1 basics of counting cs/apma 202 rosen section 4.1 aaron bloomfield

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Basics of CountingBasics of Counting

CS/APMA 202CS/APMA 202

Rosen section 4.1Rosen section 4.1

Aaron BloomfieldAaron Bloomfield

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The product ruleThe product rule

If there are If there are nn11 ways to do task 1, and ways to do task 1, and nn22 ways to ways to

do task 2do task 2 Then there are Then there are nn11nn22 ways to do both tasks in ways to do both tasks in

sequencesequence This applies when doing the “procedure” is made up This applies when doing the “procedure” is made up

of separate tasksof separate tasks We must make one choice AND a second choiceWe must make one choice AND a second choice

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Product rule exampleProduct rule example

Rosen, section 4.1, question 1 (a)Rosen, section 4.1, question 1 (a) There are 18 math majors and 325 CS majorsThere are 18 math majors and 325 CS majors How many ways are there to pick one math How many ways are there to pick one math

major major andand one CS major? one CS major?

Total is 18 * 325 = 5850Total is 18 * 325 = 5850

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Product rule exampleProduct rule example

Rosen, section 4.1, question 22 (a) and (b)Rosen, section 4.1, question 22 (a) and (b)How many strings of 4 decimal digits…How many strings of 4 decimal digits…

a)a) Do not contain the same digit twice?Do not contain the same digit twice?We want to chose a digit, then another that is not the same, We want to chose a digit, then another that is not the same, then another…then another…

First digit: 10 possibilitiesFirst digit: 10 possibilitiesSecond digit: 9 possibilities (all but first digit)Second digit: 9 possibilities (all but first digit)Third digit: 8 possibilitiesThird digit: 8 possibilitiesFourth digit: 7 possibilitiesFourth digit: 7 possibilities

Total = 10*9*8*7 = 5040Total = 10*9*8*7 = 5040

b)b) End with an even digit?End with an even digit?First three digits have 10 possibilitiesFirst three digits have 10 possibilitiesLast digit has 5 possibilitiesLast digit has 5 possibilitiesTotal = 10*10*10*5 = 5000Total = 10*10*10*5 = 5000

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The sum ruleThe sum rule

If there are If there are nn11 ways to do task 1, and ways to do task 1, and nn22 ways to ways to

do task 2do task 2 If these tasks can be done at the same time, then…If these tasks can be done at the same time, then… Then there are Then there are nn11++nn22 ways to do one of the two tasks ways to do one of the two tasks We must make one choice OR a second choiceWe must make one choice OR a second choice

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Sum rule exampleSum rule example

Rosen, section 4.1, question 1 (b)Rosen, section 4.1, question 1 (b) There are 18 math majors and 325 CS majorsThere are 18 math majors and 325 CS majors How many ways are there to pick one math How many ways are there to pick one math

major major oror one CS major? one CS major?

Total is 18 + 325 = 343Total is 18 + 325 = 343

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Sum rule exampleSum rule example

Rosen, section 4.1, question 22 (c)Rosen, section 4.1, question 22 (c)How many strings of 4 decimal digits…How many strings of 4 decimal digits…

c)c) Have exactly three digits that are 9s?Have exactly three digits that are 9s? The string can have:The string can have:

The non-9 as the first digitThe non-9 as the first digitOR the non-9 as the second digitOR the non-9 as the second digitOR the non-9 as the third digitOR the non-9 as the third digitOR the non-9 as the fourth digitOR the non-9 as the fourth digitThus, we use the sum ruleThus, we use the sum rule

For each of those cases, there are 9 possibilities for For each of those cases, there are 9 possibilities for the non-9 digit (any number other than 9)the non-9 digit (any number other than 9)

Thus, the answer is 9+9+9+9 = 36Thus, the answer is 9+9+9+9 = 36

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Quick surveyQuick survey

I’m feeling good with the sum rule I’m feeling good with the sum rule and the product rule…and the product rule…

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

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More complex counting problemsMore complex counting problems

We combining the product rule and the We combining the product rule and the sum rulesum rule

Thus we can solve more interesting and Thus we can solve more interesting and complex problemscomplex problems

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Wedding pictures exampleWedding pictures example

Rosen, section 4.1, question 38Rosen, section 4.1, question 38Consider a wedding picture of 6 peopleConsider a wedding picture of 6 people

