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Understanding Combinations: Calculating nCr for Different Scenarios

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fiona
40 diapositives · 683 Views · Aug 06, 2014

This guide explains combinations of n objects taken r at a time, where the order does not matter. The formula for combinations is nCr = n!/(r!(n-r)!), where n is the total number of objects and r is the number of objects chosen. For example, if 16 construction companies are bidding for a job, you can calculate the different combinations of 4 companies using this formula, resulting in 1820 unique combinations. This guide also includes a calculator method for quick determination of combinations.

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  1. 3.4C-Combinations • Combinationsof n objects taken r at a time • Order does NOT matter • nCr = n!/(n-r)!n! n=# objects r=# taken • Calculator: n, MATH, → PRB, 3:nCr, r, ENTER • Ex: 16 construction companies are bidding to do the job. How many different combinations of 4 companies can be chosen? .

  2. 3.4C-Combinations • Combinationsof n objects taken r at a time • Order does NOT matter • nCr = n!/(n-r)!n! n=# objects r=# taken • Calculator: n, MATH, → PRB, 3:nCr, r, ENTER • Ex: 16 construction companies are bidding to do the job. How many different combinations of 4 companies can be chosen? no particular order .

  3. 3.4C-Combinations • Combinationsof n objects taken r at a time • Order does NOT matter • nCr = n!/(n-r)!n! n=# objects r=# taken • Calculator: n, MATH, → PRB, 3:nCr, r, ENTER • Ex: 16 construction companies are bidding to do the job. How many different combinations of 4 companies can be chosen? no particular order . ₁₆C₄ =

  4. 3.4C-Combinations • Combinationsof n objects taken r at a time • Order does NOT matter • nCr = n!/(n-r)!n! n=# objects r=# taken • Calculator: n, MATH, → PRB, 3:nCr, r, ENTER • Ex: 16 construction companies are bidding to do the job. How many different combinations of 4 companies can be chosen? no particular order . ₁₆C₄ = 16!/(16-4)!4!

  5. 3.4C-Combinations • Combinationsof n objects taken r at a time • Order does NOT matter • nCr = n!/(n-r)!n! n=# objects r=# taken • Calculator: n, MATH, → PRB, 3:nCr, r, ENTER • Ex: 16 construction companies are bidding to do the job. How many different combinations of 4 companies can be chosen? no particular order . ₁₆C₄ = 16!/(16-4)!4! = 16!/(12!4!)

  6. 3.4C-Combinations • Combinationsof n objects taken r at a time • Order does NOT matter • nCr = n!/(n-r)!n! n=# objects r=# taken • Calculator: n, MATH, → PRB, 3:nCr, r, ENTER • Ex: 16 construction companies are bidding to do the job. How many different combinations of 4 companies can be chosen? no particular order . ₁₆C₄ = 16!/(16-4)!4! = 16!/(12!4!)=1820

  7. More Examples: • There are 16 employees. In how many ways can an advisory committee of 3 be chosen? • A word has 1 M, 4 I’s, 4 S’s & 2 P’s. What is the probability they can be arranged to spell MISSISSIPPI? .

  8. More Examples: • There are 16 employees. In how many ways can an advisory committee of 3 be chosen? order NOT important • A word has 1 M, 4 I’s, 4 S’s & 2 P’s. What is the probability they can be arranged to spell MISSISSIPPI? .

  9. More Examples: • There are 16 employees. In how many ways can an advisory committee of 3 be chosen? order NOT important ₁₆C₃ = • A word has 1 M, 4 I’s, 4 S’s & 2 P’s. What is the probability they can be arranged to spell MISSISSIPPI? .

  10. More Examples: • There are 16 employees. In how many ways can an advisory committee of 3 be chosen? order NOT important ₁₆C₃ = 16!/(16-3)!3! • A word has 1 M, 4 I’s, 4 S’s & 2 P’s. What is the probability they can be arranged to spell MISSISSIPPI? .

  11. More Examples: • There are 16 employees. In how many ways can an advisory committee of 3 be chosen? order NOT important ₁₆C₃ = 16!/(16-3)!3! = 16!/(13!3!) • A word has 1 M, 4 I’s, 4 S’s & 2 P’s. What is the probability they can be arranged to spell MISSISSIPPI? .

  12. More Examples: • There are 16 employees. In how many ways can an advisory committee of 3 be chosen? order NOT important ₁₆C₃ = 16!/(16-3)!3! = 16!/(13!3!) = 560 • A word has 1 M, 4 I’s, 4 S’s & 2 P’s. What is the probability they can be arranged to spell MISSISSIPPI? .

