07Ma6aSol2

07Ma6aSol2 - Math 6a - Solution Set 2 Problem # 1 Is Liu,...

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Math 6a - Solution Set 2 Problem # 1 Is Liu, Example 2.14 correct? Solution. In fact, it is. Problem # 2: Liu 2.32 Find the number of permutations of the letters a,b,c,d,e,f,g so that neither the pattern beg nor the pattern cad appears. ( Hint : Use the principle of inclusion and exclusion.) Solution. Let A be the set of all permutations on the above set of letters, let B be the set of all such permutations containing the pattern beg, and let C be the set of all such permutations containing the pattern cad. Then, the number of all such permutations containing neither pattern is | A - ( B C ) | = | A |-| B C | (since B C is a proper subset of A ).By the principle of inclusion and exclusion, we have | B C | = | B | + | C | - | B C | . Further, it is clear that | A | = 7! = 5040 . To count | B | note that we can consider B to be the set of permutations on the 5 symbols a,c,d,f,beg, ie we count the pattern beg as its own symbol, as those three letters must appear together and in that order. Thus | B | = 5! = 120 and similarly | C | = 5! = 120 . We count | B C | by counting permutations on the symbols f,beg,cap of which there are 3! = 6 . Thus we have: | A | - | B C | = 5040 - 120 - 120 + 6 = 4806 . Problem # 3: Liu 3.22 Let P be an arbitrary partially ordered set, and L be a chain of two elements. Let Q denote the cartesian product P × L. Let A be an antichain in Q. Let B be the largest possible subset of P such that B does not contain a chain of length exceeding 2 . Show that | A | ≤ | B | . Solution. Note that an arbitrary element q Q is of the form q = ( a,b ) where a P and b L. Consider the function f : Q P defined by f ( a,b ) = a. Then, define the set A 0 P by A 0 = { f ( q ) : q A } . 1
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2 Then, I claim that | A | = | A 0 | . To prove this, note that it is sufficient to show that for any q,q 0 A with q 6 = q 0 we have f ( q ) 6 = f ( q 0 ) . For if this is true, we have a bijective correspondence between the elements of A and A 0 (simply correspond q with f ( q )). So assume it is not true, ie that we have q,q 0 A with q 6 = q 0 but f ( q ) = f ( q 0 ) . Then, if q = ( a,b ) and q 0 = ( a 0 ,b 0 ) , as f ( q ) = a and f ( q 0 ) = a 0 we have a = a 0 . Since q 6 = q 0 we must have b 6 = b 0 . But as b,b 0 L and L is a chain, b,b 0 must be comparable. So, without loss of generality, b < b 0 (relabel q,q 0 if necessary). But then, we have ( a,b ) < ( a,b 0 ) and thus q < q 0 , contradicting the fact that A is an antichain. Thus, | A | = | A 0 | . So now I claim that A 0 does not contain a chain of length exceeding 2 . So again, assume this is false. Then we have a 1 ,a 2 ,a 3 A 0 with a 1 < a 2 < a 3 (since if we have any chain of length greater than 2 it contains a sub-chain of length 3). Then, as each a i A 0 we must have q 1 ,q 2 ,q 3 A with f ( q i ) = a i , for each i ∈ { 1 , 2 , 3 } . So define b i by q i = ( a i ,b i ) . Note that A is an antichain, so the three elements ( a 1 ,b 1 ) , ( a 2 ,b 2
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This note was uploaded on 04/28/2010 for the course MATH 6A taught by Professor Wilson during the Fall '07 term at Caltech.

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07Ma6aSol2 - Math 6a - Solution Set 2 Problem # 1 Is Liu,...

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