16 combinatroics-2
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The document summarizes key topics covered in a discrete mathematics lecture: 1. The binomial theorem for expanding (a+b)n and calculating coefficients using binomial coefficients. 2. Counting combinations with repetition using partitioning and binomial coefficients. 3. An introduction to finite probability, defining sample spaces, events, and calculating probabilities as ratios of event sizes to sample space sizes.
16 combinatroics-2
- 1. JC Liu MACM101 Discrete Mathematics 1 Lecture 16: Combinations (2) The Binomial Theorem Combinations with Repetition Finite Probability (Intro.)
- 2. JC Liu MACM101 Discrete Mathematics 2 16.1. Binomial Theorem Consider the n-fold product (a+b)n : (a+b)(a+b)…(a+b) = an + n⋅an–1 b + n(n–1)/2⋅an–2 b2 +…+ bn ∑= − ⋅ =+ n 0k kknn ba k n )ba(Binomial Theorem (Thm 1.1): The coefficient of ak bn–k is C(n,k) Example: (a+b)4 = a4 +4⋅a3 b+6⋅a2 b2 +4⋅ab3 +b4 Binomial coefficient
- 3. JC Liu MACM101 Discrete Mathematics 3 16.1. Binomial Theorem Method 1: Given set A with n objects • C(n,k) is the number of subsets with k objects • Total number of subsets of A is 2n Method 2: Binomial theorem: • (1+1)n = 0 ( , ) ( ,0) ( ,1) ... ( , ) 2 n n k C n k C n C n C n n = = + + + =∑ ∑= − ⋅ =+ n 0k kknn ba k n )ba(Binomial Theorem (Thm 1.1): The coefficient of ak bn–k is C(n,k)
- 4. JC Liu MACM101 Discrete Mathematics 4 16.1. Binomial Theorem 0 ( 1) ( , ) ( ,0) ( ,1) ( ,2) ... ( 1) ( , ) ? n k n k C n k C n C n c n C n n = − = − + − + − =∑ ∑= − ⋅ =+ n 0k kknn ba k n )ba(Binomial Theorem (Thm 1.1): The coefficient of ak bn–k is C(n,k) •(1-1)n =
- 5. JC Liu MACM101 Discrete Mathematics 5 16.2. More Combinations Q1: How many 11-letter words out of “m i s s i s s i p p i”? Hint: There are 1 “m”, 4 “i”, 4 “s”, and 2 “p”; calculate the answer by positioning the letters by type: 11!/(1! 4! 4! 2!) = 34650. Q2: How many 11-letter words using “m”,”i”,”s”,”p”? -- like Q1, but no limit on the number of each letter -- duplication allowed -- order matters (mmmisssimpi <> mimmsssimip) Q3: How many 11-letter strings using “m”,”i”,”s”,”p” if ignoring order? -- mmmisssimpi <> mimmsssimip -- so we can consider alphabetized strings only
- 6. JC Liu MACM101 Discrete Mathematics 6 16.3. Combinations with Repetitions Question: How many alphabetized strings are there of length 3 containing letters a,b,c ? • Duplication allowed • Order doesn’t matter Solution 1: 10 • aaa, aab, aac, abb, abc, acc, bbb, bbc, bcc, and ccc. Solution 2: Partition • The only thing that matters is the numbers of each: a,b,c. • So, introduce 3 nonnegative counters na nb and nc. • New problem: How many combinations are there such that na+nb+nc=3? -- easy ? • Answer is 10: (3,0,0) (2,1,0) … (0,0,3).
