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MATH 106 Combinatorics

MATH 106 Combinatorics. “In how many ways is it possible to …?”. Such as …. How many solutions in positive integers are there to the equation a + b + c + d = 52? In how many ways can you make change for a dollar?

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MATH 106 Combinatorics

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  1. MATH 106Combinatorics “In how many ways is it possible to …?” MATH 106, Section 1

  2. Such as … • How many solutions in positive integers are there to the equation a + b + c + d = 52? • In how many ways can you make change for a dollar? • You are taking your friend home to meet your parents. How many possible seating arrangements are there around the dinner table? • How many seating arrangements are there if you and your friend are seated across from each other? MATH 106, Section 1

  3. When faced with a problem we’ve never seen before … • We can try to answer the problem directly with some formula, or “recipe.” • If that doesn’t work, we can ask ourselves if the problem is similar to one we already know how to do? • If these attempts fail, try to solve an easier version of the problem. If successful, try to work up to the original problem. Look for what’s known as a “general solution” – that is, one that will work in all cases of this problem. • Let’s try this third method … MATH 106, Section 1

  4. Building a narrow staircase Problem: A narrow staircase, one foot wide, is to be built out of concrete blocks. Each block is a one foot cube, and the space underneath the steps is to be filled in as a massive wall of concrete blocks. How many blocks are necessary to construct a staircase with ten steps? MATH 106, Section 1

  5. Let’s start with a small case How many would it take for 1 step? 2 steps? 3 steps? and so on? For 10 steps … 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 = 55 MATH 106, Section 1

  6. Can we “generalize” that result? • How many blocks would it take for n steps? • 1 + 2 + 3 + … + (n – 1) + n • That’s the right idea, but what if we have 100 steps? That’s a lot of addition. Can we do better? • Some of you may know a formula for getting the sum of integers from 1 to n • Let’s see if we can derive it from this problem … MATH 106, Section 1

  7. Try it for 5 steps n + 1 This is  . 2 Change the steps into a rectangle: 3 by 5 That gives us 15 again. That’s good! We’ve computed the answer by a different method and got the same answer. This is an example of a combinatorial proof. This is n. MATH 106, Section 1

  8. Change the steps into a rectangle: 3 by 7 Work in groups to come up with a general rule for n steps. Hint: Do the cases where n is odd and n is even, separately. Try it for 6 steps n This is  . 2 This is n + 1. MATH 106, Section 1

  9. So what formula did you come up with. How many blocks are needed for 25 steps? 100 steps? n is odd n is even any n n = 25 n = 100 n+1 ——n 2 n — (n+1) 2 n(n+1) ——— 2 325 5050 Let’s try this method on other problems … MATH 106, Section 1

  10. #1 Each step in a spiral staircase is to be painted with one of the two colors red (R) or blue (B). How many different color arrangements are possible with two steps? three steps? four steps? RR RB BR BB 4 arrangements RRR RRB RBR RBB BRR BRB BBR BBB 8 arrangements RRRR RRRB RRBR RRBB RBRR RBRB RBBR RBBB BRRR BRRB BRBR BRBB BBRR BBRB BBBR BBBB 16 arrangements Come up with a general rule for n steps. 2n arrangements MATH 106, Section 1

  11. How many different color arrangements are possible with eight steps? 28 = 256 arrangements MATH 106, Section 1

  12. #2 Different colored flags are to be flown on a flagpole with one color on top, a different color underneath, down to the last different color on the bottom. How many different arrangements are possible with two different colored flags? three different colored flags? four different colored flags? RB BR 2 arrangements PRB RPB RBP PBR BPR BRP 6 arrangements GPRB PGRB PRGB PRBG GRPB RGPB RPGB RPBG …etc. 24 arrangements Come up with a general rule for n different colored flags. n(n–1)(n–2)…(3)(2)(1) arrangements MATH 106, Section 1

  13. How many different arrangements are possible with eight different colored flags? (8)(7)(6)(5)(4)(3)(2)(1) = 40320 arrangements MATH 106, Section 1

  14. Homework Hints: In Section 1 Homework Problem #2, In Section 1 Homework Problem #4, construct a two-column table, where the first column is the number of people attending and the second column is the number of handshakes. note the similarity with #2 on the Section#1 class handout. MATH 106, Section 1

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