There are 10 people, including the bride and groomThere are 10 people, including the bride and groom

a)a) How many possibilities are there if the bride must be in How many possibilities are there if the bride must be in the picturethe picture

Product rule: place the bride AND then place the rest of the Product rule: place the bride AND then place the rest of the partypartyFirst place the brideFirst place the bride

She can be in one of 6 positionsShe can be in one of 6 positions

Next, place the other five people via the product ruleNext, place the other five people via the product ruleThere are 9 people to choose for the second person, 8 for the There are 9 people to choose for the second person, 8 for the third, etc.third, etc.Total = 9*8*7*6*5 = 15120Total = 9*8*7*6*5 = 15120

Product rule yields 6 * 15120 = 90,720 possibilitiesProduct rule yields 6 * 15120 = 90,720 possibilities

1111




b)b) How many possibilities are there if the bride and groom How many possibilities are there if the bride and groom must both be in the picturemust both be in the picture

Product rule: place the bride/groom AND then place the rest of Product rule: place the bride/groom AND then place the rest of the partythe partyFirst place the bride and groomFirst place the bride and groom

She can be in one of 6 positionsShe can be in one of 6 positionsHe can be in one 5 remaining positionsHe can be in one 5 remaining positionsTotal of 30 possibilitiesTotal of 30 possibilities

Next, place the other four people via the product ruleNext, place the other four people via the product ruleThere are 8 people to choose for the third person, 7 for the fourth, There are 8 people to choose for the third person, 7 for the fourth, etc.etc.Total = 8*7*6*5 = 1680Total = 8*7*6*5 = 1680

Product rule yields 30 * 1680 = 50,400 possibilitiesProduct rule yields 30 * 1680 = 50,400 possibilities

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c)c) How many possibilities are there if only one of the bride and How many possibilities are there if only one of the bride and groom are in the picturegroom are in the picture

Sum rule: place only the brideSum rule: place only the brideProduct rule: place the bride AND then place the rest of the partyProduct rule: place the bride AND then place the rest of the partyFirst place the brideFirst place the bride She can be in one of 6 positionsShe can be in one of 6 positions

Next, place the other five people via the product ruleNext, place the other five people via the product ruleThere are 8 people to choose for the second person, 7 for the third, etc.There are 8 people to choose for the second person, 7 for the third, etc.

We can’t choose the groom!We can’t choose the groom!Total = 8*7*6*5*4 = 6720Total = 8*7*6*5*4 = 6720

Product rule yields 6 * 6720 = 40,320 possibilitiesProduct rule yields 6 * 6720 = 40,320 possibilities OR place only the groomOR place only the groom

Same possibilities as for bride: 40,320Same possibilities as for bride: 40,320

Sum rule yields 40,320 + 40,320 = 80,640 possibilitiesSum rule yields 40,320 + 40,320 = 80,640 possibilities

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Alternative means to get the answerAlternative means to get the answer

c)c) How many possibilities are there if only one of the bride and How many possibilities are there if only one of the bride and groom are in the picturegroom are in the picture

Total ways to place the bride (with or without groom): 90,720 Total ways to place the bride (with or without groom): 90,720 From part (a)From part (a)

Total ways for both the bride and groom: 50,400Total ways for both the bride and groom: 50,400From part (b)From part (b)

Total ways to place ONLY the bride: 90,720 – 50,400 = 40,320Total ways to place ONLY the bride: 90,720 – 50,400 = 40,320Same number for the groomSame number for the groomTotal = 40,320 + 40,320 = 80,640Total = 40,320 + 40,320 = 80,640

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I’m feeling good with these sum and I’m feeling good with these sum and product rule examples…product rule examples…





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The inclusion-exclusion principleThe inclusion-exclusion principle

When counting the possibilities, we can’t When counting the possibilities, we can’t include a given outcome more than once!include a given outcome more than once!

|A|A11U AU A22| = |A| = |A11| + |A| + |A22| - |A| - |A11∩ A∩ A22|| Let ALet A11 have 5 elements, A have 5 elements, A22 have 3 elements, have 3 elements,

and 1 element be both in Aand 1 element be both in A11 and A and A22

Total in the union is 5+3-1 = 7, not 8Total in the union is 5+3-1 = 7, not 8

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Inclusion-exclusion exampleInclusion-exclusion example

Rosen, section 4.1, example 16Rosen, section 4.1, example 16How may bit strings of length eight start with 1 or end with 00?How may bit strings of length eight start with 1 or end with 00?

