There are 6 letters and 6 houses. Find Probability that
1. no house receives correct letter.
2. at least one house gets a correct letter.

• Correct above solution
  1. P(no house receive correct letter)=
  6!(1/1!-1/2!+1/3!-1/4!+1/5!-1/6!)/6!=455/720
  2. P(at least one house get a correct letter)=1-455/720=265/720
• 7 years agoHelpfull: Yes(11) No(0)
• The formula for the derangement can be used directly. The number of derangements is
  n!(1/2!−1/3!+1/4!+…+(−1)n/n!)
  
  Ans 1) 265/720
  Ans 2) 1- 265/720 = 455/720
• 7 years agoHelpfull: Yes(5) No(1)
• 2)at least one house gets correct letter----->1/1! - 1/2! + 1/3! - 1/4! + 1/5! - 1/6!
  1)no house= 1- at least one= 1/2! - 1/3! + 1/4! - 1/5! + 1/6!
• 7 years agoHelpfull: Yes(2) No(7)
• hello guys i am yaswanth can any one help out in cracking the elitmus i am taking the test in December. please send me any material if you have to yaswanth.1424@gmail.com . Please guys , please help me. thanks in advance :-)
• 7 years agoHelpfull: Yes(1) No(15)
• 1. P( no. house receive correct letter) =
  1/ 10!
  2. P(at least one house get a correct letter)= 1-1/10!
• 7 years agoHelpfull: Yes(0) No(3)
• 2) atleast one gets correct letter = 6!/1!*5! + 6!/2!*4! + 6!/3!*3! + 6!/4!*2! + 6!/5!*1! + 6!/6!*0! / 6! =6+15+20+15+6+1/720 = 7/80
  1) No house= 1-atleast = 1-7/80
• 7 years agoHelpfull: Yes(0) No(2)
• D(n)=n![1-1/1!+1/2!-1/3! .......(-1)^n 1/n!]
  
  p(no house receive correct letter)=D(6)
  6!(1-1/1!+1/2!-1/3!+1/4!-1/5!+1/6!)/6!=265/720
  
  p(atleast one)=1-265/720=464/720
• 7 years agoHelpfull: Yes(0) No(0)
• (1) 145
  (2) 575
• 7 years agoHelpfull: Yes(0) No(0)