Raju and Reenu are standing in a row there are 9 peoples. what is the probability that at least 3 people will stand between Raju and Reenu

• there are at least 3 people between Raju and Reena... people can be 4,5,6,7
  when there are 3 people between them... in 5*7!*2!
  when there are 4 people between them... in 4*7!*2!
  when there are 5 people between them... in 3*7!*2!
  when there are 6 people between them... in 2*7!*2!
  when there are 7 people between them... in 1*7!*2!
  so total 15*7!*2!....
  total no of possibilities... 9!
  probability will be 15*7!*2!/9!=5/12
• 9 years agoHelpfull: Yes(12) No(6)
• 9C2=36
  3 people between Raju and Reena 7+5+3=15
  15/36=5/12 Ans
• 9 years agoHelpfull: Yes(10) No(3)
• A shopkeeper has three types of rice pocket.all pocket is multiple of one kg. there are 4 persons A, B,C,D, each take exact two packet. and take 11kg,12kg,16kg, and 17kg respectively then
  
  Q1.how much maximum weight taken by any person
  
  Q2. how much minimum weight taken by any person
  ASKED in yesterday elitmus Exam
• 9 years agoHelpfull: Yes(10) No(0)
• 11 = 6 + 5
  12 = 6 + 6
  16 = 5 + 11
  17 = 6 + 11
  So, Max Weight = 11 + 11 = 22
  And Min Weight = 5 + 5 =10
• 9 years agoHelpfull: Yes(7) No(3)
• when 3 people b/w them. then - 5*7!*2!
  when 4 people b/w them. then- 4*7!*2!
  when 5 people b/w them.then- 3*7!*2!
  when 6 people b/w them.then- 2*7!*2!
  when 7 people b/w them.then- 1*7!*2!
  total no of possibilities... 9!
  then Ans will be 15*7!*2!/9!=5/12
  
• 9 years agoHelpfull: Yes(4) No(0)
• link ---
  https://math.stackexchange.com/questions/1397099/what-is-the-probability-that-raj-and-rana-have-atleast-3-persons-between-them-gi
• 6 years agoHelpfull: Yes(2) No(0)
• 9 people can be arranged in 9! ways. Excluding Raj and Rana, out of 7 people.
  
  Case 1: I select 3 people 7C3 who can be arranged in 3! ways.
  
  Remaining (9−(3+2))=4 people can be arranged in 4! ways
  
  Therefore required arrangement=7C3∗3!∗4!∗2! ----(i)
  Case 2: I select 4 people 7C4 who can be arranged in 4! ways
  
  Remaining (9−(4+2))=3 people can be arranged in 3! ways
  
  Therefore required arrangement=7C4∗4!∗3!∗2! ----(ii)
  Case 3: I select 5 people 7C5 who can be arranged in 5! ways
  
  Remaining (9−(5+2))=2 people can be arranged in 2! ways
  
  Therefore required arrangement=7C3∗5!∗2!∗2! ----(iii)
  Case 4: I select 6 people 7C6 who can be arranged in 6! ways
  
  Remaining (9−(6+2))=1 people can -be arranged in 1! way.
  
  Therefore required arrangement=7C6∗6!∗1!∗2! ----(iv)
  Case 5: I select 7 people 7C7 which can be arranged in 7! ways
  
  Remaining (9−(7+2))=0 people can be arranged in 1! ways
  
  Therefore required arrangement=7C7∗7!∗1!∗2! ----(v)
  Adding equations (i)+(ii)+(iii)+(iv)+(v)9! we get the required probability.
  
  Is it a correct approach and if there exists any shorter method, could you please tell me?
  
  Correct answer is 512.
• 5 years agoHelpfull: Yes(1) No(0)
• in this question we didnt need to care about the positioning of raju n reena ryt??? like whether raju is placed first or reena?
• 9 years agoHelpfull: Yes(0) No(1)
• what does 5 , 4 , 3... signify in the ans. why arent we using 5! , 4!...?
• 8 years agoHelpfull: Yes(0) No(0)