Count bit strings that start with 1Count bit strings that start with 1 Rest of bits can be anything: 2Rest of bits can be anything: 277 = 128 = 128 This is |AThis is |A11||

Count bit strings that end with 00Count bit strings that end with 00 Rest of bits can be anything: 2Rest of bits can be anything: 266 = 64 = 64 This is |AThis is |A22||

Count bit strings that both start with 1 and end with 00Count bit strings that both start with 1 and end with 00 Rest of the bits can be anything: 2Rest of the bits can be anything: 255 = 32 = 32 This is This is |AThis is This is |A11∩ A∩ A22||

Use formula |AUse formula |A11U AU A22| = |A| = |A11| + |A| + |A22| - |A| - |A11∩ A∩ A22||

Total is 128 + 64 – 32 = 160Total is 128 + 64 – 32 = 160

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Bit string possibilitiesBit string possibilities

Rosen, section 4.1, question 42Rosen, section 4.1, question 42

How many bit strings of length 10 contain How many bit strings of length 10 contain either 5 consecutive 0s or 5 consecutive either 5 consecutive 0s or 5 consecutive 1s?1s?

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Bit string possibilitiesBit string possibilities

Consider 5 consecutive 0s firstConsider 5 consecutive 0s first

Sum rule: the 5 consecutive 0’s can start at position 1, 2, 3, 4, 5, or 6Sum rule: the 5 consecutive 0’s can start at position 1, 2, 3, 4, 5, or 6 Starting at position 1Starting at position 1

Remaining 5 bits can be anything: 2Remaining 5 bits can be anything: 255 = 32 = 32 Starting at position 2Starting at position 2

First bit must be a 1First bit must be a 1 Otherwise, we are including possibilities from the previous case!Otherwise, we are including possibilities from the previous case!

Remaining bits can be anything: 2Remaining bits can be anything: 244 = 16 = 16 Starting at position 3Starting at position 3

Second bit must be a 1 (same reason as above)Second bit must be a 1 (same reason as above)

First bit and last 3 bits can be anything: 2First bit and last 3 bits can be anything: 244 = 16 = 16 Starting at positions 4 and 5 and 6Starting at positions 4 and 5 and 6

Same as starting at positions 2 or 3: 16 eachSame as starting at positions 2 or 3: 16 each Total = 32 + 16 + 16 + 16 + 16 + 16 = 112Total = 32 + 16 + 16 + 16 + 16 + 16 = 112

The 5 consecutive 1’s follow the same pattern, and have 112 possibilitiesThe 5 consecutive 1’s follow the same pattern, and have 112 possibilities

There are two cases counted twice (that we thus need to exclude): There are two cases counted twice (that we thus need to exclude): 0000011111 and 11111000000000011111 and 1111100000

Total = 112 + 112 – 2 = 222Total = 112 + 112 – 2 = 222

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Tree diagramsTree diagrams

We can use tree diagrams to enumerate We can use tree diagrams to enumerate the possible choicesthe possible choices

Once the tree is laid out, the result is the Once the tree is laid out, the result is the number of (valid) leavesnumber of (valid) leaves

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Tree diagrams exampleTree diagrams example

Rosen, section 4.1, question 48Rosen, section 4.1, question 48

Use a tree diagram to find the number of bit strings of Use a tree diagram to find the number of bit strings of length four with no three consecutive 0slength four with no three consecutive 0s

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An example closer to home…An example closer to home…

winlose

winwin

loselose

win

winwin

win

lose

lose

lose

lose

How many ways can the Cavs finish the How many ways can the Cavs finish the season 9 and 2?season 9 and 2?

Playing Miami

Playing GA Tech

Playing VA Tech

(7,1)

(7,2) (8,1)

(7,3)

(7,4) (8,3) (7,3)

(8,2)

(9,2)

(8,2) (9,1)

(8,3) (9,2) (9,2) (10,1)

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I felt I understood the material in this I felt I understood the material in this slide set…slide set…





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The pace of the lecture for this The pace of the lecture for this slide set was…slide set was…

a)a) FastFast

b)b) About rightAbout right

c)c) A little slowA little slow

d)d) Too slowToo slow

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How interesting was the material in How interesting was the material in this slide set? Be honest!this slide 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

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Beware!!!Beware!!!

1 basics of counting cs/apma 202 rosen section 4.1 aaron bloomfield

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