1506 binomial-coefficients
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1506 binomial-coefficients
- 1. 11 Binomial CoefficientsBinomial Coefficients CS/APMA 202CS/APMA 202 Rosen section 4.4Rosen section 4.4 Aaron BloomfieldAaron Bloomfield
- 2. 22 Binomial CoefficientsBinomial Coefficients It allows us to do a quick expansion ofIt allows us to do a quick expansion of ((xx++yy))nn Why it’s really important:Why it’s really important: It provides a good context to presentIt provides a good context to present proofsproofs Especially combinatorial proofsEspecially combinatorial proofs
- 3. 33 LetLet nn andand rr be non-negative integers withbe non-negative integers with rr ≤≤ nn. Then. Then CC((nn,,rr) =) = CC((nn,,n-rn-r)) Or,Or, Proof (from last slide set):Proof (from last slide set): [ ]!)()!( ! ),( rnnrn n rnnC −−− =− )!(! ! rnr n − = )!(! ! ),( rnr n rnC − = Review: corollary 1Review: corollary 1 from section 4.3from section 4.3 − = rn n r n
- 4. 44 Review: combinatorial proofReview: combinatorial proof AA combinatorial proofcombinatorial proof is a proof that usesis a proof that uses counting arguments to prove a theorem,counting arguments to prove a theorem, rather than some other method such asrather than some other method such as algebraic techniquesalgebraic techniques Essentially, show that both sides of theEssentially, show that both sides of the proof manage to count the same objectsproof manage to count the same objects Usually in the form of an English explanationUsually in the form of an English explanation with supporting formulaewith supporting formulae
- 5. 55 Polynomial expansionPolynomial expansion Consider (Consider (xx++yy))33 :: Rephrase it as:Rephrase it as: When choosingWhen choosing xx twice andtwice and yy once, there areonce, there are C(3,2) = C(3,1) = 3 ways to choose where theC(3,2) = C(3,1) = 3 ways to choose where the xx comes fromcomes from When choosingWhen choosing xx once andonce and yy twice, there aretwice, there are C(3,2) = C(3,1) = 3 ways to choose where theC(3,2) = C(3,1) = 3 ways to choose where the yy comes fromcomes from 32233 33)( yxyyxxyx +++=+ [ ] [ ] 32222223 ))()(( yxyxyxyyxyxyxxyxyxyx +++++++=+++
- 6. 66 Polynomial expansionPolynomial expansion ConsiderConsider To obtain theTo obtain the xx55 termterm Each time you multiple by (Each time you multiple by (xx++yy), you select the), you select the xx Thus, of the 5 choices, you chooseThus, of the 5 choices, you choose xx 5 times5 times C(5,5) = 1C(5,5) = 1 Alternatively, you chooseAlternatively, you choose yy 0 times0 times C(5,0) = 1C(5,0) = 1 To obtain theTo obtain the xx44 yy termterm Four of the times you multiply by (Four of the times you multiply by (xx++yy), you select the), you select the xx The other time you select theThe other time you select the yy Thus, of the 5 choices, you chooseThus, of the 5 choices, you choose xx 4 times4 times C(5,4) = 5C(5,4) = 5 Alternatively, you chooseAlternatively, you choose yy 1 time1 time C(5,1) = 5C(5,1) = 5 To obtain theTo obtain the xx33 yy22 termterm C(5,3) = C(5,2) = 10C(5,3) = C(5,2) = 10 Etc…Etc… 543223455 510105)( yxyyxyxyxxyx +++++=+
- 7. 77 Polynomial expansionPolynomial expansion For (For (xx++yy))55 543223455 0 5 1 5 2 5 3 5 4 5 5 5 )( yxyyxyxyxxyx + + + + + =+ 543223455 510105)( yxyyxyxyxxyx +++++=+
- 8. 88 Polynomial expansion:Polynomial expansion: The binomial theoremThe binomial theorem For (For (xx++yy))nn The book calls this Theorem 1The book calls this Theorem 1 nnnnn yx n yx n yx n n yx n n yx 011110 011 )( + ++ − + =+ −− ∑= − = n j jjn yx j n 0 nnnn yx n n yx n n yx n yx n 011110 110 + − ++ + = −−
- 9. 99 ExamplesExamples What is the coefficient ofWhat is the coefficient of xx1212 yy1313 in (in (xx++yy))2525 ?? What is the coefficient ofWhat is the coefficient of xx1212 yy1313 in (2in (2xx-3-3yy))2525 ?? Rephrase it as (2x+(-3y))Rephrase it as (2x+(-3y))2525 The coefficient occurs whenThe coefficient occurs when jj=13:=13: 300,200,5 !12!13 !25 12 25 13 25 == = ( ) ∑= − − =−+ 25 0 2525 )3()2( 25 )3(2 j jj yx j yx 00,545,702,433,959,763)3(2 !12!13 !25 )3(2 13 25 13121312 −=−=−
