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.
Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:
Have you ever wondered how to score a PERFECT 805 on the GMAT? Meet Julia, a banking professional who used the Target Test Prep course to achieve this incredible feat. Julia's story is nothing short of an inspiration.
Think a 100% GMAT Verbal score is out of your reach? Target Test Prep will make you think again! Our course uses techniques such as topical study and spaced repetition to maximize knowledge retention and make studying simple and fun.
GMAT Club 12 Days of Christmas is a 4th Annual GMAT Club Winter Competition based on solving questions. This is the Winter GMAT competition on GMAT Club with an amazing opportunity to win over $40,000 worth of prizes!
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
(N/A)
Question Stats:
0%
(00:00)
correct
0%
(00:00)
wrong
based on 0
sessions
History
Date
Time
Result
Not Attempted Yet
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?
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.
Still interested in this question? Check out the "Best Topics" block below for a better discussion on this exact question, as well as several more related questions.
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
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.
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.
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.