组合数学 (Fall 2019)/Problem Set 1: Difference between revisions
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== Problem 3 == | == Problem 3 == | ||
In each of the following cases, firstly determine <math>a_n</math>, and then directly give the generating function | In each of the following cases, firstly determine <math>a_n</math>, and then directly give the closed form for generating function <math>A(x)=\sum_{n\ge 0}a_n x^n</math>. | ||
#<math>a_n</math> is the number of ways of distributing <math>n</math> identical objects into 4 distinct boxes; | #<math>a_n</math> is the number of ways of distributing <math>n</math> identical objects into 4 distinct boxes; | ||
#<math>a_n</math> is the number of ways of distributing <math>n</math> identical objects into 4 distinct boxes so that no box is empty; | #<math>a_n</math> is the number of ways of distributing <math>n</math> identical objects into 4 distinct boxes so that no box is empty; | ||
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== Problem 4 == | == Problem 4 == | ||
Let <math>(a_n)</math> be a sequence of numbers satisfying the recurrence relation: <math> a_n-p \cdot a_{n-1}+(p-q)\cdot q \cdot a_{n-2}=0</math> with <math>a_0=1</math> and <math>a_1=p</math>, where <math>p</math> and <math>q</math> are distinct nonzero constants. Solve the recurrence relation. | Let <math>(a_n)</math> be a sequence of numbers satisfying the recurrence relation: | ||
:<math> a_n-p \cdot a_{n-1}+(p-q)\cdot q \cdot a_{n-2}=0</math> | |||
with initial condition <math>a_0=1</math> and <math>a_1=p</math>, where <math>p</math> and <math>q</math> are distinct nonzero constants. Solve the recurrence relation. | |||
== Problem 5 == | == Problem 5 == |
Revision as of 04:00, 17 September 2019
Under construction
- 每道题目的解答都要有完整的解题过程。中英文不限。
Problem 1
Suppose that there are [math]\displaystyle{ n }[/math] red balls and another [math]\displaystyle{ m }[/math] balls which are distinct and not red. Balls with the same color are indistinguishable. Find the number of ways to select [math]\displaystyle{ r }[/math] balls simultaneously from these [math]\displaystyle{ n+m }[/math] balls, in each of the following cases:
- [math]\displaystyle{ r\leq m,\ r\leq n }[/math];
- [math]\displaystyle{ n\leq r\leq m }[/math];
- [math]\displaystyle{ m\leq r\leq n }[/math].
Problem 2
李雷和韩梅梅竞选学生会主席,韩梅梅获得选票 [math]\displaystyle{ p }[/math] 张,李雷获得选票 [math]\displaystyle{ q }[/math] 张,[math]\displaystyle{ p\gt q }[/math]。我们将总共的 [math]\displaystyle{ p+q }[/math] 张选票一张一张的点数,有多少种选票的排序方式使得在整个点票过程中,韩梅梅的票数一直高于李雷的票数?等价地,假设选票均匀分布的随机排列,以多大概率在整个点票过程中,韩梅梅的票数一直高于李雷的票数。
Problem 3
In each of the following cases, firstly determine [math]\displaystyle{ a_n }[/math], and then directly give the closed form for generating function [math]\displaystyle{ A(x)=\sum_{n\ge 0}a_n x^n }[/math].
- [math]\displaystyle{ a_n }[/math] is the number of ways of distributing [math]\displaystyle{ n }[/math] identical objects into 4 distinct boxes;
- [math]\displaystyle{ a_n }[/math] is the number of ways of distributing [math]\displaystyle{ n }[/math] identical objects into 4 distinct boxes so that no box is empty;
- [math]\displaystyle{ a_n }[/math] is the number of ways of distributing [math]\displaystyle{ n }[/math] identical objects into 4 identical boxes so that no box is empty;
- [math]\displaystyle{ a_n }[/math] is the number of ways of distributing [math]\displaystyle{ n }[/math] identical objects into 4 identical boxes.
Problem 4
Let [math]\displaystyle{ (a_n) }[/math] be a sequence of numbers satisfying the recurrence relation:
- [math]\displaystyle{ a_n-p \cdot a_{n-1}+(p-q)\cdot q \cdot a_{n-2}=0 }[/math]
with initial condition [math]\displaystyle{ a_0=1 }[/math] and [math]\displaystyle{ a_1=p }[/math], where [math]\displaystyle{ p }[/math] and [math]\displaystyle{ q }[/math] are distinct nonzero constants. Solve the recurrence relation.
Problem 5
Consider the decimal expansion [math]\displaystyle{ 1/9899=0.00010203050813213455\cdots }[/math]
Why do the Fibonacci numbers 1,2,3,5,8,13,21,34,55 appear?
Problem 6
Let [math]\displaystyle{ f(n,r,s) }[/math] denote the number of subsets [math]\displaystyle{ S }[/math] of [math]\displaystyle{ [2n] }[/math] consisting of [math]\displaystyle{ r }[/math] odd and [math]\displaystyle{ s }[/math] even integers, with no two elements of [math]\displaystyle{ S }[/math] differing by [math]\displaystyle{ 1 }[/math]. Give a combinatorial proof that [math]\displaystyle{ f(n,r,s)=\binom{n-r}{s}\binom{n-s}{r} }[/math].