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ketkoag: Legendary Member; Posts: 876; Joined: Thu Apr 10, 2008 8:14 am; Thanked: 13 times

arrange

by ketkoag » Thu Apr 30, 2009 10:38 am

A photographer will arrange 6 people of 6 different heights for photograph by placing
them in two rows of three so that each person in the first row is standing in front of
someone in the second row. The heights of the people within each row must increase
from left to right, and each person in the second row must be taller than the person
standing in front of him or her. How many such arrangements of the 6 people are
possible?
A. 5
B. 6
C. 9
D. 24
E. 36

please explain how to do this question, i would have count all the possibilities but its very time consuming and is not accurate. please tell me a standard method how to go for this problem and such problems..

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Source: Beat The GMAT — Problem Solving | Thu Apr 30, 2009 10:38 am

Uri: Master | Next Rank: 500 Posts; Posts: 229; Joined: Tue Jan 13, 2009 6:56 am; Thanked: 8 times; GMAT Score:700

by Uri » Thu Apr 30, 2009 11:10 am

confused.....what is the OA?

Last edited by Uri on Thu Apr 30, 2009 12:01 pm, edited 1 time in total.

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kanha81: Master | Next Rank: 500 Posts; Posts: 431; Joined: Sat Jan 10, 2009 9:32 am; Thanked: 16 times; Followed by:1 members

Re: arrange

by kanha81 » Thu Apr 30, 2009 11:11 am

ketkoag wrote:A photographer will arrange 6 people of 6 different heights for photograph by placing
them in two rows of three so that each person in the first row is standing in front of
someone in the second row. The heights of the people within each row must increase
from left to right, and each person in the second row must be taller than the person
standing in front of him or her. How many such arrangements of the 6 people are
possible?
A. 5
B. 6
C. 9
D. 24
E. 36

please explain how to do this question, i would have count all the possibilities but its very time consuming and is not accurate. please tell me a standard method how to go for this problem and such problems..

I think. Not sure. [spoiler][C][/spoiler]
heights of 2 ppl will remain fixed- h1, h6
height of remaining 4 ppl can be arranged in 2 ways-
ie height of each h2, h3, h4, h5 has two options to choose where to stand by obeying the constraints mentioned in the problem.

hence total number of ways this can be achieved = 4*2 + 1 = 9

here 4 represents h2, h3, h4, h5

2 represents number of places can chose

1 represents h1, h6 who has only one place to choose.

hope it helps.

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Always do what you're afraid to do. Whoooop GMAT

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mike22629: Master | Next Rank: 500 Posts; Posts: 322; Joined: Fri Mar 27, 2009 3:56 pm; Thanked: 24 times; GMAT Score:710

by mike22629 » Thu Apr 30, 2009 11:30 am

I think it is 36.

5 has 2 possible spots

2 has 2 spots

3 has 3 spots

4 has 3 spots

2*2*3*3 = 36

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mike22629: Master | Next Rank: 500 Posts; Posts: 322; Joined: Fri Mar 27, 2009 3:56 pm; Thanked: 24 times; GMAT Score:710

by mike22629 » Thu Apr 30, 2009 11:44 am

Ha. That doesnt work because I did not count for overlaps. And 4 can only go in two spots anyways.

Probably just got to count them out.

Would be interested if there is a way to do it using algebra, but if there is, I'm sure its more difficult than just counting them out.

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kanha81: Master | Next Rank: 500 Posts; Posts: 431; Joined: Sat Jan 10, 2009 9:32 am; Thanked: 16 times; Followed by:1 members

by kanha81 » Thu Apr 30, 2009 11:46 am

mike22629 wrote:I think it is 36.

5 has 2 possible spots

2 has 2 spots

3 has 3 spots

4 has 3 spots

2*2*3*3 = 36

Yupe! I revisited the problem and the calc. above certainly adds up! Were you able to solve it under 2 min? How did you approach?

Want to Beat GMAT.
Always do what you're afraid to do. Whoooop GMAT

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Uri: Master | Next Rank: 500 Posts; Posts: 229; Joined: Tue Jan 13, 2009 6:56 am; Thanked: 8 times; GMAT Score:700

by Uri » Thu Apr 30, 2009 12:16 pm

at first, i thought the answer would be 6; had some logic behind that. but later could not find more than 5 combinations, using brute force. can any one of you please try to check if more than 6 combinations are possible?

the combinations that i got are:

Combination1:
f=>1 2 3
b=>4 5 6

Combination 2:
f=>1 2 4
b=>3 5 6

Combination 3:
f=>1 2 5
b=>3 4 6

Combination 4:
f=>1 3 4
b=>2 5 6

Combination 5:
f=>1 3 5
b=>2 4 6

it would be great to get the input from some expert as well as the OA.

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Stuart@KaplanGMAT: GMAT Instructor; Posts: 3225; Joined: Tue Jan 08, 2008 2:40 pm; Location: Toronto; Thanked: 1710 times; Followed by:614 members; GMAT Score:800

by Stuart@KaplanGMAT » Thu Apr 30, 2009 1:45 pm

The key is to realize that 1 and 6 are fixed and that once you select the other two members of the back (i.e. taller) row, the front row is fixed.

We have 2, 3, 4 and 5 to play with.

At first, it may seem that the solution is simply 4C2 = 6, since we're selecting 2 out of the 4 remaining people. However, this solution ignores the conflict of selecting both 4 and 5 for the "tall" row, since that will lead to both 2 and 3 in the front row and 3 (a taller person) standing in front of 4 (a shorter person), i.e.:

1 4 5
2 3 6

So, ignoring 1, who automatically goes in back, we can have:

2 3
2 4
2 5
3 4
3 5

in the "tall" row, giving us a total of 5 possible arrangements.

Stuart Kovinsky | Kaplan GMAT Faculty | Toronto

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artistocrat: Master | Next Rank: 500 Posts; Posts: 152; Joined: Wed Mar 12, 2008 4:36 pm; Thanked: 8 times; Followed by:2 members

by artistocrat » Thu Apr 30, 2009 2:06 pm

I think it is misleading to think that each solution for a GMAT problem has a shortcut or formula. Some problems have so many conditions that the usual methods are cumbersome and time consuming. For this question I decided to create a mini-chart that accounts for the fixed 1 and 6 (the shortest and tallest person, respectively)

1 _ _
_ _ 6

then I filled in my silly chart with the combos.

123
456

134
256

124
356

135
246

125
346

For a total of five combos.

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dumb.doofus: Master | Next Rank: 500 Posts; Posts: 435; Joined: Sat Sep 27, 2008 2:02 pm; Location: San Jose, CA; Thanked: 43 times; Followed by:1 members; GMAT Score:720

by dumb.doofus » Thu Apr 30, 2009 3:46 pm

Yup.. there's no shortcut here.. but then such questions in which the number of possibilities is quite low demands that you solve it by the crude method..

As others have also pointed out, once you fix the shortest and tallest guy, its actually pretty easy to see that there are not many choices that we are left with..

Others have explained well.. answer is 5.

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