Thread: Arrangements

A photographer is taking a photograph of 10 people. He has to arrange them, all of different heights, in two rows of five, one behind the other. Each person at the back must be taller than the person directly in front of them. Along the rows the heights must increase from left to right. In how many ways can the people be arranged.

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What I got so far:

the arrangement would be something like this

F G H I J
A B C D E

let's says the 10 people is named according to their heights. From 1~ 10, 1 is the shortest, 10 is the tallest.

i) Where A must be 1 and J must be 10
ii)2 must be put at either B or F
iii)9 must be put at either I or E
iv)listing the possible arrangement is out of question, too many ways

that is as far as I can think

Any help are appreciated.
Thanks in advance.

Start with simple cases;

No. of people 2 4 6 8
No of possibilities 1 2 5 14

Start with cases with need choices case:6 people
at 6, when its,
3 x 6
1 2 [ ] <- 2 possible answers
when its,
2 x 6
1 3 [ ] <- 2 possible answers ;; as 2/3 change placed its considered as +1 possibility

at 6 people (3+2)=5

at 8 people !refer people on the 1st row: (4+3+2)+(3+2)=14

at 10 people: (5 + 4 + 3 + 2) + (4 + 3 + 2) + (3 + 2) +(4 + 3 + 2) + (3 + 2) = 14 +9 +5 + 9 + 5 = 42

at 12 people FYI: (6 + 5 + 4 + 3 + 2) + (5 + 4 + 3 + 2) + (4 + 3 + 2) + ( 3 + 2) + (5 + 4 + 3 + 2) + (4 + 3 + 2) + ( 3 + 2) + (4 + 3 + 2) + ( 3 + 2) + (5 + 4 + 3 + 2) + (4 + 3 + 2) + ( 3 + 2) + (4 + 3 + 2) + ( 3 + 2) = 48 + 28 +14 + 28 + 14 = 132

If you would like to read how one primary school worked on the problem then find an article in the Mathematics Teaching number 188 (published by the Association of Teachers of Mathematics).

Thanks for the explanations and helps.

Well, I assume this is the article you're talking about.
http://www.atm.org.uk/journal/archiv...T188-34-37.pdf

Is it suppose to be a shame for a college student to hardly comprehend the solution while some primary students actually solved it.