In how many ways can final eleven be selected from 15 cricket players if Show
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ExampleOut of 5 boys and 4 girls, a committee of 5 is to be formed so as to include no more than 2 girls. In how many ways can it be done? SolutionWe may choose 5 boys and no girl in 5C5.4C0 = 1×1 = 1 way only. Example
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ExampleThere are 15 points in a plane, no three of which are in the same straight line except 4 which are collinear. Find the number of
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ExampleFind the number of (i) combinations (ii) permutations of four letters taken from the word EXAMINATION. SolutionThe word examination consists of 11 letters - (AA), (II), (NN), E, X, M, T, O. The following combinations are possible: (a) 2 alike, 2 alike:³C2 = 3 ways (b) 2 alike, 2 different:³C1×7C2 = 63 ways (c) all 4 different: 8C4 = 70 ways Hence required number of combinations = 3 +63 +70 = 136. To find the number of permutations: In (a), the number of permutations = 3×4!/[2!2!] = 18 In (b), the number of permutations = 63×4!/2! = 756 In (c), the number of permutations = 70×4!= 1680 Hence required number of permutations = 18 +756 +1680 = 2454. Exercise
Answers1. 4 2. 715 3. 200 4. 65. (i) 60 (ii) 3255 6. 252 7. 7800 8. (i) 60 (ii) 120 9. 420 10. 52!/(13!)4 11. 15 12. 209 13. (i) 90 (ii) 185 14. 63 15. 365 16. 15 17. 12 18. 16 men, 5 women 19. (i) 57 (ii) 20 20. [n (n -3)]/2 21. (i) 2454 (ii) 2454 |