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Photographer

by aplavakarthik » Fri Sep 09, 2011 4:49 am
A photographer will arrange 10 people of 10 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 10 people are possible?

a.41
b.42
c.38
d.78
e.91
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by saketk » Fri Sep 09, 2011 9:40 am
aplavakarthik wrote:A photographer will arrange 10 people of 10 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 10 people are possible?

a.41
b.42
c.38
d.78
e.91
I guess this question involves lot of calculations: If i were you, I would have made a wise guess and moved on.

I tried to do it this way --

let's assume heights of 10 people as -- 1,2,3,4,5,6,7,8,9,10

We need to arrange people ilke this

P1 P2 P3 (1st row) [ increasing from left to right.. this means p3>p2>p1
P4 P5 P6 (2nd row) p6>p5>p4

Now , we have to see that p6 can only have heights from 6.. to 10. Reason- he should be the tallest among the 6, and if we choose a value, say 5 then we would not be able to complete the arrangements here.

Now there will be several cases..

First - Let p6 = 6

then we can only have one arrangement

1 2 3
4 5 6

Case 2: P6 = 7

then arrangements can be :

p5 can have 2 values 5 or 6
similarly p4, p3, p2 and p1 all can be filled in 2 ways. total 2*5 = 10 ways


Case 3: p6 = 8

then arrangements can be

p5 can have 3 values 5,6,7 = 3 ways

Similarly p4 can have 3 values 4,5,6 = 3 ways..
likewise for p3, p2 and p1 can be filled in 3 ways each

total 3*5=15

Case 4: p6= 9

then arrangements can be

4*5 = 20

and for case 5: p6 =10

then arrangements can be

5*5 = 25

total 1+10+15+20+25 = 71

I am getting 71, which is not in the options.

Experts please help with this one
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by saketk » Fri Sep 09, 2011 10:04 am
i think i understand my mistake...

when we choose p=6

we can have 5 and not 1 arrangement

1 2 3
4 5 6

1 2 4
3 5 6

1 2 5
3 4 6

1 3 4
2 5 6

1 3 5
2 4 6

Likewise we can have many combinations for rest of the cases..


I think the correct answer should be E
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by aplavakarthik » Fri Sep 09, 2011 10:19 am
- I start with 1 2 3 4 in the first line, and there are 5 possibilities for the last position: 5, 6, 7, 8 and 9.
- If 4 goes on the second line, than in the first there will be 1 2 3 5, and there are 4 possibilities for the last place: 6, 7, 8 and 9.
- If 5 goes besides 4 on the second line, than in the first there will be 1 2 3 6 and there are 3 possibilities for the last place: 7, 8 and 9.
- If 6 goes also on the second, there will be 1 2 3 7 8 or 1 2 3 7 9 in the first. So, for the sequence 1 2 3 in the first line, there are (5 + 4 + 3 + 2) possibilities.
I change 3 on the second line.
- If on the first line there is the sequence 1 2 4 5, there are 4 possibilities for the last place: 6, 7, 8 and 9.
- If 4 goes on the second line, than on the first there will be 1 2 5 6 followed by 7, 8 or 9 '?? 3 possibilities
- If 5 goes too on the second line, than 1 2 6 7 will be followed by 8 or 9 '?? 2 possibilities.
So, for the group 3 x x x 10 1 2 x x x , there are (4 + 3 + 2) possibilities. Studying all the possibilities further, I arrived at the conclusion that for 10 people, I find the following number of possibilities:
(5 + 4 + 3 + 2) + (4 + 3 + 2) + (3 + 2) +(4 + 3 + 2) + (3 + 2) = 14 +9 +5 + 9 + 5 = 42

I know how to do this but I am looking for a different approach. and by the way the answer is 42
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by saketk » Fri Sep 09, 2011 10:28 am
aplavakarthik wrote:- I start with 1 2 3 4 in the first line, and there are 5 possibilities for the last position: 5, 6, 7, 8 and 9.
- If 4 goes on the second line, than in the first there will be 1 2 3 5, and there are 4 possibilities for the last place: 6, 7, 8 and 9.
- If 5 goes besides 4 on the second line, than in the first there will be 1 2 3 6 and there are 3 possibilities for the last place: 7, 8 and 9.
- If 6 goes also on the second, there will be 1 2 3 7 8 or 1 2 3 7 9 in the first. So, for the sequence 1 2 3 in the first line, there are (5 + 4 + 3 + 2) possibilities.
I change 3 on the second line.
- If on the first line there is the sequence 1 2 4 5, there are 4 possibilities for the last place: 6, 7, 8 and 9.
- If 4 goes on the second line, than on the first there will be 1 2 5 6 followed by 7, 8 or 9 '?? 3 possibilities
- If 5 goes too on the second line, than 1 2 6 7 will be followed by 8 or 9 '?? 2 possibilities.
So, for the group 3 x x x 10 1 2 x x x , there are (4 + 3 + 2) possibilities. Studying all the possibilities further, I arrived at the conclusion that for 10 people, I find the following number of possibilities:
(5 + 4 + 3 + 2) + (4 + 3 + 2) + (3 + 2) +(4 + 3 + 2) + (3 + 2) = 14 +9 +5 + 9 + 5 = 42


I know how to do this but I am looking for a different approach. and by the way the answer is 42

Either your question is posted incorrectly or you are solving a different question.

Read your question once again, please! Your question says 3 in each row.
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by gmatboost » Fri Sep 09, 2011 4:12 pm
Each set of 6 people is basically like any other set of 6 people:
The six can be ranked in order of height 1, 2, 3, 4, 5, 6

So we need to solve this for 1-2-3-4-5-6 and then multiply by "10 choose 6" because that's how many different groups there are of 6 people.

For any group the shortest must be in the front left and the tallest in the back right:

_ _ 6
1 _ _

Now, from there, there are two options for the second shortest:

_ _ 6 2 _ 6
1 2 _ 1 _ _

In the first one, we can fill in either 3, 4, or 5 in the slot next to 2, so there are 3 possibilities.

In the second one, we must put 3 next to 1, and then we can fill in 4 and 5 in the other two slots, so there are two possibilities.

So, for each set of 6 people, there are 5 possibilities.

Since 10C6 = 10*9*8*7/(4*3*2*1) = 10*9*7/(3) = 10*3*7/(1) = 210, the total number of possibilities is 1050.

I don't happen to believe that this question is a realistic GMAT question, but I guess it tests the same concepts.

All of the answers aren't even close, so obviously something is wrong with the phrasing or the answers.
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by saketk » Fri Sep 09, 2011 8:38 pm
Hi Greg, thanks a lot. I totally agree with you. The number of possibilities is indeed way too
Different from the options given.
That's what in pointed out in my last post when the original poster posted some unusual solution.
Just couple of mins back I googled for this question and found that this question is not at all posted correctly.

The original question says there are only 6 men and they have to be arranged in 2 rows of 3 each

The next line is exactly the same. Increasing from left to right and the each person standing in the 2nd row is taller than the person standing in front of him or her.

This question says 10 people instead of 6.
The original question is much easier. And like I said in my previous post for case 1 there are total of 5 arrangements possible.

The correct answer for that question is also 5.
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by aplavakarthik » Sat Sep 10, 2011 3:37 am
Small change,

A photographer will arrange 10 people of 10 different heights for photograph by placing them in two rows of five 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 10 people are possible?

a.41
b.42
c.38
d.78
e.91
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by saketk » Sat Sep 10, 2011 7:15 am
That small change is known as a very important change.
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