Basics of CountingThe product ruleProduct rule exampleSlide 4The sum ruleSum rule exampleSlide 7More complex counting problemsWedding pictures exampleSlide 10Slide 11Slide 12The inclusion-exclusion principleInclusion-exclusion exampleBit string possibilitiesSlide 16Tree diagramsTree diagrams exampleAn example closer to home…1Basics of CountingCS/APMA 202Epp sections 6.2 & 6.3Aaron Bloomfield2The product rule•Also called the multiplication rule•If there are n1 ways to do task 1, and n2 ways to do task 2–Then there are n1n2 ways to do both tasks in sequence–This applies when doing the “procedure” is made up of separate tasks–We must make one choice AND a second choice3Product rule example•Sample question–There are 18 math majors and 325 CS majors–How many ways are there to pick one math major and one CS major?•Total is 18 * 325 = 58504Product rule exampleMore sample questions…•How many strings of 4 decimal digits…a) Do not contain the same digit twice?We want to chose a digit, then another that is not the same, then another…•First digit: 10 possibilities•Second digit: 9 possibilities (all but first digit)•Third digit: 8 possibilities•Fourth digit: 7 possibilitiesTotal = 10*9*8*7 = 5040b) End with an even digit?First three digits have 10 possibilitiesLast digit has 5 possibilitiesTotal = 10*10*10*5 = 50005The sum rule•Also called the addition rule•If there are n1 ways to do task 1, and n2 ways to do task 2–If these tasks can be done at the same time, then…–Then there are n1+n2 ways to do one of the two tasks–We must make one choice OR a second choice6Sum rule example•Sample question–There are 18 math majors and 325 CS majors–How many ways are there to pick one math major or one CS major?•Total is 18 + 325 = 3437Sum rule exampleMore sample questions•How many strings of 4 decimal digits…•Have exactly three digits that are 9s?–The string can have:•The non-9 as the first digit•OR the non-9 as the second digit•OR the non-9 as the third digit•OR the non-9 as the fourth digit•Thus, we use the sum rule–For each of those cases, there are 9 possibilities for the non-9 digit (any number other than 9)–Thus, the answer is 9+9+9+9 = 368More complex counting problems•We combining the product rule and the sum rule•Thus we can solve more interesting and complex problems9Wedding pictures example•Consider a wedding picture of 6 people–There are 10 people, including the bride and grooma) How many possibilities are there if the bride must be in the pictureProduct rule: place the bride AND then place the rest of the partyFirst place the bride•She can be in one of 6 positionsNext, place the other five people via the product rule•There are 9 people to choose for the second person, 8 for the third, etc.•Total = 9*8*7*6*5 = 15120Product rule yields 6 * 15120 = 90,720 possibilities10Wedding pictures example•Consider a wedding picture of 6 people–There are 10 people, including the bride and groomb) How many possibilities are there if the bride and groom must both be in the pictureProduct rule: place the bride/groom AND then place the rest of the partyFirst place the bride and groom•She can be in one of 6 positions•He can be in one 5 remaining positions•Total of 30 possibilitiesNext, place the other four people via the product rule•There are 8 people to choose for the third person, 7 for the fourth, etc.•Total = 8*7*6*5 = 1680Product rule yields 30 * 1680 = 50,400 possibilities11Wedding pictures example•Consider a wedding picture of 6 people–There are 10 people, including the bride and groomc) How many possibilities are there if only one of the bride and groom are in the pictureSum rule: place only the bride•Product rule: place the bride AND then place the rest of the party•First place the brideShe can be in one of 6 positions•Next, place the other five people via the product ruleThere are 8 people to choose for the second person, 7 for the third, etc.»We can’t choose the groom!Total = 8*7*6*5*4 = 6720•Product rule yields 6 * 6720 = 40,320 possibilitiesOR place only the groom•Same possibilities as for bride: 40,320Sum rule yields 40,320 + 40,320 = 80,640 possibilities12Wedding pictures example•Consider a wedding picture of 6 people–There are 10 people, including the bride and groom•Alternative means to get the answerc) How many possibilities are there if only one of the bride and groom are in the pictureTotal ways to place the bride (with or without groom): 90,720 •From part (a)Total ways for both the bride and groom: 50,400•From part (b)Total ways to place ONLY the bride: 90,720 – 50,400 = 40,320Same number for the groomTotal = 40,320 + 40,320 = 80,64013The inclusion-exclusion principle•When counting the possibilities, we can’t include a given outcome more than once!•|A1U A2| = |A1| + |A2| - |A1∩ A2|–Let A1 have 5 elements, A2 have 3 elements, and 1 element be both in A1 and A2–Total in the union is 5+3-1 = 7, not 814Inclusion-exclusion example•How may bit strings of length eight start with 1 or end with 00?•Count bit strings that start with 1–Rest of bits can be anything: 27 = 128–This is |A1|•Count bit strings that end with 00–Rest of bits can be anything: 26 = 64–This is |A2|•Count bit strings that both start with 1 and end with 00–Rest of the bits can be anything: 25 = 32–This is |A1∩ A2|•Use formula |A1U A2| = |A1| + |A2| - |A1∩ A2|•Total is 128 + 64 – 32 = 16015Bit string possibilities•How many bit strings of length 10 contain either 5 consecutive 0s or 5 consecutive 1s?16Bit string possibilities•Consider 5 consecutive 0s first•Sum rule: the 5 consecutive 0’s can start at position 1, 2, 3, 4, 5, or 6–Starting at position 1•Remaining 5 bits can be anything: 25 = 32–Starting at position 2•First bit must be a 1–Otherwise, we are including possibilities from the previous case!•Remaining bits can be anything: 24 = 16–Starting at position 3•Second bit must be a 1 (same reason as above)•First bit and last 3 bits can be anything: 24 = 16–Starting at positions 4 and 5 and 6•Same as starting at positions 2 or 3: 16 each–Total = 32 + 16 + 16 + 16 + 16 + 16 = 112•The 5 consecutive 1’s follow the same pattern, and have 112 possibilities•There are two cases counted twice