an elevator starts with 5 passengers and stop with 5 diff floors of an apartment .find the probability of all 5 passengers are of diff floor?

• ans should be 5!/5^5
  sol : as each person can reach a floor in 5 different ways independently .so there are a total of 5^5 ways to reach i.e. sample space.
  and to reach at separate floor 1st person can reach in 5 ways,,2nd can in 4 ways,,3rd in 3 ways ,,4th in 2 ways and 5th can in 1 way ...i. e. n(e) = 5!
  therefore p(e) = n(e)/sampl space = 5!/5^5
  
• 10 years agoHelpfull: Yes(44) No(0)
• when anyone can go on any floor then
  1st can go to any floor in n ways.
  2nd can go to any floor in n ways.
  3rd can go to any floor in n ways.
  4th can go to any floor in n ways.
  5th can go to any floor in n ways.
  so total no. of ways passenger alighting at floors = n*n*n*n*n = n^5
  
  now when they go to different floor then :-
  1st can go to any floor in n ways.
  2nd can go to any floor in n-1 ways.
  3rd can go to any floor in n-2 ways.
  4th can go to any floor in n-3 ways.
  5th can go to any floor in n-4 ways.
  so probabolity required will be = n*(n-1)*(n-2)*(n-3)*(n-4)/n^5
• 10 years agoHelpfull: Yes(25) No(1)
• total no. of possible outcomes=5^5=3125
  no. of favourable outcomes=5c1*4c1*3c1*2c1*1c1=120
  so, p(e)=no. of favourable outcome/total no. of possible outcomes
  i.e 120/3125
  
• 10 years agoHelpfull: Yes(12) No(2)
• Yes, 5!/5^5 is correct answer !!!
• 10 years agoHelpfull: Yes(0) No(3)