Women posing differently, can somebody help?

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indir0ver: Senior | Next Rank: 100 Posts; Posts: 45; Joined: Fri Jun 12, 2009 2:42 am; Location: UK

Women posing differently, can somebody help?

by indir0ver » Wed Jul 22, 2009 10:48 am

Eight women of eight different heights are to pose for a photo in two rows of four. Each woman in the second row must stand directly behind a shorter woman in the first row. In addition, all of the women in each row must be arranged in order of increasing height from left to right. Assuming that these restrictions are fully adhered to, in how many different ways can the women pose?

(A) 2
(B) 14
(C) 15
(D) 16
(E) 18

Ans is 14

looks like its 700+ types question definitely, can somebody also explain how they solved. thanks in advance

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Source: Beat The GMAT — Problem Solving | Wed Jul 22, 2009 10:48 am

m&m: Master | Next Rank: 500 Posts; Posts: 134; Joined: Sun Feb 15, 2009 7:44 pm; Thanked: 14 times

by m&m » Wed Jul 22, 2009 1:28 pm

let's order women by numbers from smallest=1 to tallest = 8

5 6 7 8
1 2 3 4

so 1 cannot move (because it is always smallest)
and 8 cannot more (because it is always tallest)

so lets tackle by looking at the number of changes if we move 1 person, 2 persons, 3, persons, etc

if we move 1 person we can get following combination

2 6 7 8
1 3 4 5

or

3 6 7 8
1 2 4 5

or

4 6 7 8
1 2 3 5

notice how lower numbers always go to top left and higher numbers always go to bottom right so in general with 1 move we can have 6 possible combinations. 6C1

With 2 moves we can have 6C2 moves = 10

we cannot move 3 or more women at same time...try it you'll see you cannot

so answer is 6 + 10 = 16 = D

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El Cucu: Master | Next Rank: 500 Posts; Posts: 236; Joined: Fri Nov 28, 2008 9:16 am; Thanked: 9 times

Re: Women posing differently, can somebody help?

by El Cucu » Wed Jul 22, 2009 1:57 pm

indir0ver wrote:Eight women of eight different heights are to pose for a photo in two rows of four. Each woman in the second row must stand directly behind a shorter woman in the first row. In addition, all of the women in each row must be arranged in order of increasing height from left to right. Assuming that these restrictions are fully adhered to, in how many different ways can the women pose?

(A) 2
(B) 14
(C) 15
(D) 16
(E) 18

Ans is 14

looks like its 700+ types question definitely, can somebody also explain how they solved. thanks in advance

Very interesting...

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indir0ver: Senior | Next Rank: 100 Posts; Posts: 45; Joined: Fri Jun 12, 2009 2:42 am; Location: UK

by indir0ver » Thu Jul 23, 2009 3:21 am

@m&m,
I understand your explaination but the answer is 14. It is from MGMAT doc a friend sent me the question and i couldnt solve it. Answer if offical.

Can somebody please solve it for me?

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m&m: Master | Next Rank: 500 Posts; Posts: 134; Joined: Sun Feb 15, 2009 7:44 pm; Thanked: 14 times

by m&m » Thu Jul 23, 2009 4:39 am

My answer above is off, to start 6C2 is not 10 but 15.

We know that if we move 1 digit only, a 5 or 4 will get bumped down or up respectively. 5 if the digit is in the bottom row and 4 if the digit is in the top row. Let's go through possible answers

2 <-> 5
3 <-> 5
4 <-> 5
6 <-> 4
7 <-> 4

so 5 swaps when bumping digit 4&5 to bottom OR 3C1+3C1 - 1 (common) = 5

we can also swap 1 and force other digits (not 5 or 4) to bottom ie 2 <-> 6
2 <-> 6
2 <-> 7
3 <-> 6
3 <-> 7

so 4 swaps or 2C1 + 2C1 = 4

when moving 2 digits we cannot move any digit which does not have 2 smaller digits (front row)/higher digits (back row) beside it. So cannot move 2&3 to top and 6&7 to bottom. Let's enumerate possibilities

2&4 <-> 5&6
2&4 <-> 5&7
3&4 <-> 5&6
3&4 <-> 5&7

So we have done every single move, double move, and we cannot move 3 pieces. Only additional combination is if we have 0 moves (original setup).

so let's add them up

5+4+4+1(original) = 14

This method works but is prone to errors when working quick, I'm sure there must be a FASTER way too.

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kaulnikhil: Master | Next Rank: 500 Posts; Posts: 338; Joined: Fri Apr 17, 2009 1:49 am; Thanked: 9 times; Followed by:3 members

by kaulnikhil » Thu Jul 23, 2009 5:55 am

let us consider the ages as
20 2122 23
24 252627
this is the first arrangement we have
now replace 23 by 24 so we have
20 21 22 24
23 25 26 27
similarly replace 22 by 24
20 21 23 24
22 25 26 27
simlarly replace 21 by 24
20 22 23 24
21 25 26 27

u cant replace 20 by 24 as it wont satisfy the conditions
repeat the same steps for 25 . and 26
so we will have 1 +3+3+3 = 10
now replace two numbers in upper row
let us start with 24 and 25 ..replace 22 and 23
20 22 24 25
22 23 26 27
now replace 21 and 23
20 22 24 25
21 23 26 27

we cant replace the same steps for 25 and 26 as it wont satisfy the copnditions
now choose 24 and 26--replace 22 and 23
20 21 24 26
22 23 25 27
now replace 21 and 23
20 22 24 26
21 23 25 27
so we have 10 + 2 +2 = 14
hope i make sense

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tohellandback: Legendary Member; Posts: 752; Joined: Sun May 17, 2009 11:04 pm; Location: Tokyo; Thanked: 81 times; GMAT Score:680

by tohellandback » Thu Jul 23, 2009 6:03 am

I guess MGMAT is way tougher than the real thing.

anyways, the solution
let the heights in increasing order are 1..to 8

Y1 Y2 Y3 Y4
X1 X2 X3 X4

now its clear that X1=1, Y4=8 because no number is greater than 8 and no number is smaller that 1.

Y1 Y2 Y3 8
1 X2 X3 X4
Now Y3 must be >1,Y1,Y2,X2,X3
and Y3 must be <8
greater than 5 numbers and less than 8. so Y3 must be 6 or 7

Y2>1,X2,Y1
y2<8,Y3
Y2 can be 4,5,6

Y1>1
Y1<Y2,Y3,8
Y1 can be 2,3,4,5

so the cases are
when Y3=7:
Y2 can be 4,5,or 6
when Y2=6,Y1=2,3,4or 5 -total=4
when Y2=5,Y1=2,3,or 4 total=3
when Y2=4 ,Y1=2 or 3 total =2
when Y3=6:
Y2 can be 4 or 5
using the above counting: total=5

total no of ways=4+3+2+5=14

The powers of two are bloody impolite!!