In a locality there are ten houses in a row. On a particular night a thief planned to steal from three houses of locality. In how many ways can he plan such that no two houses are next to each other?
Options are:
56, 73, 80, 100

• The answer would be 56.
  Lets say we have removed the three said houses. Now we are left with 7 houses.
  If we arrange these 7 houses we have 8 places to put the other 3 houses at.
  
  _ H1 _ H2 _ H3 _ H4 _ H5 _ H6 _ H7 _
  
  so 8C3 = 56
• 8 years agoHelpfull: Yes(93) No(3)
• I think ,it will be 56...
  assume that the thief start from 1st house then 3rd house and then 5th house to any of remaining 5th houses..
  1st step:the thief can stole from 1st house ,3rd and from 5 house; there are 6+5+4+3+2+1=21 ways
  2nd step:the thief can stole from 2nd house,4th house and from 6th house ;there are 5+4+3+2+1=15 ways
  3rd step:the thief can stole from 3rd ,5th and from 7th; there are 4+3+2+1=10 ways
  4th step: the thief can stole from 4th , 6th and from 8th ;there are 3+2+1=6 ways
  5th step:the thief can stole from 5th ,7th and from 9th ;there are 2+1=3 ways
  6th step:the thief can stole from 6th , 8th and from 10th ;there is 1 way
  
  So total(1+3+6+10+15+21)=56
  
• 8 years agoHelpfull: Yes(16) No(10)
• The answer would be 56.
  Lets say we have removed the three said houses. Now we are left with 7 houses.
  If we arrange these 7 houses we have 8 places to put the other 3 houses at.
  
  _ H1 _ H2 _ H3 _ H4 _ H5 _ H6 _ H7 _
  
  so 8C3 = 56
• 4 years agoHelpfull: Yes(0) No(2)