  13. More Examples: • There are 16 employees. In how many ways can an advisory committee of 3 be chosen? order NOT important ₁₆C₃ = 16!/(16-3)!3! = 16!/(13!3!) = 560 • A word has 1 M, 4 I’s, 4 S’s & 2 P’s. What is the probability they can be arranged to spell MISSISSIPPI? order IS important (distinguishable permutation) .

  14. More Examples: • There are 16 employees. In how many ways can an advisory committee of 3 be chosen? order NOT important ₁₆C₃ = 16!/(16-3)!3! = 16!/(13!3!) = 560 • A word has 1 M, 4 I’s, 4 S’s & 2 P’s. What is the probability they can be arranged to spell MISSISSIPPI? order IS important (distinguishable permutation) . 11!/(1!4!4!2!)

  15. More Examples: • There are 16 employees. In how many ways can an advisory committee of 3 be chosen? order NOT important ₁₆C₃ = 16!/(16-3)!3! = 16!/(13!3!) = 560 • A word has 1 M, 4 I’s, 4 S’s & 2 P’s. What is the probability they can be arranged to spell MISSISSIPPI? order IS important (distinguishable permutation) . 11!/(1!4!4!2!) = 34650 possible & 1 favorable 1/34650

  16. More Examples: • There are 16 employees. In how many ways can an advisory committee of 3 be chosen? order NOT important ₁₆C₃ = 16!/(16-3)!3! = 16!/(13!3!) = 560 • A word has 1 M, 4 I’s, 4 S’s & 2 P’s. What is the probability they can be arranged to spell MISSISSIPPI? order IS important (distinguishable permutation) . 11!/(1!4!4!2!) = 34650 possible & 1 favorable 1/34650 = .000029

  17. More Examples • A word has 1 L, 2 E’s, 2 T’s, & 1R. What is the probability they will be arranged to spell LETER? • Find the probability of being dealt 5 diamonds in a row from a standard deck of 52. .

  18. More Examples • A word has 1 L, 2 E’s, 2 T’s, & 1R. What is the probability they will be arranged to spell LETER? Order IS important • Find the probability of being dealt 5 diamonds in a row from a standard deck of 52. .

  19. More Examples • A word has 1 L, 2 E’s, 2 T’s, & 1R. What is the probability they will be arranged to spell LETER? Order IS important Distinguishable permutation • Find the probability of being dealt 5 diamonds in a row from a standard deck of 52. .

  20. More Examples • A word has 1 L, 2 E’s, 2 T’s, & 1R. What is the probability they will be arranged to spell LETER? Order IS important Distinguishable permutation 6!/(1!2!2!1!) = • Find the probability of being dealt 5 diamonds in a row from a standard deck of 52. .

  21. More Examples • A word has 1 L, 2 E’s, 2 T’s, & 1R. What is the probability they will be arranged to spell LETER? Order IS important Distinguishable permutation 6!/(1!2!2!1!) = 180 possible, 1 desirable • Find the probability of being dealt 5 diamonds in a row from a standard deck of 52. .

  22. More Examples • A word has 1 L, 2 E’s, 2 T’s, & 1R. What is the probability they will be arranged to spell LETER? Order IS important Distinguishable permutation 6!/(1!2!2!1!) = 180 possible, 1 desirable = 1/180 • Find the probability of being dealt 5 diamonds in a row from a standard deck of 52. .

  23. More Examples • A word has 1 L, 2 E’s, 2 T’s, & 1R. What is the probability they will be arranged to spell LETER? Order IS important Distinguishable permutation 6!/(1!2!2!1!) = 180 possible, 1 desirable = 1/180 = .0056 • Find the probability of being dealt 5 diamonds in a row from a standard deck of 52. .

  24. More Examples • A word has 1 L, 2 E’s, 2 T’s, & 1R. What is the probability they will be arranged to spell LETER? Order IS important Distinguishable permutation 6!/(1!2!2!1!) = 180 possible, 1 desirable = 1/180 = .0056 • Find the probability of being dealt 5 diamonds in a row from a standard deck of 52. Order NOT important .

  25. More Examples • A word has 1 L, 2 E’s, 2 T’s, & 1R. What is the probability they will be arranged to spell LETER? Order IS important Distinguishable permutation 6!/(1!2!2!1!) = 180 possible, 1 desirable = 1/180 = .0056 • Find the probability of being dealt 5 diamonds in a row from a standard deck of 52. Order NOT important P(desirable)/P(Total) .

