Praveen Prasad an elevator starts with 5 passengers and stops at 5 diff floors of an apartment find the probability of all 5 passengers are of diff floors
option
a. 24/109
b. 55/412
c. 105/512
d. none

• there are 5^5 possible cases . now the first person can get off in 5 ways and the second will have as 4 ways and so on
  the probability will be
  (5*4*3*2*1)/5^5 ... so the answer will be 24/625
• 10 years agoHelpfull: Yes(36) No(0)
• 5!/5^5
  ans. none
• 10 years agoHelpfull: Yes(12) No(1)
• no.of cases where each person is of a "different" floor = 5!= 5*4*3*2*1= 120.
  total no. of cases of 5 persons belonging to any of the 5 floors=5*5*5*5*5=3125
  Hence the required probability= 120/3125.
  
• 10 years agoHelpfull: Yes(4) No(0)
• the probality should be 1 as 5 passengers are from different floors(5/5)abd there are 5 floors
  
• 10 years agoHelpfull: Yes(1) No(3)
• maximum case - if can go out of lift at same floor ie 5^5
  now if different floors then for 1st there are 5 cases for 2nd there are 4 3rd there are 3 for 4th there are 2 and for last 1 hence 5*4*3*2*1/5^5 hence none
• 10 years agoHelpfull: Yes(1) No(0)
• answer would be none as per the formula of combination and permutation......
  5!/r!*n-r!
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• a. 24/109
  
• 10 years agoHelpfull: Yes(0) No(1)