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number theory

by priyankagumber » Wed Jul 22, 2009 9:06 am
Thirty six numbers are filled in the cells of a matrix as shown in the figure given below. Six numbers are chosen from the matrix such that no two numbers belong to the same row or the same column. In how many ways can the numbers be chosen?

1 2 3 4 5 6
7 8 9 10 11 12
13 14 15 16 17 18
19 20 21 22 23 24
25 26 27 28 29 30
31 32 33 34 35 36
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by shibal » Wed Jul 22, 2009 9:10 am
IMO 6C1*5C1*4C1*3C1*2C1*1C1
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by real2008 » Thu Jul 23, 2009 10:49 am
answer is 36*25*16*9*4*1
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by pradeepsarathy » Thu Jul 23, 2009 5:28 pm
IMO - 6 X 5 X 4 X 3 X 2 X 1 = 720 ways
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by PussInBoots » Fri Jul 24, 2009 11:08 pm
36 combinations to pick 1st number * 25 combinations to pick 2nd number = 900 already. Most of people in this thread are wrong. Answer is 36 * 25 * 16 * 9 * 4 * 1
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by prindaroy » Sat Jul 25, 2009 12:05 am
PussinBoots is correct.

Think of it this way; First number you pick, you have 36 choices. Next number you pick, you only have 25 choices since the second number can't be from row x or column x, third pick only 16, and so on so forth you have

36*25*16*9*4*1....

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by Ian Stewart » Sat Jul 25, 2009 4:17 am
As I read the question, the answer should be 6!, and not (6!)^2 (which is the same as 6^2 * 5^2 * 4^2 * 3^2 * 2^2 * 1^2). If you arrive at the answer (6!)^2, then you're assuming the order of the selection matters; that is, you're considering the selection (6, 11, 16, 21, 26, 31) to be different from the selection (31, 26, 21, 16, 11, 6), and you're counting them twice. Of course, there are 6! orders in which you could sequence each set of numbers, so the answer (6!)^2 is too large by a factor of 6!. Divide by 6!, and you'll get the right answer if order does not matter: 6!.

It's easier just to fix a particular order - we have six choices we can make from the first row, five from the second, etc, which gives the answer 6!.
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