- 10. 1010 Rosen, section 4.4, question 4Rosen, section 4.4, question 4 Find the coefficient ofFind the coefficient of xx55 yy88 in (in (xx++yy))1313 Answer:Answer: 1287 8 13 5 13 = =
- 12. 1212 Pascal’s IdentityPascal’s Identity By Pascal’s identity: or 21=15+6By Pascal’s identity: or 21=15+6 LetLet nn andand kk be positive integers withbe positive integers with nn ≥≥ kk.. ThenThen or C(or C(nn+1,+1,kk) = C() = C(nn,,kk-1) + C(-1) + C(nn,,kk)) The book calls this Theorem 2The book calls this Theorem 2 We will prove this via two ways:We will prove this via two ways: Combinatorial proofCombinatorial proof Using the formula forUsing the formula for + − = + k n k n k n 1 1 + = 5 6 4 6 5 7 k n
- 13. 1313 Combinatorial proof of Pascal’sCombinatorial proof of Pascal’s identityidentity Prove C(Prove C(nn+1,+1,kk) = C() = C(nn,,kk-1) + C(-1) + C(nn,,kk)) Consider a set T ofConsider a set T of nn+1 elements+1 elements We want to choose a subset ofWe want to choose a subset of kk elementselements We will count the number of subsets ofWe will count the number of subsets of kk elements via 2 methodselements via 2 methods Method 1: There are C(Method 1: There are C(nn+1,+1,kk) ways to choose such a subset) ways to choose such a subset Method 2: LetMethod 2: Let aa be an element of set Tbe an element of set T Two casesTwo cases aa is in such a subsetis in such a subset There are C(There are C(nn,,kk-1) ways to choose such a subset-1) ways to choose such a subset aa is not in such a subsetis not in such a subset There are C(There are C(nn,,kk) ways to choose such a subset) ways to choose such a subset Thus, there are C(Thus, there are C(nn,,kk-1) + C(-1) + C(nn,,kk) ways to choose a subset of) ways to choose a subset of kk elementselements Therefore, C(Therefore, C(nn+1,+1,kk) = C() = C(nn,,kk-1) + C(-1) + C(nn,,kk))
- 14. 1414 Rosen, section 4.4, question 19:Rosen, section 4.4, question 19: algebraic proof of Pascal’s identityalgebraic proof of Pascal’s identity + − = + k n k n k n 1 1 )!)(1()1( !)1()!1( )!(*)1()!1( knknkn nnn knknkn −+−=+− +=+ −−+=−+ 11 +=+ nn )!(! ! ))!1(()!1( ! )!1(! )!1( knk n knk n knk n − + −−− = −+ + )!()!1( ! )!)(1()!1( ! )!)(1()!1( !)1( knkk n knknk n knknkk nn −− + −+−− = −−+− + kknknk n 1 )1( 1 )1( )1( + +− = −+ + 11 +−+=+ knkn )1( )1( )1()1( )1( +− +− + +− = −+ + knk kn knk k knk n Substitutions:
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- 16. 1616 Pascal’s trianglePascal’s triangle 0 1 2 3 4 5 6 7 8 n = 1 2 4 8 16 32 64 128 256 sum = = 2n
- 17. 1717 Proof practice: corollary 1Proof practice: corollary 1 LetLet nn be a non-negative integer. Thenbe a non-negative integer. Then Algebraic proofAlgebraic proof ∑= = n k n k n 0 2 ∑= − = n k knk k n 0 11 nn )11(2 += ∑= = n k k n 0 ∑= − =+ n j jjnn yx j n yx 0 )(
- 18. 1818 Proof practice: corollary 1Proof practice: corollary 1 LetLet nn be a non-negative integer. Thenbe a non-negative integer. Then Combinatorial proofCombinatorial proof A set withA set with nn elements has 2elements has 2nn subsetssubsets By definition of power setBy definition of power set Each subset has either 0 or 1 or 2 or … orEach subset has either 0 or 1 or 2 or … or nn elementselements There are subsets with 0 elements, subsets with 1There are subsets with 0 elements, subsets with 1 element, … and subsets withelement, … and subsets with nn elementselements Thus, the total number of subsets isThus, the total number of subsets is Thus,Thus, ∑= = n k n k n 0 2 n n k k n 2 0 = ∑= 0 n ∑= n k k n 0 1 n n n
- 20. 2121 Proof practice: corollary 2Proof practice: corollary 2 LetLet nn be a positive integer. Thenbe a positive integer. Then Algebraic proofAlgebraic proof This implies thatThis implies that n 00 = ∑= = − n k k k n 0 0)1( + + + =+ + + 531420 nnnnnn ∑= − − = n k knk k n 0 1)1( ( )n 1)1( +−= k n k k n )1( 0 − = ∑=
- 21. 2222 Proof practice: corollary 3Proof practice: corollary 3 LetLet nn be a non-negative integer. Thenbe a non-negative integer. Then Algebraic proofAlgebraic proof ∑= − = n k kkn k n 0 21 ∑= = n k nk k n 0 32 nn )21(3 += ∑= = n k k k n 0 2