  26. More Examples • A word has 1 L, 2 E’s, 2 T’s, & 1R. What is the probability they will be arranged to spell LETER? Order IS important Distinguishable permutation 6!/(1!2!2!1!) = 180 possible, 1 desirable = 1/180 = .0056 • Find the probability of being dealt 5 diamonds in a row from a standard deck of 52. Order NOT important P(desirable)/P(Total)=P(5 of 13 D)/P(5 of 52) .

  27. More Examples • A word has 1 L, 2 E’s, 2 T’s, & 1R. What is the probability they will be arranged to spell LETER? Order IS important Distinguishable permutation 6!/(1!2!2!1!) = 180 possible, 1 desirable = 1/180 = .0056 • Find the probability of being dealt 5 diamonds in a row from a standard deck of 52. Order NOT important P(desirable)/P(Total)=P(5 of 13 D)/P(5 of 52) . ₁₃C₅/₅₂C₅

  28. More Examples • A word has 1 L, 2 E’s, 2 T’s, & 1R. What is the probability they will be arranged to spell LETER? Order IS important Distinguishable permutation 6!/(1!2!2!1!) = 180 possible, 1 desirable = 1/180 = .0056 • Find the probability of being dealt 5 diamonds in a row from a standard deck of 52. Order NOT important P(desirable)/P(Total)=P(5 of 13 D)/P(5 of 52) . ₁₃C₅/₅₂C₅ = 1287/2598960

  29. More Examples • A word has 1 L, 2 E’s, 2 T’s, & 1R. What is the probability they will be arranged to spell LETER? Order IS important Distinguishable permutation 6!/(1!2!2!1!) = 180 possible, 1 desirable = 1/180 = .0056 • Find the probability of being dealt 5 diamonds in a row from a standard deck of 52. Order NOT important P(desirable)/P(Total)=P(5 of 13 D)/P(5 of 52) . ₁₃C₅/₅₂C₅ = 1287/2598960 = .0005

  30. More Examples: • A group contains 5 men and 7 women. If 3 are selected at random, what is the probability that all 3 are men?

  31. More Examples: • A group contains 5 men and 7 women. If 3 are selected at random, what is the probability that all 3 are men? Order is NOT important

  32. More Examples: • A group contains 5 men and 7 women. If 3 are selected at random, what is the probability that all 3 are men? Order is NOT important P(3 M)·P(0W)

  33. More Examples: • A group contains 5 men and 7 women. If 3 are selected at random, what is the probability that all 3 are men? Order is NOT important P(3 M)·P(0W) = ₅C₃· ₇C₀ =

  34. More Examples: • A group contains 5 men and 7 women. If 3 are selected at random, what is the probability that all 3 are men? Order is NOT important P(3 M)·P(0W) = ₅C₃· ₇C₀ = 10·1

  35. More Examples: • A group contains 5 men and 7 women. If 3 are selected at random, what is the probability that all 3 are men? Order is NOT important P(3 M)·P(0W) = ₅C₃· ₇C₀ = 10·1 =10 P(choose 3 from 12)

  36. More Examples: • A group contains 5 men and 7 women. If 3 are selected at random, what is the probability that all 3 are men? Order is NOT important P(3 M)·P(0W) = ₅C₃· ₇C₀ = 10·1 =10 P(choose 3 from 12) = ₁₂C₃ =

  37. More Examples: • A group contains 5 men and 7 women. If 3 are selected at random, what is the probability that all 3 are men? Order is NOT important P(3 M)·P(0W) = ₅C₃· ₇C₀ = 10·1 =10 P(choose 3 from 12) = ₁₂C₃ = 220

  38. More Examples: • A group contains 5 men and 7 women. If 3 are selected at random, what is the probability that all 3 are men? Order is NOT important P(3 M)·P(0W) = ₅C₃· ₇C₀ = 10·1 =10 P(choose 3 from 12) = ₁₂C₃ = 220 P(all men) = P(3 M)·P( 0 W) / P(3 from 12)

  39. More Examples: • A group contains 5 men and 7 women. If 3 are selected at random, what is the probability that all 3 are men? Order is NOT important P(3 M)·P(0W) = ₅C₃· ₇C₀ = 10·1 =10 P(choose 3 from 12) = ₁₂C₃ = 220 P(all men) = P(3 M)·P( 0 W) / P(3 from 12) 10/220 =

  40. More Examples: • A group contains 5 men and 7 women. If 3 are selected at random, what is the probability that all 3 are men? Order is NOT important P(3 M)·P(0W) = ₅C₃· ₇C₀ = 10·1 =10 P(choose 3 from 12) = ₁₂C₃ = 220 P(all men) = P(3 M)·P( 0 W) / P(3 from 12) 10/220 = .045