of height increase from left to right. Providing the restrictions mentioned above how many different photos could be taken?


I don't know if there is a formula, so I will try to solve this logically:

First, let's say the formation looks like this:

0000
0000

where the heights go from 1 to 8 and the unknowns are 0's.

We know that at a minimum, it must look like this:

0008
1000

Now, consider the person top row 2nd from left. He must have one person shorter than him both to his left and below. So at a minimum, he must be a 4. Similarly, the person in the top row, 3rd from the left must have 5 shorter than him (two on his left, one below plus the two to the left of the one below) so his minimum height is 6.

Now we can count. Let's designate the arrangement as:

XYZ8
1000

Y can only be 4,5, or 6.

If Y is 4, Z can be 6 or 7 and X can be 2 or 3. 4 ways
If Y is 5, Z can be 6 or 7 and X can be 2, 3, or 4. 6 ways.
If Y is 6, Z must be 7 and X can be 2, 3, 4, or 5. 4 ways.

For each of the combinations above, the bottom row is determined (we have 3 numbers left all of which must be in order).

Hence, there are 4 + 6 + 4 = 14 ways.

KL, is there a better way to solve this?



Archived Topic
Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Where to now? Join ongoing discussions on thousands of quality questions in our Data Sufficiency (DS) Forum
Still interested in this question? Check out the "Best Topics" block above for a better discussion on this exact question, as well as several more related questions.
Thank you for understanding, and happy exploring!
Moderators:
Math Expert
97797 posts
Senior Moderator - Masters Forum
3116 posts
GMAT Tutor
1930 posts