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Probability-III (Permutations and Combinations)

Probability-III (Permutations and Combinations). QSCI 381 – Lecture 11 (Larson and Farber, Sect 3.4). Permutations-I. A is an ordered arrangement of objects. The number of different arrangements (permutations) of n different objects is n ! ( n -factorial).

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Probability-III (Permutations and Combinations)

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  1. Probability-III(Permutations and Combinations) QSCI 381 – Lecture 11 (Larson and Farber, Sect 3.4)

  2. Permutations-I • A is an ordered arrangement of objects. The number of different arrangements (permutations) of n different objects is n! (n-factorial). • Key note: We are interested in the order of things. Permutation

  3. Permutations-I 1 2 3 • 3! = 3 x 2 x 1 = 6

  4. Permutations-II • The number of permutations of n distinct objects taken r (r <n) at a time is: Returning to our earlier example; selecting 4 of 10 animals to age. Here n=10 and r=4 so

  5. Permutations-II • There are forty-three republican candidates hoping to be in the top three candidates at each primary election. In each state, how many ways can the candidates finish first, second, and third? • nPr = 43P3 = 43!/(43-3)! = 43!/40! = 43 x 42 x 41 74,046 primary standing options

  6. Permutations-III • The number of of n objects when n1 are of one type, n2 are of another type, etc. is: Distinguishable permutations = 6!/(3! x 2! X 1!) = (6 x 5 x 4)/(2 x 1) = 120/2 = 60

  7. Permutations-III(Example) • You have five sockeye, four pink and four chum salmon. How many ways can they be arranged?

  8. Permutations-III(Example) • You have five sockeye, four pink and four chum salmon. How many ways can they be arranged?

  9. Combinations-I Combination • A is a selection of r objects from a group of n objects without regard to order and is denoted by . The number of combinations of r objects selected from a group of n objects is: • Key note: We are not interested in the order of things.

  10. Combinations-II • Example. We have fish from three species (A, B, C) and wish to select two of them. How many ways can we do this: • (A,B), (B,C), (A,C). 3C2=3 • It is not • (A,B),(B,C),(A,C),(B,A),(C,B),(C,A) Because order does not count. 3!/(3-2)!2! = 6/2 = 3

  11. Combinations-III(Example) • You wish to test the impact of five treatments. However, your experiment can only consider three treatments at once. How many experiments do you need to conduct so that all combinations of the three treatments are examined? • Hint: after the lecture write down all the combinations after naming them A, B, C, D, E.

  12. Combinations -IV • The number of combinations of r objects chosen from n distinct objects allowing for repetitions is:

  13. Combinations vs Permutations-I • Permutations relate to the number of ways of selecting r objects from nwhen the order of the items matters. • Combinations relate to the number of ways of selecting r objects from nwhen the order does not matter.

  14. Combinations vs Permutations-II

  15. Relationship to Probability-I • If there are n people together - what is the probability that all have different birthdays: • Total number of permutations of birthdays: 365n (fundamental counting rule 365x365…) • Number of ways that n people can have different birthdays: 365Pn=365!/(365-n)! • For n=23, the ratio 365P23 / 36523 =0.493. Permutation without repetition

  16. More on Birthdays and Probability Number of people

  17. Relationship to Probability-II • A subgroup of four people is selected at random from a group of 5 married couples. What is the probability that the selection consists of two married couples: • Number of ways of selecting 4 people from 10 is: 10!/(6!.4!) • Number of ways of selecting 2 couples from 5 couples is : 5!/(3!.2!) • The ratio of these is 0.0476. Combination without repetition

  18. Relationship to Probability-III • There are 33 candidates in an election to a Committee of 3. What is the probability that Smith, Jones and Williams are elected (assume that elections are random): • We are interested in one combination of all possible combinations. • There 33C3 combinations of 3 selected from 33 so the probability 1/33C3=0.000183. • Hint. The factorial function in EXCEL is “Fact(x)” and there are also functions “Combin(A,B)” and “Permut(A,B)”.

  19. Relationship to Probability-IV • An experiment involves three counting devices. Each has a probability of failure of 0.02. • What is the probability that one fails? • What is the probability that none fails? • What is the probability at least two don’t fail?

  20. Common Mistakes • Assuming probabilities have a memory: You toss a (fair) coin and get eight heads – what was the probability of this and what is the probability of an eight head? 0.004 0.5 0.25 • Adding probabilities incorrectly. Why is the probability of surveying a pregnant female bowhead whale not 0.7 if the probability of encountering a female is 0.5 and 20% of the population is pregnant?

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