07Ma6aSol5

07Ma6aSol5 - Math 6a - Solution Set 5 Problem # 1 Show that...

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Unformatted text preview: Math 6a - Solution Set 5 Problem # 1 Show that S ( n, 2) = 2 n- 1- 1 and S ( n, 3) = 1 2 (3 n- 1- 2 n + 1) for n 1 . Solution. Note that S ( n, 2) counts the number of partitions of an n element set S into 2 non-empty subsets. So if we call the sets A and B, each element in S can either be placed in A or B, yielding 2 n partitions. However, this counts the partition where A is empty and B = S and the partition where B is empty and A = S, both of which are not partitions into two non-empty subsets (all other partitions have both A and B non-empty). Further, each partition will be counted twice, as we can relabel A and B and still have the same partition. Thus S ( n, 2) = 1 2 (2 n- 2) = 2 n- 1- 1 . We prove the second result by induction. Note that if n = 1 , we clearly cannot partition an n element set into 3 non-empty subsets, so S ( n, 3) = 0 = 1 2 (3 n- 1- 2 n +1) . So assume that it is true for n = m. Then consider the case n = m + 1 . Note that we have the formula (from class) S ( n,k ) = kS ( n- 1 ,k ) + S ( n- 1 ,k- 1) so, by induction and our result in part (a), S ( m + 1 , 3) = 3 S ( m, 3) + S ( m, 2) = 3( 1 2 (3 m- 1- 2 m + 1)) + 2 m- 1- 1 S ( m + 1 , 3) = 1 2 (3 m )- 3(2 m- 1 ) + 2 m- 1 + 3 2- 1 = 1 2 (3 m )- 2 m + 1 2 S ( m + 1) = 1 2 (3 m +1- 1- 2 m +1 + 1) which finishes the induction and shows, for all n 1 , that S ( n, 3) = 1 2 (3 n- 1- 2 n + 1) . Problem # 2 A partition has type 1 k 1 2 k 2 ...r k r if it has k i blocks of size i for i = 1 ,...,r. Show that the number of partitions of an n-element set of type 1 k 1 2 k 2 ...r k r is n ! Q r m =1 ( m !) k m ( k m )! . 1 2 Solution. We count the number of pairs ( S,T ) where S is a permutation of { 1 , 2 ,...,n } written as a sequence ( a 1 ,a 2 ,...,a n ) of distinct elements of { 1 , 2 ,...,n } and T is the parti- tion of { 1 , 2 ,...,n } of type 1 k 1 2 k 2 ...r k r formed by taking the first k 1 of the a i as singleton sets, the next 2 k 2 of the a i consecutively as pairs of the form { a i ,a i +1 } , the next 3 K 3 of the a i consecutively in triples of the form a i ,a i +1 ,a i +2 , etc (in general we take the next jk j of the a i consecutively in j-element sets of the form { a i ,a i +1 ,...,a i + j- 1 } ). So, for example, if n = 10 and if k 1 = 3 , k 2 = 2 , and k 3 = 1 , if S = (4 , 1 , 3 , 2 , 5 , 8 , 6 , 9 , 7 , 10) we will have T = {{ 4 } , { 1 } , { 3 } , { 2 , 5 } , { 8 , 6 } , { 9 , 7 , 10 }} . Thus, if N is the number of all such pairs ( S,T ) it follows that, since for each S we have a unique T, and there are n ! such permutations S we must have N = n ! . So consider any T of type 1 k 1 2 k 2 ...r k r . To get S, we first choose an order on each of the elements of each block. A block with m elements can be ordered in m ! ways, and there are k m such blocks for each m, so the number of ways to order the elements in each block is r Y m =1 ( m !) k m ....
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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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07Ma6aSol5 - Math 6a - Solution Set 5 Problem # 1 Show that...

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