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Basic counting principles, day 1

To decide how many ways something can occur, you can draw a tree diagram. . Basic counting principles, day 1. Note that this only shows half of the tree – the automatic transmission. Possible choices = 2 * 3 * 4 = 24 choices transmission * music * color

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Basic counting principles, day 1

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  1. To decide how many ways something can occur, you can draw a tree diagram. Basic counting principles, day 1 Note that this only shows half of the tree – the automatic transmission

  2. Possible choices = 2 * 3 * 4 = 24 choices transmission * music * color • Basic counting principle: If an event can occur in p ways, and another event in q ways, then there are p * q ways both events can occur. • From now on we will use multiplication and “fill the slots” as follows.

  3. How many different batting orders are there in a 9-person softball team? Fill the slots for each position

  4. 9! = 362,880 ways to write the batting order.

  5. How many 7 digit phone numbers are there if the first digit cannot be 0 or 1? ____*____*___* ____*____*____*____

  6. 8,000,000 ways

  7. How many 7-digit phone numbers begin with 867? ____*____ *____*____*____*____*____

  8. 1 * 1 * 1 * 10 * 10 * 10 * 10 these four digits can be 0-9 (10 choices) one choice one choice one choice 10,000 different phone numbers

  9. Using letters from the word “MATRIX” How many 4-letter patterns can be formed? # ways to fill slots: _____ * _____ * _____ * _____

  10. # ways to fill slots: 6 * 5 * 4 * 3 = 360 6 letters to choose from 5 letters left . . etc.

  11. Still using letters from the word “MATRIX”, what if the first letter must be a vowel? _____ * _____ * _____ * _____

  12. # ways to fill slots: 2 * 5 * 4 * 3 = 120 2 vowels to choose from 5 letters left . . etc.

  13. What if we fill only 4 slots and the first and last letters must be consonants? ______ * ______ * ______ * ______

  14. # ways to fill slots: 4 * 4 * 3 * 3 = 144 4 conso-nents to choose from 3 conso-nents left to choose from fill these two first 4 letters left . . 3 letters left . . fill less important slots last

  15. How many 3-digit palindromes are possible? Note that a number does not usually begin with a zero . . . ______ * ______ * ______

  16. 9 * 10 * 1 = 90 Can’t begin with a zero and be 3-digit fill first must be the same as the first digit (one way to fill) fill second can be any digit then fill the last one

  17. In a 5-card poker hand, the 1st 3 cards were red face cards, the last 2 were black non-face cards. How many ways can this happen? 5 slots to fill: _____ * ____ * _____ * _____* _____

  18. 6 * 5 * 4 * 20 * 19 = 45,600 ways first 3 cards choices: K, Q, J hearts or diamonds; 1st slot, 2nd slot, 3rd slot A – 10 and black: 20 cards; 1st slot, 2nd slot (of that type)

  19. Mutually Exclusive events • To get to school, Rita can either walk or ride the bus. If she walks, she can take 3 routes, if she rides the bus, there are 2 routes. Rita’s choices are mutually exclusive - she can’t do both, therefore, the possibilities are added to each other. • 3 ways to walk • 2 ways to take a bus • total possible routes: 3 + 2 = 5

  20. How many odd numbers between 10 and 1000 start and end with the same digit? We have two mutually exclusive events: ___ * ___ two digit choices + ___ * ___ * ___ three digit choices

  21. two digit choices: just count multiples of 11 that are odd: (11, 33, 55, 77, 99) 5 choices three digit choices: calculate like our palindrome problem (3 slots to fill) 1 * 10 * 5 50 choices total = 50 + 5 = 55 odd numbers between 10 and 1000 that have the same first and last digit.

  22. An example that is not mutually exclusive… (each time you have the same number of choices) An ID label has 4 letters. How many different labels are possible? ____ * _____ * _____ * _____

  23. From before, we know it is 26* 26 * 26 *26 = 264 = 456,976 • You can also look at it like: 264where 26 (base) = number of different choices 4 (exponent) = number of times you make that choice

  24. On a multiple choice test with 15 questions and 4 answer choices per question, how many answer combinations are there? 415=1,073,741,824!!!! 4 = number of different choices 15 = number of times you make that choice

  25. How many license plates of 2 symbols (letters and digits) can be made using at least one letter in each? • cases: • one letter: L D or D L 2(26*10) • two letters: • L L 262 • total: 2(26*10) + 262 = 1196 license plates

  26. How many ways can you make a 3 symbol license plate using at least one letter? What are the cases?

  27. Cases if the license plate has at least one letter: • one letter: • LDD or DLD or DDL 3(26*102) • two letters: + • L L D or L D L or D L L 3(262*10) • three letters: + • L L L 263

  28. Permutations • A permutation is an ordered arrangement: ORDER MATTERS (KAT is different than TAK). Our batting order problem and use of the letters of matrix have been permutations.

  29. How many ways can letters in SPRING be arranged? 6 * 5 * 4 * 3 * 2 * 1 = 6! = 720 ways to arrange the 6 letters when order matters. In general, there are n! permutations of n objects.

  30. What if we want to use the letters in SPRING, but only want to know how many 2 letter arrangements can be made? Using the “filling the slots” idea, we have 6 * 5 = 30 ways to fill two slots with 6 letter choices.

  31. Another way to look at it is similar to what we did for Pascal’s triangle: In general, the number of permutations of n objects taken r at a time is:

  32. From 1000 contest entries, how many ways can 1st, 2nd, and 3rd place prizes be awarded? Does order matter? We can fill in the slots: _____ * _____ * _____ Or use our formula:

  33. How does it change if the values are not all unique? How many 6-letter permutations of ACOSTA are there? note: there are two A’s that will not be distinguishable from each other in a word. A1A2COST = A2A1COST the number of permutations will be cut in half.

  34. Is the bottom just “2” because there are 2 A’s? What if there were more? Consider using the letters of PARABOLA:

  35. Consider PARABOLA the number of arrangements of the 3 A’s that will be equivalent are: A1A2A3PRBOL A1A3A2PRBOL A2A1A3PRBOL A2A3A1PRBOL A3A1A2PRBOL A3A2A1PRBOL This shows that 3! arrangements would be identical (the number of ways you can arrange 3 objects in different order). Thus, we need to divide by 3!, not just 3.

  36. This generalizes to objects that have more than one duplicate. If all objects are used, the number of permutations of n objects of which p and q represent the number of items that are alike is:

  37. How many permutations of the letters in the word POSSIBILITY using only 5 letters are there?

  38. where the denominator represents the 2 S’s and the 3 I’s.

  39. That’s all folks Have a counting, permutable kind of day 

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