A Stirling Encounter with Harmonic Numbers - HMC Math
... becomes the distinguished cycle (b1 · · · b j ). The distinguished cycle is then inserted among the cycles Ck−1 , . . . C2 , C1 , which are generated one at a time as follows: C1 (the rightmost cycle) begins with a1 followed by a2 and so on until we encounter a number ai that is less than a1 . Assum ...
... becomes the distinguished cycle (b1 · · · b j ). The distinguished cycle is then inserted among the cycles Ck−1 , . . . C2 , C1 , which are generated one at a time as follows: C1 (the rightmost cycle) begins with a1 followed by a2 and so on until we encounter a number ai that is less than a1 . Assum ...
Permutations and combinations
... When you multiply a whole number by all the whole numbers below it, it’s called the factorial of that number. Factorials are used to compute permutations. In the previous example, we saw that 4 letters could be permuted 4 x 3 x 2 x 1 ways, or 24 ways. Six things can be permuted 6 x 5 x 4 x 3 x 2 x 1 ...
... When you multiply a whole number by all the whole numbers below it, it’s called the factorial of that number. Factorials are used to compute permutations. In the previous example, we saw that 4 letters could be permuted 4 x 3 x 2 x 1 ways, or 24 ways. Six things can be permuted 6 x 5 x 4 x 3 x 2 x 1 ...
Permutations and determinants Math 130 Linear Algebra
... σ, denoted sgn σ, is defined to be 1 if σ is an even the determinant of A. Thus, reading down the first permutation, and −1 if σ is an odd permutation. column, we see that the determinant starts out with the following six terms: Construction of the determinant. The de+a11 a22 a33 a44 − a11 a22 a34 a ...
... σ, denoted sgn σ, is defined to be 1 if σ is an even the determinant of A. Thus, reading down the first permutation, and −1 if σ is an odd permutation. column, we see that the determinant starts out with the following six terms: Construction of the determinant. The de+a11 a22 a33 a44 − a11 a22 a34 a ...
REU 2006 · Discrete Math · Lecture 2
... Take a permutation π uniformly at random. Let `1 be the length of the cycle containing 1. Since there are n possible places that 1 can go are all equally likely, we get Pr(`1 = 1) = n1 . Now let us consider the, Pr(`1 = n). There are n−1 choices for π(1), because 1 is excluded. = n1 . Then there are ...
... Take a permutation π uniformly at random. Let `1 be the length of the cycle containing 1. Since there are n possible places that 1 can go are all equally likely, we get Pr(`1 = 1) = n1 . Now let us consider the, Pr(`1 = n). There are n−1 choices for π(1), because 1 is excluded. = n1 . Then there are ...
Permutations with Inversions
... Clearly the number of permutations with no inversions, I n (0), is 1 for all n, and in particular I1 (0) = 1 = Φ1 (x). So the formula given in the theorem is correct for n = 1. Consider a permutation of n − 1 elements. We insert the nth element in the jth position, j = 1, 2, . . . , n, choosing the ...
... Clearly the number of permutations with no inversions, I n (0), is 1 for all n, and in particular I1 (0) = 1 = Φ1 (x). So the formula given in the theorem is correct for n = 1. Consider a permutation of n − 1 elements. We insert the nth element in the jth position, j = 1, 2, . . . , n, choosing the ...
MA 3260 Lecture 14 - Multiplication Principle, Combinations, and
... Answers: Quiz 14B: 1) 161,700. 2) These are all of the 8-element subsets, all the 7-element subsets, etc. In other words, these are all the subsets of an 8-element set. The answer is 28 = 256. Homework 14: 1a) 479,001,600. b) 9. c) 9. d) 9900. e) 480700. 2) We’re missing 9 C9 and 9 C0 , which are bo ...
... Answers: Quiz 14B: 1) 161,700. 2) These are all of the 8-element subsets, all the 7-element subsets, etc. In other words, these are all the subsets of an 8-element set. The answer is 28 = 256. Homework 14: 1a) 479,001,600. b) 9. c) 9. d) 9900. e) 480700. 2) We’re missing 9 C9 and 9 C0 , which are bo ...
Solutions. - University of Bristol
... are both even permutations (since they are cycles of odd length. Then στ = (1, 2)(3, 4), but τ σ = (1, 3)(2, 4), so στ 6= τ σ and hence An is non-abelian. So An is abelian if and only if n ≤ 3. 2. For which integers d does the alternating group A8 have elements of order d? Justify your answer. Solut ...
... are both even permutations (since they are cycles of odd length. Then στ = (1, 2)(3, 4), but τ σ = (1, 3)(2, 4), so στ 6= τ σ and hence An is non-abelian. So An is abelian if and only if n ≤ 3. 2. For which integers d does the alternating group A8 have elements of order d? Justify your answer. Solut ...
Basic Counting
... [n] to [k] that exclude i from their range}. For instance, A1 is the set of functions that map nothing to 1. Consequently the intersection A2A4 is the set of functions that map nothing to 2 or 4. Can you calculate |S|, |A1|, |A2A4|, and, in general, |A1A2…Ai|? (2) We note that in general the size of ...
... [n] to [k] that exclude i from their range}. For instance, A1 is the set of functions that map nothing to 1. Consequently the intersection A2A4 is the set of functions that map nothing to 2 or 4. Can you calculate |S|, |A1|, |A2A4|, and, in general, |A1A2…Ai|? (2) We note that in general the size of ...
Even and Odd Permutations
... N (π) + 1 if a and b belong to the same cycle of π N ((a, b)π) = N (π) − 1 if a and b belong to different cycles of π. Proof. If a and b belong to one cycle, then (a, b) (a, c1 , c2 , c3 , . . . , ck , b, d1 , d2 , . . . , dℓ ) = (a, c1 , c2 , c3 , . . . , ck ) (b, d1 , d2 , . . . , dℓ ), thus incre ...
... N (π) + 1 if a and b belong to the same cycle of π N ((a, b)π) = N (π) − 1 if a and b belong to different cycles of π. Proof. If a and b belong to one cycle, then (a, b) (a, c1 , c2 , c3 , . . . , ck , b, d1 , d2 , . . . , dℓ ) = (a, c1 , c2 , c3 , . . . , ck ) (b, d1 , d2 , . . . , dℓ ), thus incre ...