Shortcut of Permutation-1

Shortcut 91: Events   Dependent=multiplication   Independent =addition Question: There 3 busses from city A to city B and there are 5 busses from city B to city C. In how many ways a person can travel from city A to C through B? Answer: Choosing a bus from city B depends on choosing a bus from city A. Number of ways = 3 x 5 = 15 —————————————————————– Question: There 3 busses from city A to city B and there are 5 busses from city A to city C. In how many ways a person can travel from city A to C or B? Answer: Choosing a bus to city B or C are not dependent. Number of ways = 3 + 5 = 8 Shortcut 92: Arrangement with repetition [latex]n^n;n^r[/latex] Question: In how many ways the letters of the word “ORANGE” can be arranged with repetition? Answer: n = 6   (n = total number of elements) Since all the elements are taken, Number of arrangements = 66 —————————————————————- Question: In how many ways three letters from the word “ORANGE” can be arranged with repetition? Answer: n = 6; r = 3    (r = number of elements taken for arrangement) Number of arrangements = 63 = 216 Shortcut 93: Arrangement without repetition      [latex]n!;nP_r=n!/(n-r)![/latex] Question: In how many ways the letters of the word “MANGO” can be arranged without repetition? Answer: n = 5 Since all the elements are taken for arrangement, Number of elements = n! = 5! = 120 Question: In how many ways any three letters of the word ‘MANGO” can be arranged without repetition? Answer: n= 5; r = 3 nPr = 5!/(5 – 3)! = 120/2 = 60 ways Shortcut 94: Elements occurring together    2!*(n-1)    3!*(n-2)  Question: In how many ways letters of the word “ORANGE” arranged so that the vowels will always occur together? Answer: n = 6 Three letters ‘O, A and E’ should occur together. Since three letters occur together 3! x (6 – 2)! = 6 x 24 = 144 Note: If there are 4 letters occurring together, then 4!(6 – 3)! If there are 5 letters occurring together, then 5!(6 – 4)!]]>

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