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The Art of Combinatorics: Permutations, Combinations, and Binomials

Explore the fundamental concepts of combinatorics, including permutations, combinations, and binomials. Learn how to count the ways to order and select elements, and discover the properties of binomials and Pascal's Triangle.

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The Art of Combinatorics: Permutations, Combinations, and Binomials

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  1. We are going to count: The Art of Combinatorics Thus far: multiplying events adding events inclusion – exclusion principle Today: Permutations and Combinations Binomials, Pascal’s Triangle Last Week

  2. How many ways can you order n elements? Answer: n(n–1)(n–2)…1 = n! (say “n factorial”) Note: 0!=1, 1!=1, 2!=2, 3!=6, 4!=24, … How many ways can you order r out of n elements? Answer: n(n–1)(n–2)…(n–r+1) = n! / (n–r)! Sometimes denoted by P(n,r) Permutations / Combinations

  3. Let there be 15 different people, in how many wayscan you line up 10 of them? Answer: 1514…6 = P(15,10) = 10,897,286,400 Note how fast n! grows: n! = (2n) Example:

  4. How many ways can you select r elements out of n?(without concern for the order of selection) Answer: You can order the r elements in n!/(n–r)! ways Dividing out the r! orderings, gives you Binomial Say: “n choose r”, or “the binomial of n over r”

  5. Let there be 15 different people, in how many wayscan you select 10 of them? Answer: C(15,10) = 15! / (10!5!) = 3003 I’m flipping a coin 20 times, in how many ways can I have the outcome “heads” 10 times? Answer: C(20,10) = 20! / (10!10!) = 184 756 … the outcome “heads” k times? Answer: C(20,k) = 20! / ((20–k)!k!) Examples:

  6. Because: we have Also: For constant c and variable n: Properties of Binomials

  7. Consider the n-fold product (a+b)n: (a+b)(a+b)…(a+b) = an + nan–1b + n(n–1)/2an–2b2 +…+ bn Binomial Theorem (8.39): Immediate consequence: Binomial Theorem Example: (a+b)4 = a4+4a3b+6a2b2+4ab3+b4

  8. Consider all bit strings of length n (all 2n of them). The number of strings with k “ones” is C(n,k). Hence indeed C(n,0)+C(n,1)+C(n,2)+…+C(n,n) = 2n. We expect that ‘most’ strings will have  n/2 ones. Indeed, when we plot C(20,k) for k=0,…,20,we get the bell shaped binomial distribution: Bit Strings

  9. Theorem 8.43: For all n,kN: - - æ ö æ ö æ ö n n 1 n 1 ç ÷ ç ÷ ç ÷ = + ç ÷ ç ÷ ç ÷ - k k 1 k è ø è ø è ø AlgebraicProof: Binomial Identity Proof by interpretation: “How can we pick k out of n?…consider n–1…” et cetera (see book).

  10. Because of the relation C(n,k) = C(n–1,k–1)+C(n–1,k),we can write the binomials in a triangle: On the ‘inside’ we have Triangle of Binomials Where on the ‘outside edges’we have C(n,0)=C(n,n)=1 This gives…

  11. For large n, the valueson a row (like 1,4,6,4,1)behave like a smooth bell curve: the Gaussian Pascal’s Triangle By filling in the triangle of binomial values, we get Pascal’s Triangle:

  12. Consider a random n bit string x1,…,xn with xj{0,1}. The probability of this string is 2–n,the probability of k “ones” in the string is C(n,k)/2n The binomial distribution for large n teaches us that, although we expect half of the n bits to be “one”, the probability of k=½ n will be: Binomials and Bits Instead, we can expect the number of “ones” to be between k = ½ n – n and ½ n + n

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