10 boys stand in a queue.what is the probability that 5 boys stand between 2 particular boys?

• 2 boys can be arranged in 4 ways leaving 5 spaces between them
  5 boys can be arranged in 8c5 ways
  remaining 3 boys can be arranged in 3c3 ways
  4*8c5*3c3=672
  But total arrangement possible is 8!
  So the probabilty is
  672/8! Which is equal to 1/60
  So ans is 1/60
• 7 years agoHelpfull: Yes(10) No(1)
• 2*8C5*6=672: 2 particular boys are arranged in 2 ways and 5 boys between them in 8C5 ways and remaining 3 are arranged in 3! ways
• 7 years agoHelpfull: Yes(7) No(4)
• 2 boys can be arranged in 4 ways leaving 5 spaces between them
  5 boys can be arranged in 8c5 ways
  remaining 3 boys can be arranged in 3c3 ways
  4*8c5*3c3=672
• 7 years agoHelpfull: Yes(6) No(2)
• no of occurrence is 5!*2!*4!
  and total occurrence is 10!
  so probability =(5!*2!*4!)/10!=1/630.
• 7 years agoHelpfull: Yes(4) No(3)
• let 10 boys stand as
  1 2 3 4 5 6 7 8 9 10
  if 5 boys are in between two boys
  then pairs are:
  1-7
  2-8
  3-9 and
  4-10
  therefore total no of pairs are 4
  total no of stutends=10
  therefore P=4/10=.4
• 6 years agoHelpfull: Yes(2) No(0)
• 2*8C5*5!*6 2 particular boys are arranged in 2 ways and 5 boys between them are selected by 8C5 ways and those 5 are arranged in 5! ways and remaining 3 are arranged in 3! ways
• 7 years agoHelpfull: Yes(1) No(1)
• 10! is the answer,i guess
• 7 years agoHelpfull: Yes(0) No(4)
• 2 boys can be arranged in 4 ways leaving 5 spaces between them
  5 boys can be arranged in 8c5 ways
  remaining 3 boys can be arranged in 3c3 ways
  4*8c5*3c3=672
• 7 years agoHelpfull: Yes(0) No(1)
• (8c5*5!*2!*3!)/10!
• 6 years agoHelpfull: Yes(0) No(0)
• (8*8!)/10!
• 6 years agoHelpfull: Yes(0) No(0)
• 2!*5!*3!*4(ways)/10!(total ways)=1/630
  ans=1/630.
  2!-fixed two boys
  5!-between two boys
  3!-remaining three boys
  4 ways as 1-7,2-8,3-9,4-10
  so the ans is 1/630.
• 4 years agoHelpfull: Yes(0) No(0)