- 7. JC Liu MACM101 Discrete Mathematics 7 16.3. Combinations with Repetitions Question: How many alphabetized strings are there of length 3 containing letters a,b,c (note: duplication allowed)? Solution 1: List • aaa, aab, aac, abb, abc, acc, bbb, bbc, bcc, and ccc. • Total 10 Solution 2: Partition 3 • (3,0,0) (2,1,0) … (0,0,3). • Visualize it – 3 spaces, divided by 2 separators = (3,0,0) and = (2,1,0) … and = (1,1,1) … and = (0,0,3). • Answer: C(5,2) = 10
- 8. JC Liu MACM101 Discrete Mathematics 8 16.3. Combinations with Repetitions Consider a list of k letters, out of a alphabet of n letters. How many different, alphabetized lists are possible? • Each letter can be used multiple times • Order doesn’t matter Let the 1st letter occur k1 times, the 2nd letter occur k2 times,…, the nth occur kn times. • New problem: How many different ways to have k1+…+kn = k? Consider k places for letters “”, with n–1 separators “” • Say, k=5, n=3, then 3 (k1) +0 (k2) +2 (k3) = 5 is represented by Answer: In general, there are C(n+k–1,k) ways.
- 9. JC Liu MACM101 Discrete Mathematics 9 16.3. Combinations with Repetitions Consider a list of k letters, out of a alphabet of n letters. How many different, alphabetized lists are possible? • New problem: How many different ways to have k1+…+kn = k? Answer: In general, there are C(n+k–1,k) ways. Question: How many different non-negative solutions for x1+x2+x3+x4=7? Answer: C(4+7-1,7)=120 k=7, n=4
- 10. JC Liu MACM101 Discrete Mathematics 10 16.3. Combinations with Repetitions Consider a list of k letters, out of a alphabet of n letters. How many different, alphabetized lists are possible? • New problem: How many different ways to have k1+…+kn = k? Answer: In general, there are C(n+k–1,k) ways. Question: A donut shop offers 20 kinds of donuts. Assume that there are at least 12 of each kind. Then how many ways we can select 12 donuts ? Answer: C(20+12-1,12) = 141,120,525 k=12, n=20 – so k can be < n, but still with repetition
- 11. JC Liu MACM101 Discrete Mathematics 11 Summarizing When counting the number if selections of k elements out of 1,…,n consider two crucial questions: Repetition: Can an element be picked more than once? Order: Does the order of picking matter? Ordering matters? Repetition allowed? C(n,k),k)1kC(nNo P(n,k)nYes NoYes k −+
- 12. JC Liu MACM101 Discrete Mathematics 12 “How to Solve It” Important observation: The previous 4 cases are not answers by themselves. Instead they are tools that you have to learn to use. It is often not immediately clear which tool to use; it is a skill to recognize which case you are dealing with. Tricks of the trade: When in doubt, try small examples. Verify your answers (look at extremes like k=0 or n=k). Practice, practice, practice.
- 13. JC Liu MACM101 Discrete Mathematics 13 “How to Solve It” Important observation: The previous 4 cases are not answers by themselves. Instead they are tools that you have to learn to use. It is often not immediately clear which tool to use; it is a skill to recognize which case you are dealing with. Tricks of the trade: When in doubt, try small examples. Verify your answers (look at extremes like k=0 or n=k). Practice, practice, practice.
- 14. JC Liu MACM101 Discrete Mathematics 14 16.4. Intro. Finite Probability Experiment: toss a fair coin, roll a fair die Sample space S: all possible outcomes • Toss coin 1 time: {H, T} • Toss coin 2 times: {HH, HT, TH, TT} • Same (or equal) likelihood of occurrence of each outcome Event: A subset A of S • Say, both are H • Can be empty set Then • Pr(A) = The probability that A occurs = |A|/|S| • For each a in S, we have Pr({a})= The probability of each outcome = 1/| S| • The equal likelihood assumption
- 15. JC Liu MACM101 Discrete Mathematics 15 16.4. Intro. Finite Probability Recall: • Suppose you flip a fair coin n times. How many different ways can you get • no heads? • exactly one head? • exactly two heads? • exactly r heads? • at least 2 heads? • Probability of • no heads? • exactly one head? • exactly two heads? • exactly r heads? • at least 2 heads? • What is the experiment ? • What is sample space S ? • What is event A ? • What is outcome a ? • Think about lottery
- 16. JC Liu MACM101 Discrete Mathematics 16 9.4. Further Readings 1.3. Binomial Theorem 1.4. Combinations with Repetition 3.4. Finite probability 4.5 Fundamental theorem of arithmetic (side topic)