Permutation and combination - Math Statistic
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The document discusses permutations and combinations. It defines permutations as arrangements where order is important, and provides an example of finding the number of permutations of selecting 2 letters from a set of 3 letters. Combinations are arrangements where order is not important, and gives an example of finding the number of combinations of selecting 2 letters from a set of 3. It also presents the factorial formulas for permutations and combinations and provides examples of using them to calculate permutation and combination problems. Finally, it highlights the difference between permutations and combinations and provides 4 example problems to practice calculating permutations and combinations.
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Permutation and combination - Math Statistic
- 1. Math Statistic Cambridge UniversityPress By: danica p Permutation and Combination
- 2. Permutation, what sit? Permutation is an arrangement of elements of a set in which the order of the elements considered. Means, it explain the number of ways in which a subset of objects can be selected from a given set of objects, where order is important. Example: Given the set of three letters, {A, B, C}, how many possibilities are there for selecting any two letters where order is important? Answer: (AB, AC, BC, BA, CA, CB) there is 6 arrangements are possible. from A: A-B, A-C (2 arrangements) From B: B-A, B-C (2 arrangements) From C: C-A, C-B (2 arrangements) Total: 2*3 = 6 arrangements Notice! PERMUTATION (P) = ARRANGEMENT !
- 3. Factorial Formula ofPermutation ! . ( )! n r n P n r Notation of n = n! n! = n*(n-1)*(n-2)*.......*2*1 e.g. 9!=9*8*7*6*5*4*3*2*1 * = multiply Example: 15 runners are hoping to take part in a marathon competition, but the track has only 4 lines. How many ways can 4 of the 15 runers be assigned to lanes? Answer: Using Factorial formula of P, 15P4 = = 32760 1*2*3*4 1*2*3*4*5*6*7*8*9*10*11*12*13*14*15 )!1115( !15 You can using your scientific calcuator, it simple, you enter number n, then shift-x-then enter the number of r, try it! 15-shift-x-P- then, 4, so it became to 15P4 = 32760
- 4. Then, what abouta combination? Combination is an arrangement of elements of a set in which the order of elements is not considered. Means: The number of ways in which a subset of objects can be selected from a given set of objects, where order is not important. Example: Given the set of three letters, {A, B, C}, how many possibilities are there for selecting any two letters where order is not important? Answer: (AB, AC, BC) 3 combinations. each alphabet has different couple combination, and that is: A can together with B, A can together with C, B can together with C, total: 3 Notice! COMBINATION (C) = CHOOSE!
- 5. Factorial Formula forCombinations ! . ! !( )! n r n r P n C r r n r Example: The manager of a football team has a squad of 18 players. He needs to choose 11 to play in a match. How many possible teams can be choosen? Answer: Using Factional formula of C, 18C11= = 31824 !7!*11 !18 )!1118(!11 !18 !11 1118P You can using your scientific calcuator, it simple, you enter number n, then shift-( )-then enter the number of r, try it! 18-shift-( )-P- then, 11 so it became to 18C11= 31824
- 6. Difference between Permutationand Combination
- 7. Exercises: 1. From agroup of 7 men and 6 women, five persons are to be selected to form a committee so that at least 3 men are there on the committee. In how many ways can it be done? A.564 B.645 C.735 D.756 E.None of these Answer: Option D Explanation: We may have (3 men and 2 women) or (4 men and 1 woman) or (5 men only). Required number of ways= (7C3 x 6C2) + (7C4 x 6C1) + (7C5)=7 x 6 x 5x6 x 5+ (7C3 x 6C1) + (7C2)3 x 2 x 12 x 1= 525 +7 x 6 x 5x 6+7 x 63 x 2 x 12 x 1= (525 + 210 + 21)= 756. 2. A box contains 2 white balls, 3 black balls and 4 red balls. In how many ways can 3 balls be drawn from the box, if at least one black ball is to be included in the draw? A.32 B.48 C.64 D.96 E.None of these Answer: Option C Explanation: We may have(1 black and 2 non-black) or (2 black and 1 non-black) or (3 black). Required number of ways= (3C1 x 6C2) + (3C2 x 6C1) + (3C3)=3 x6 x 5+3 x 2x 6+ 12 x 12 x 1= (45 + 18 + 1)= 64.
- 8. 3. In howmany different ways can the letters of the word 'LEADING' be arranged in such a way that the vowels always come together? A.360 B.480 C.720 D.5040 E.None of these Answer: Option C Explanation: The word 'LEADING' has 7 different letters. When the vowels EAI are always together, they can be supposed to form one letter. Then, we have to arrange the letters LNDG (EAI). Now, 5 (4 + 1 = 5) letters can be arranged in 5! = 120 ways. The vowels (EAI) can be arranged among themselves in 3! = 6 ways. Required number of ways = (120 x 6) = 720. 4. In how many ways can the letters of the word 'LEADER' be arranged? A.72 B.144 C.360 D.720 E.None of these Answer: Option C Explanation: The word 'LEADER' contains 6 letters, namely 1L, 2E, 1A, 1D and 1R. Required number of ways = = 360 )!1)(!1)(!1)(!2)(!1( !6
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