Question 1199570: How many ways are there to arrange the first ten positive integers such that the multiples of 2 appear in increasing order, and the multiples of 3 appear in decreasing order?
Answer by greenestamps(13200)   (Show Source):
You can put this solution on YOUR website!
I'll take a crack at this and wait to see if another tutor can find a different answer and point out a fault in my analysis.
(1) 6 is the only one of the first ten positive integers that is a multiple of both 2 and 3.
(2) The integers 3, 8, and 10 must be after the 6; they can be in any of 3 orders -- (a) 3,8,10 (b) 8,3,10 (c) 8,10,3
(3) Similarly the integers 2, 4, and 9 must be before the 6; they can be in any of 3 orders -- (a) 2,4,9 (b) 2,9,4 (c) 9,2,4
So there are 3*3 = 9 ways to arrange the 7 integers 2, 3, 4, 6, 8, 9, and 10.
After those integers are arranged, the integers 1, 5, and 7 remain. There are no restrictions on where those integers can be in the sequence. So...
(4a) There are 3 ways to choose one of the 3 remaining integers to be the next one to be placed in the sequence; since there are currently 7 integers in the sequence, there are 7+1 = 8 places in the sequence where this integer can be placed.
(4b) There are 2 ways to choose one of the 2 remaining integers to be the next one to be placed in the sequence; since there are now 8 integers in the sequence, there are 8+1 = 9 places in the sequence where this integer can be placed.
(4c) There is only 1 way to choose remaining integer to be the next (last) one to be placed in the sequence; since there are now 9 integers in the sequence, there are 9+1 = 10 places in the sequence where this integer can be placed.
So the number of ways to place the integers 1, 5, and 7 in the sequence is
(3*8)(2*9)(1*10) = 4320
And so the number of ways to arrange the first ten positive integers such that the multiples of 2 appear in increasing order and the multiples of 3 appear in decreasing order is
ANSWER: 9*4320 = 38880
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