Binomial Theorem. Combinations with repetition.

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Donald Jackson
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1 Biomial..

2 Permutatios ad combiatios Give a set with elemets. The umber of permutatios of the elemets the set: P() =! = ( 1) ( 2)... 1 The umber of r-permutatios of the set: P(, r) =! = ( 1) ( 2)... ( r + 1) ( r)! }{{} product of r umbers The umber of uordered r-combiatios ( choose r ): P(, r)! = = r P(r) ( r)! r! p. 2

3 What are the properties of r? How does r chage with r? 0 0 = 0! 0! 0! = = 1! 1! 0! = 1, 1 1 = 1! 0! 1! = = 2! 2! 0! = 1, 2 1 = 2! 1! 1! = 2, 2 2 = 2! 0! 2! = = 3! 3! 0! = 1, 3 1 = 3! 2! 1! = 3, 3 2 = 3! 1! 2! = 3, 3 3 = 3! 0! 3! = 1. p. 3

4 p. 4

5 The umbers i are the coefficiets of the polyomials of the form (x + y) : (x + y) 0 = 1 (x + y) 1 = 1x + 1 y (x + y) 2 = 1x 2 + 2x y + 1 y 2 (x + y) 3 = 1x 3 + 3x 2 y + 3x y y 3 (x + y) 4 = 1x 4 + 4x 3 y + 6x 2 y 2 + 4x y 3 + 1y 4 (x + y) = x + x 1 y + x 2 y y (x + y) = x k y k k p. 5

6 The umbers i are the coefficiets of the polyomials of the form (x + y) : (x + y) 0 = 1 (x + y) 1 = 1x + 1 y (x + y) 2 = 1x 2 + 2x y + 1 y 2 (x + y) 3 = 1x 3 + 3x 2 y + 3x y y 3 (x + y) 4 = 1x 4 + 4x 3 y + 6x 2 y 2 + 4x y 3 + 1y 4 (x + y) = x + x 1 y + x 2 y y (x + y) = x k y k k p. 6

7 The umbers i are the coefficiets of the polyomials of the form (x + y) : (x + y) 0 = 1 (x + y) 1 = 1x + 1 y (x + y) 2 = 1x 2 + 2x y + 1 y 2 (x + y) 3 = 1x 3 + 3x 2 y + 3x y y 3 (x + y) 4 = 1x 4 + 4x 3 y + 6x 2 y 2 + 4x y 3 + 1y 4 (x + y) = x + x 1 y + x 2 y y (x + y) = x k y k k p. 7

8 The umbers i are the coefficiets of the polyomials of the form (x + y) : (x + y) 0 = 1 (x + y) 1 = 1x + 1 y (x + y) 2 = 1x 2 + 2x y + 1 y 2 (x + y) 3 = 1x 3 + 3x 2 y + 3x y y 3 (x + y) 4 = 1x 4 + 4x 3 y + 6x 2 y 2 + 4x y 3 + 1y 4 (x + y) = x + x 1 y + x 2 y y (x + y) = x k y k k p. 8

9 Coefficiets of (x + y) Let s prove (x + y) = x + x 1 y + x 2 y y Example: (x + y) 3 =(x + y)(x + y)(x + y) = x x x+ x x y + x y x + y x x+ x y y + y x y + y y x+ y y y 2 = 2 3 = 8 terms i total. The same as the umber of the bit strigs of legth 3. p. 9

10 Coefficiets of (x + y) (x + y) = (x + y) (x + y)... (x + y) =? }{{} times What is happeig whe we multiply (x + y) times? We get the sum of p. 10

11 Coefficiets of (x + y) (x + y) = (x + y) (x + y)... (x + y) =? }{{} times What is happeig whe we multiply (x + y) times? We get the sum of x x x x x... x+ yx x x x... x+ xyx x x... x+ yyx x x... x+ x xyx x... x y y y y y... y p. 11

12 Coefficiets of (x + y) (x + y) = (x + y) (x + y)... (x + y) = }{{} times } x x {{... x } + =x (x }. {{.. x } ) y y (x... x)+ =x 1 (x }. {{.. x } ) (y y) (y y) (x }. {{.. x } =x y y... y }{{} =y =x 2 )+ p. 12

