Tag Info

Hot answers tagged

6

First, assume you choose the cards in order of pair 1, pair 2, non pair card. The first card in pair 1 can be any of 52 cards. The second card in pair 2 can be one of 3 cards (the other cards of the same number). The first card of pair 2 can be one of the 48 cards which are not of the number in pair 1. The second card of pair 2 is one of the 3 cards of the ...


1

Supposing your constituent values are $0,1,2,\ldots,q$ the answer is given by the following Polya Enumeration formula: $$[z^N] Z(S_k)\left(\sum_{m=0}^q z^m\right).$$ In the example we have $k=3$ and $N=6$ so we obtain the generating function $$f(z) = [z^6] Z(S_3)\left(\sum_{m=0}^4 z^m\right).$$ Now $$Z(S_3) = ...


1

You’ve made a pretty good start. You’re right that an $n$-cycle has $n$ spanning trees. Another way to explain it is to notice that deleting one edge leaves $n$ vertices and $n-1$ edges, so you have a tree; clearly that tree spans the cycle, and there are $n$ possible edges to remove, so there are $n$ spanning trees. With $K_4$, the tetrahedron, you got $4$ ...


1

For $A\subseteq L$ let $U(A)=\{u\in U:\ell<u\text{ for some }\ell\in A\}$. Let $$r=\max\{|A|-|U(A)|:A\subseteq L\}\;;\tag{1}$$ then $|U(A)|\ge|A|-r$ for all $A\subseteq L$, so there is an $A\subseteq L$ such that $|A|=|L|-r$, and $\{U(a):a\in A\}$ has a transversal (SDR). Let $T$ be a transversal for $\{U(a):a\in A\}$, and for each $a\in A$ let $u_a$ be ...


1

Recall the species of set partitions with the constituents marked which is $$\mathfrak{P}(\mathcal{U}\mathfrak{P}_{\ge 1}(\mathcal{Z}))$$ which gives the generating function $$G(z, u) = \exp(u(\exp(z)-1)).$$ It follows that the exponential generating function of Bell numbers is given by $$G(z) = \exp(\exp(z)-1).$$ Suppose we are trying to compute ...



Only top voted, non community-wiki answers of a minimum length are eligible