Grades 11-12 Video Solutions 2012-2012_11-12

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How many permutations x1, x2, x3, x4 of the set of integers 1, 2, 3, 4 have the property that the sum x1, x2, plus x2, x3, plus x3, x4, plus x4, x1 is divisible by 2?  So first we see that these two terms share an x2, and these two terms share an x4, and so this thing factors as x1 plus x3 times x2 plus x4.  Now we want this thing to be a multiple of 3, so that means at least one of the factors has to be a multiple of 3.  So let's see how many ways there are to take two integers out of here and have them be a multiple of 3.  Well one way to do that is to do 1 plus 2, and the other way to do that is 2 plus 4.  Notice that when we do this actually, when we do 1 plus 2, that means the other sum has to be 3 plus 4,  which isn't a multiple of 3. And when we do 2 plus 4, the other sum has to be 1 plus 3, which also isn't a multiple of 3. That means exactly one of these factors right here has to  be a multiple of 3. So I guess without loss of generality, let's just make it the first one. At the end of the day we're going to multiply it by 2 if it was in the opposite order.  So that means x1 x3 has to be either 1 2 or 2 4, and in that case x2 x4 has to be either 3 4 or 1 3.  So now we can make this happen in two ways, with x1 equals 1 and x3 equals 2, or x1 equals 2 and x3 equals 1.  So we have two options for this, and then similarly we have two options for this. So we multiply them together to get four options in the first way.  Or we could do it the second way to get four more options, so that's eight ways in total. And now we multiply by two, because who said that x1 x3 is the multiple of 3? What if this  term was the multiple of 3? Then everything would just be swapped around. So in total we have 16 permutations that do the trick.

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