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Last time we discussed: The rule of sum The rule of product

Last time we discussed: The rule of sum The rule of product Permutations = arrangements when the order is important all objects are distinct/not all are distinct without repetition/ with repetitions. Combinations =selections when the order does not matter . 2. 6.

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Last time we discussed: The rule of sum The rule of product

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  1. Last time we discussed: • The rule of sum • The rule of product • Permutations = arrangements when the order is important • all objects are distinct/not all are distinct • without repetition/ with repetitions • Combinations =selections when the order does not matter

  2. 2 6 Suppose two dice are rolled, one blue and one red. • How many outcomes are possible? Product rule: 66=36 • How many outcomes are doubles? The rule of sum: 6 • How many outcomes give the sum 7 or the sum 11? 1+6 2+5 3+4 4+3 5+2 6+1 5+6 6+5

  3. How many outcomes have the blue die showing 2? 6 • How many outcomes have exactly one die showing 2? 5+5 • How many outcomes have at least one die showing 2? 10+1 • How many outcomes have neither die showing 2? 55 3611

  4. How many outcomes give an even sum? Possible even sums: 12, 10, 8, 6, 4, 2. 10=5+5 4+6 6+4 8=6+2 5+3 4+4 3+5 2+6 12=6+6 6=5+1 4+2 3+3 2+4 1+5 4=3+1 2+2 1+3 2=1+1

  5. ABC DEF GHI JKL There are 3! ways to move people in ABC around without changing an assignment. The same hold for all other groups. Suppose you have 12 students in the recitation class who are supposed to break up into 4 groups of 3 students each. How many different group assignments are possible? (There is no “numbering” of the groups. All that matters is who collaborates with whom). There are 12! different lists (permutations), but not all of them result in a different group assignment. How many different permutations represent the same group assignment? We can also permute different groups in 4! ways.

  6. So, there are 12! different lists, and each assignment is produced by (3!)44! different lists. Ans: 12!/((3!)44!)

  7. After that you select 3 from 6 : Alternative solution. You can select 3 people out of 12 for the first group in Select 3 people for the second group out of remaining 9 in different ways. different ways. By the product rule the total number of different assignments is And you need to take into account that groups are not distinguishable.

  8. We can represent this problem as the arrangements of 4 identical objects and one separator |: {a, a, a, a} ****| {a, a, a, b} ***|* {a, a, b, b} **|** {a, b, b, b} *|*** {b, b, b, b} |**** 5-permutations of 5 objects if 4 of the them are identical: Combinations with repetitions. Consider 4-combinations of {a,b}: {a, a, a, a}, {a, a, a, b}, {a, a, b, b}, {a, b, b, b}, {b, b, b, b}

  9. One possible arrangement: Combinations with repetitions. Donut shop has 5 types of donuts. In how many ways we can select ten donuts? This problem can be represented as an equivalent arrangement of ten donuts into 5 boxes. All possible “distributions” Can be considered as “permutations” of a dozen of donuts and 4 separators between boxes:

  10. 14! = C (14, 4) From another side, any arrangement can be viewed asa selection of 4 numbers out of 14 (or 10 out of 14) 10!4! 1 2 3 4 5 6 7 8 9 1011 12 13 14 We need to count the number of permutations of 10 donuts and 4 separators. So, we have 14 objects, 4 of which are identical and 10 are identical.

  11. The number of r-combinations of n objects that can be repeated (any number of times) can be considered as the number of arrangements of r identical objects and n-1 separators (bars).

  12. x | y | z ** |*****|*** |**********| *******||*** How many distinct integer solutions of the equation x+y+z=10 exist if x, y, z0? Any solution corresponds to some distribution of 10 between three distinct boxes: How to count all possible arrangements of 10 stars and 2 bars?

  13. Pigeonhole Principle. If k +1 or more pigeons fly into k pigeonholes, then there is at least one pigeonhole containing two or more pigeons. Examples. Among any group of 366 people there must be at least two with the same birthday, because there are only 365 possible birthdays. In any group of 27 words there must be at least two that begin with the same letter. How many students must be in class to guarantee that at least two students receive the same score on the final (exam is graded on a scale from 0 to 100)? Ans. 102

  14. Generalized Pigeonhole principle. If n objects are placed into k boxes, then there is at least one box containing at least n/k objects. Here n/k denotes the largest integer less then n/k+1. Among 100 people there are at least 100/12= 8.333… =9 who were born in the same month. How many students, each of whom comes from one of 50 states, should be enrolled in a university to ensure that there are at least 10 coming from the same state? 451

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