13 Coefficiets of (x + y) (x + y) = (x + y) (x + y)... (x + y) = }{{} times 1 x + x 1 y + x 2 y y p. 13

14 (x + y) = (x + y) (x + y)... (x + y) = }{{} times x + x 1 y + x 2 y y. Shorter otatio for the same thig: (x + y) = x k y k. k This result is called, ad this is why the coefficiets, k, are also called the biomial coefficiets. p. 14

15 Let s prove that = (1 + 1) = 1 k 1 k = k 1 = k. k Therefore, = 2. k p. 15

16 Let s prove that = (1 + 1) = 1 k 1 k = k 1 = k. k Therefore, = 2. k p. 16

17 = Results like this are ot very obvious. Recall that So,! = k ( k)! k!!! 0! +! ( 1)! 1! +! ( 2)! 2! ! 0!! = 2. I this form, the result seems to be much harder to prove. However, ca you thik of aother way to prove this idetity? (You ca try to use double coutig) p. 17

18 Usig the biomial theorem (x + y) = x k y k, k prove that 2 k = k = p. 18

19 Coutig routes You ca go oly North ad East. Cout the umber of paths from (0, 0) to (5, 3). Aswer: p. 19

20 Coutig routes You ca go oly North ad East. Cout the umber of paths from (0, 0) to (5, 3). Aswer: p. 20

21 Agai p. 21

22 Pascal s Idetity + 1 = + k k 1 k Every umber i Pascal s triagle is equal to the sum of the two umbers that are immediately above. p. 22

23 Aother Idetity Prove that for r ad r m: m + = r r m r k k p. 23

24 Vadermode s Idetity Prove that for r ad r m: m + = r r m r k k We split the iitial set of m + objsects ito two arbitrary subsets of m ad objects. After that, we ca choose r objects from the two subsets i the followig ways: k subset of size m subset of size 0 choose r choose oe 1 choose r 1 choose 1 2 choose r 2 choose 2... r choose 0 choose r m m + + r 0 r 1 1 m m r r r m = r k k p. 24

25 r-combiatios without Recall that without s, this is r. For example, you have books, but do t have time to read all of them, ad have to select oly r books to read. I how may ways ca you do so?! = r ( r)! r! There are r ways to make the choice. p. 25

26 r-combiatios with There is a vedig machie with 3 types of driks, $1 each drik. You have to sped $5. $ $ $ $ $ p. 26

27 r-combiatios with There is a vedig machie with 3 types of driks, $1 each drik. You have to sped $5. $ $ $ $ $ p. 27

28 r-combiatios with There is a vedig machie with 3 types of driks, $1 each drik. You have to sped $5. $ }{{} 1 drik $ $ }{{} 2 driks $ $ }{{} 2 driks p. 28

29 r-combiatios with There is a vedig machie with 3 types of driks, $1 each drik. You have to sped $5. $ $ $ $ $ p. 29

30 r-combiatios with There is a vedig machie with 3 types of driks, $1 each drik. You have to sped $5. $ $ }{{} 2 drik } $ {{ $ $ } 3 driks }{{} oe p. 30

31 r-combiatios with So, there are 5 + (3 1) places that stad for 5 dollars ad (3 1) separators betwee the driks types. _ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ 7 7 = = 21 ways to buy 5 driks 2 5 p. 31

32 r-combiatios with So, there are 5 + (3 1) places that stad for 5 dollars ad (3 1) separators betwee the driks types. _ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ 7 7 = = 21 ways to buy 5 driks 5 2 p. 32

33 r-combiatios with To select r objects out of with s, there are r + 1 r + 1 = ways. r 1 * * * * * * (r objects ad 1 separator) I other words, this is the umber of r-combiatios with from the set of objects. p. 33

34 r-permutatios with We kow that the umber of r-permutatios of objects without is ( 1)( 2)... ( r + 1) =! ( r)! But, if s are allowed, it is eve easier, the simple product rule works just fie!... = r p. 34

35 Summary with s? r-combiatio r-combiatio No Yes r+ 1 r r = r+ 1 1 r-permutatio No P(, r) =! ( r)! r-permutatio Yes r p. 35

Section Mathematical Induction and Section Strong Induction and Well-Ordering

Section Mathematical Induction and Section Strong Induction and Well-Ordering Sectio 4.1 - Mathematical Iductio ad Sectio 4. - Strog Iductio ad Well-Orderig A very special rule of iferece! Defiitio: A set S is well ordered if every subset has a least elemet. Note: [0, 1] is ot well