Permutations and Combinations Permutation: The number of ways in which a subset of objects can be selected from a given set of objects, where order is important. Given the set of three letters, {A, B, C}, how many possibilities are there for selecting any two letters where order is important? (AB, AC, BC, BA, CA, CB) Combination: The number of ways in which a subset of objects can be selected from a given set of objects, where order is not important. Given the set of three letters, {A, B, C}, how many possibilities are there for selecting any two letters where order is not important? (AB, AC, BC).

Definition of n Factorial (n !) n! = n(n – 1)(n – 2)(n – 3)…1 For example, 5! = 5(4)(3)(2)(1) = 120 n! = n(n – 1)! 0! = 1 by definition. Most calculators have an n! key or the equivalent. n! grows very rapidly, which may result in overload on a calculator.

Example Vasopressin* is a small polypeptide composed of 9 amino acids in a particular order. The order of these amino acids is of critical to the proper functioning of vasopressin. If these 9 amino acids were placed in a hat and drawn out randomly one by one, how many different arrangements of these 9 amino acids are possible? *Retains water in the body and constricts blood vessels

Solution Solution: Let A, B, C, D, E, F, G, H, I symbolize the 9 amino acids. They must fill 9 slots: ___ ___ ___ ___ ___ ___ ___ ___ ___ There are 9 choices for the first position, leaving 8 choices for the second slot, 7 choices for the third slot and so on. The number of different orderings is 9(8)(7)(6)(5)(4)(3)(2)(1) = 9! = 362,880

Number of Possible Proteins Given 21 amino acids, how many different dipeptides can be formed? 21 possibilities in there first position 21 possibilities in the second position Therefore there are 21 x 21 = 441 permutations

Number of Possible Proteins in the Universe Given 21 amino acids, how many different proteins can be made that are 300 amino acids long? 21 possibilities in the first position etc Therefore there are 21300 = 4.6 x 10396 permutations By comparison, the number of all atoms in the observable universe is 1080

Permutations and Combinations Factorial Formula for Permutations The number of permutations, or arrangements, of n district things taken r at a time, where r <= n, is calculated using the expression: Factorial Formula for Combinations The number of combinations, or subsets, of n district things taken r at a time, where r <= n, is calculated using the expression:

Permutations and Combinations Example How many ways can you order the 4 objects {A, B, C, D} Solution: N = 4! = 4(3)(2)(1) = 24 Here are the orderings. ABCD ABDC ACBD ACDB ADBC ADCB BACD BADC BCAD BCDA BDAC BDCA CABD CADB CBAD CBDA CDAB CDBA DABC DACB DBAC DBCA DCAB DCBA

Permutations and Combinations Example How many permutations of size 3 can be found in the group of 5 objects {A, B, C, D, E} Solution: ABC ABD ABE ACD ACE ADE BCD BCE BDE CDE ACB ADB AEB ADC AEC AED BDC BEC BED CED BAC BAD BAE CAD CAE DAE CBD CBE DBE DCE BCA BDA BEA CDA CEA DEA CDB CEB DEB DEC CAB DAB EAB DAC EAC EAD DBC EBC EBD ECD DBA EBA DCA ECA EDA DCB ECB EDB EDC

Permutations and Combinations Find the number of permutations of 8 elements taken 3 at a time.   Find the number of permutations of the letters A, B, and C  

Using Permutations and Combinations A PhD examination includes 4 women and 3 male examiners. How many ways can the be examiners be seated in a row of 7 chairs?   If the male and female examiners have to be seated alternatively, how many combinations will there be? Sit a male in the first seat, female next etc.

Using Permutations and Combinations How many ways can you select two letters followed by three digits for an ID if repeats are not allowed? Two parts: 1. Determine the set of two letters. 2. Determine the set of three digits. 26P2 10P3         2625 1098 650 720 650720 468,000

Using Permutations and Combinations Find the number of different subsets of size 3 in the set: {m, a, t, h, r, o, c, k, s}. Find the number of arrangements of size 3 in the set: {m, a, t, h, r, o, c, k, s}. 9C3 9P3           987   504 arrangements 84 Different subsets

Using Permutations and Combinations Assume a committee of n = 10 people and we want to choose a chairperson, a vice-chairperson and a treasurer. Suppose that 6 of the members of the committee are male and 4 of the members are female. What is the probability that the three executives selected are all male? Solution: We select 3 persons (male and female) from the committee of 10 in a specific order. (Permutations of size 3 from a group of 10).The total number of ways that this can be done is: This is the size, N = n(S), of the sample space S. Assume all outcomes in the sample space are equally likely.

Using Permutations and Combinations Let E be the event that all three executives are male (6 males in total) Hence Thus if all candidates are equally likely to be selected to any position on the executive then the probability of selecting an all male executive is:

Application to Lotto 6/49 Here you choose 6 numbers from the integers 1, 2, 3, …, 47, 48, 49. Six winning numbers are chosen together with a bonus number. How many choices for the 6 winning numbers

Software Support Excel: Permutations (in EXCEL =PERMUT(6,2)) Combinations (in EXCEL =COMBIN(6,2)) Python: scipy.special.comb (N, k) scipy.special.perm (N, k)

One more thing An alternative and more common way to denote an r-combination: