RI Permutations and Combinations Probability Solns

RAFFLES INSTITUTION H2 Mathematics 9758 2023 Year 6 Term 3 Revision 13 (Summary and Tutorial) Topic: Permutations and Combinations, Probability Summary for Permutations and Combinations Definitions - A permutation is an ordered arrangement of objects - A combination is a selection of objects in which the order of selection does not matter. Permutations Given Objects taken No of Permutations n distinct objects n n ! n distinct objects r no repetitions r n P or r nC r! n distinct objects r with repetitions rn n objects, not all distinct (n1 of type 1, n2 of type 2, …, n k of type k, where 12 ... knn n n ) n 12 ! !! . . .! k n nn n n objects, not all distinct r Involves combinations & permutations See section on Combinations. Useful Techniques when dealing with restrictions: 1. Grouping or Slotting. (a) Grouping (“must be together”, “cannot be separated” etc.). Eg. No of ways to arrange the letters a, b, c, d, e, f, g such that a, b & c are together 5! 3! [ abc, d, e, f, g: 5 items  5! ways, within abc  3! ways ] (b) Slotting (“cannot be together”, “must be separated” etc.) Eg. No of ways to arrange the letters a, b, c, d, e, f, g such that a, b & c are not adjacent to each other   5 34! P [ arrange d, e, f, g first  4! ways, slot and permute a, b, c  5 3P ways ] Add or multiply? Mutually exclusive operations Doing one operation followed by another to get to the result. Add Multiply Case 1, 2, 3 etc… Step 1, 2, 3 etc…

2. Taking Complement (“two items cannot be together” or “at least 1”). Use this technique with care. Note that for Eg in 1(b) above, a, b, c not adjacent to each other is NOT the complement of abc together. Combinations Notation: r nC can also be written as       r n Note:              rn n r n = )!(! ! rnr n  . E.g.  !nn rrP Cr No of ways to select r objects Given : n distinct objects       r n Given: n objects, not all distinct No direct way to calculate. Need to consider different cases: (i) Start with case where the r selected objects are all distinct. (ii) Next, consider case(s) with pair(s) of identical objects, and so on. (iii) No of ways = Sum of the different cases Eg: To choose 3 letters (arrangements not required) Given: Letters a,b,c,d,e,f No of ways = 6 203   Given: a, a, a, b, b, c Case 1: All distinct (abc) – 1 way. Case 2: Contains an identical pair (aab, aac, bba, bbc) – 2 x 2 = 4 ways. Case 3: All identical (aaa) – 1 way. Total no of ways = 1 + 4 +1 = 6 Eg. Find the no of 3-letter codes (arrangements to be considered) Given: Letters a,b,c,d,e,f No of ways = 6 (3!) 1203   OR: 6 3 654 1 2 0P  Given: a, a, a, b, b, c Case 1: All distinct (abc) – 1x3! = 6 ways. Case 2: Contains an identical pair (aab, aac, bba, bbc) – 3!22 1 2 w a y s2!  . Case 3: All iden