- 22. 2323 Vandermonde’s identityVandermonde’s identity LetLet mm,, nn, and, and rr be non-negative integersbe non-negative integers withwith rr not exceeding eithernot exceeding either mm oror nn. Then. Then The book calls this Theorem 3The book calls this Theorem 3 ∑= − = + r k k n kr m r nm 0
- 23. 2424 Combinatorial proof ofCombinatorial proof of Vandermonde’s identityVandermonde’s identity Consider two sets, one withConsider two sets, one with mm items and one withitems and one with nn itemsitems Then there are ways to chooseThen there are ways to choose rr items from the union ofitems from the union of those two setsthose two sets Next, we’ll find that value via a different meansNext, we’ll find that value via a different means PickPick kk elements from the set withelements from the set with nn elementselements Pick the remainingPick the remaining rr--kk elements from the set withelements from the set with mm elementselements Via the product rule, there are ways to do that forVia the product rule, there are ways to do that for EACHEACH value ofvalue of kk Lastly, consider this for all values ofLastly, consider this for all values of kk:: Thus,Thus, ∑= − = + r k k n kr m r nm 0 + r nm − k n kr m ∑= − r k k n kr m 0
- 24. 2525 Review of Rosen, sectionReview of Rosen, section 4.3, question 11 (a)4.3, question 11 (a) How many bit strings of length 10 containHow many bit strings of length 10 contain exactly four 1’s?exactly four 1’s? Find the positions of the four 1’sFind the positions of the four 1’s The order of those positions does not matterThe order of those positions does not matter Positions 2, 3, 5, 7 is the same as positions 7, 5, 3, 2Positions 2, 3, 5, 7 is the same as positions 7, 5, 3, 2 Thus, the answer is C(10,4) = 210Thus, the answer is C(10,4) = 210 Generalization of this result:Generalization of this result: There are C(There are C(nn,,rr) possibilities of bit strings of length) possibilities of bit strings of length nn containingcontaining rr onesones
- 25. 2626 Yet another combinatorial proofYet another combinatorial proof LetLet nn andand rr be non-negative integers withbe non-negative integers with rr ≤≤ nn.. ThenThen The book calls this Theorem 4The book calls this Theorem 4 We will do the combinatorial proof by showingWe will do the combinatorial proof by showing that both sides show the ways to count bitthat both sides show the ways to count bit strings of lengthstrings of length nn+1 with+1 with rr+1 ones+1 ones From previous slide: achieves thisFrom previous slide: achieves this ∑= = + + n rj r j r n 1 1 + + 1 1 r n
- 26. 2727 Yet another combinatorial proofYet another combinatorial proof Next, show the right side counts the same objectsNext, show the right side counts the same objects The final one must occur at positionThe final one must occur at position rr+1 or+1 or rr+2 or … or+2 or … or nn+1+1 Assume that it occurs at theAssume that it occurs at the kkthth bit, wherebit, where rr+1 ≤+1 ≤ kk ≤≤ nn+1+1 Thus, there must be r ones in the firstThus, there must be r ones in the first kk-1 positions-1 positions Thus, there are such strings of lengthThus, there are such strings of length kk-1-1 AsAs kk can be any value fromcan be any value from rr+1 to+1 to nn+1, the total number+1, the total number of possibilities isof possibilities is Thus,Thus, − r k 1 ∑ + += −1 1 1n rk r k ∑= = n rk r k ∑= = n rj r j ∑= = + + n rj r j r n 1 1
- 27. 2828 Rosen, section 4.4, question 24Rosen, section 4.4, question 24 Show that ifShow that if pp is a prime andis a prime and kk is an integer such thatis an integer such that 1 ≤1 ≤ kk ≤≤ pp-1, then-1, then pp dividesdivides We know thatWe know that pp divides the numerator (divides the numerator (pp!) once only!) once only BecauseBecause pp is prime, it does not have any factors less thanis prime, it does not have any factors less than pp We need to show that it doesWe need to show that it does NOTNOT divide thedivide the denominatordenominator Otherwise theOtherwise the pp factor would cancel outfactor would cancel out SinceSince kk << pp (it was given that k ≤(it was given that k ≤ pp-1),-1), pp cannot dividecannot divide kk!! SinceSince kk ≥ 1, we know that≥ 1, we know that pp--kk << pp, and thus, and thus pp cannotcannot divide (divide (pp--kk)!)! Thus,Thus, pp divides the numerator but not the denominatordivides the numerator but not the denominator Thus,Thus, pp dividesdivides k p )!(! ! kpk p k p − = k p
- 28. 2929 Rosen, section 4.4, question 38Rosen, section 4.4, question 38 Give a combinatorial proof that ifGive a combinatorial proof that if nn is positiveis positive integer theninteger then Provided hint: show that both sides count theProvided hint: show that both sides count the ways to select a subset of a set ofways to select a subset of a set of nn elementselements together with two not necessarily distincttogether with two not necessarily distinct elements from the subsetelements from the subset Following the other provided hint, we expressFollowing the other provided hint, we express the right side as follows:the right side as follows: 2 0 2 2)1( − = += ∑ n n k nn k n k 12 0 2 22)1( −− = +−= ∑ nn n k nnn k n k
- 29. 3030 Rosen, section 4.4, question 38Rosen, section 4.4, question 38 Show the left side properly counts theShow the left side properly counts the desired propertydesired property = ∑= n k k n k 0 2 Choosing a subset of k elements from a set of n elements Consider each of the possible subset sizes k Choosing one of the k elements in the subset twice
- 30. 3131 Rosen, section 4.4, question 38Rosen, section 4.4, question 38 Two cases to show the right side:Two cases to show the right side: nn((n-n-1)21)2n-n-22 ++nn22nn-1-1 Pick the same element from the subsetPick the same element from the subset Pick that one element from the set ofPick that one element from the set of nn elements: total ofelements: total of nn possibilitiespossibilities Pick the rest of the subsetPick the rest of the subset As there areAs there are nn-1 elements left, there are a total of 2-1 elements left, there are a total of 2nn-1-1 possibilities to pick a givenpossibilities to pick a given subsetsubset We have to do bothWe have to do both Thus, by the product rule, the total possibilities is the product of the twoThus, by the product rule, the total possibilities is the product of the two Thus, the total possibilities isThus, the total possibilities is nn*2*2nn-1-1 Pick different elements from the subsetPick different elements from the subset Pick the first element from the set ofPick the first element from the set of nn elements: total ofelements: total of nn possibilitiespossibilities Pick the next element from the set ofPick the next element from the set of nn-1 elements: total of-1 elements: total of nn-1 possibilities-1 possibilities Pick the rest of the subsetPick the rest of the subset As there areAs there are nn-2 elements left, there are a total of 2-2 elements left, there are a total of 2nn-2-2 possibilities to pick a givenpossibilities to pick a given subsetsubset We have to do all threeWe have to do all three Thus, by the product rule, the total possibilities is the product of the threeThus, by the product rule, the total possibilities is the product of the three Thus, the total possibilities isThus, the total possibilities is nn*(n-1)*2*(n-1)*2nn-2-2 We do one or the otherWe do one or the other Thus, via the sum rule, the total possibilities is the sum of the twoThus, via the sum rule, the total possibilities is the sum of the two OrOr nn*2*2nn-1-1 ++nn*(n-1)*2*(n-1)*2nn-2-2
- 31. 32 Quick surveyQuick survey I felt I understood the material in this slide set…I felt I understood the material in this slide set… a)a) Very wellVery well b)b) With some review, I’ll be goodWith some review, I’ll be good c)c) Not reallyNot really d)d) Not at allNot at all
- 32. 33 Quick surveyQuick survey The pace of the lecture for this slide set was…The pace of the lecture for this slide set was… a)a) FastFast b)b) About rightAbout right c)c) A little slowA little slow d)d) Too slowToo slow
- 33. 34 Quick surveyQuick survey How interesting was the material in this slideHow interesting was the material in this slide set? Be honest!set? Be honest! a)a) Wow! That was SOOOOOO cool!Wow! That was SOOOOOO cool! b)b) Somewhat interestingSomewhat interesting c)c) Rather bortingRather borting d)d) ZzzzzzzzzzzZzzzzzzzzzz
- 34. 3535 Becoming an IEEE authorBecoming an IEEE author
- 35. 3636