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
n^n;n^r
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
     n!;nP_r=n!/(n-r)!
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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