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Combinations

Expert replies
by Anindya Madhudor » Thu Dec 13, 2012 8:44 am
Eight women of eight different heights are to pose for a photo in two rows of four. Each women 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

OA: B
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Source: — Problem Solving |

by GMATGuruNY » Thu Dec 13, 2012 7:27 pm
Anindya Madhudor wrote:Eight women of eight different heights are to pose for a photo in two rows of four. Each women 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

OA: B
Let the eight women be the integers 1 through 8, with 1 the shortest and 8 the tallest.

As the SHORTEST, 1 can't stand BEHIND anyone.
As the TALLEST, 8 can't stand IN FRONT OF anyone.
Thus, the positions of 1 and 8 are fixed:
1XXX
XXX8

Case 1: 2 in the front row
12XX
XXX8

A pair of women must stand to the right of 2.
From the 5 remaining women, the total number of pairs that can be formed = 5C2 = 10.
Of these 10 options, one pair -- 67 -- is not viable, since it would force 5 to stand behind 6:
1267
3458

Thus, the total number of VIABLE pairs that can stand to the right of 2 = 10-1 = 9.

Case 2: 2 in the back row, forcing 3 to stand next to 1
13XX
2XX8

A pair of women must stand to the right of 3.
From the 4 remaining women, the total number of pairs that can be formed = 4C2 = 6.
Of these 6 options, one pair -- 67 -- is not viable, since it would force 5 to stand behind 6:
1367
2458

Thus, the total number of VIABLE pairs that can stand to the right of 3 = 6-1 = 5.

Total options = 9+5 = 14.

The correct answer is B.

Here are all of the viable arrangements:

Case 1:
1234...1235...1236...1237...1245...1246...1247...1256...1257
5678...4678...4578...4568...3678...3578...3568...3478...3468

Case 2:
1345...1346...1347...1356...1357
2678...2578...2568...2478...2468
Last edited by GMATGuruNY on Wed Mar 13, 2013 12:03 pm, edited 1 time in total.
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by mariofelixpasku » Wed Mar 13, 2013 10:55 am
does the gmat come up with such questions?
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by Brent@GMATPrepNow » Wed Mar 13, 2013 11:06 am
I think this one is beyond what you should expect to see on test day.

Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
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by srcc25anu » Wed Mar 13, 2013 11:11 am
Mitch - for second Case with 2 in bottom row: the possible combinations are:

1357 1347 1346 1345 1356
2468 2568 2578 2678 2478

Otherwise great explanation!!!
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