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MG1105
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Bunuel
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

This was the last question in my Official GMAT mock, Spent 7min 33 Secs, got it right. :)
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can't have 436 in the top row!
srini123
I got 6 as answer too, manually arranging assuming 1,2,3,4,5,6 as heights (since each is different from other)

so possible positions are as below:

456 436
123 125

346 356
125 124

246 256
135 134

Kalpesh logic looks right to me... and is faster than listing possibilities
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There is something called catalan method for this given by 1/n+1* C(number of cans/number of cans in 1 row)
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Good thinking, but 6C3 = 20 overcounts because not every selection of 3 people for the front row is valid!

Simple example to see why:

Label people 1 (shortest) to 6 (tallest).

Say you pick {4, 5, 6} for the front row.
Then {1, 2, 3} must go to the back row.

Arranged by height:
Quote:
Back row: 1 — 2 — 3
Front row: 4 — 5 — 6

Now check: Is each back person taller than the front person?
- Column 1: Is 1 > 4? NO!

This selection breaks the rule, so it doesn't count.

Key insight: The 3 tallest people cannot ALL be in the front row — there'd be no one tall enough to stand behind them!

Many of your 20 selections have this same problem. When you carefully count only the valid ones, you get 5.

Answer: A (5)
MG1105
Bunuel
Basically the only option we have is to choose the three people who will stand in the front row. All other permutation will be governed by the height order and hence is 1 (since there can only be one way to line up people as per their height unless the height is same). So the answer should be 6C3, right? Just wanted to confirm if this is the correct approach.
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