there are 4 boys and 3 gals.what is the probablity that the boys and gals sit alternatively.

• b-boys,g-gals
  there are 4 boys and 3 gals can sit like
  b g b g b g b
  means at first and last place boy will sit
  so boys can sit 4! ways
  gals can sit in 3! ways
  total there are 7! ways
  probability that the boys and gals sit alternatively is
  4!*3!/7!=1/35(ans)
  
• 11 years agoHelpfull: Yes(47) No(1)
• 1/35
  
  7 persons can be arranged in 7factorial ways
  
  boys and girls can sit alternatively in 4factorial * 3factorial ways =24*6=144
  
  probability = 144/5040=1/35
• 11 years agoHelpfull: Yes(8) No(3)
• no of permutations of n things taken all at a time where p are alike and of one kind and q are alike and of one kind=n!/(p!*q!)
  No of ways in which boys and girls can sit alternatively=1 (BGBGBGB).
  So,by classical definition we have required probability= 1/(7!/4!*3!)
  =1/35
  
• 11 years agoHelpfull: Yes(3) No(5)
• boys by 4! and galz by 5p3
  24*5!/2!=1440
• 11 years agoHelpfull: Yes(1) No(29)
• there is only one possible way to arrang boys and girls alternatively
  so the probability for arranging is=1/7
• 11 years agoHelpfull: Yes(1) No(11)
• he probability is 1/35
• 10 years agoHelpfull: Yes(1) No(1)
• i think the ans is 2/7
  coz, boys can be arranged in 4! ways n we will have 5 gaps in which gals can b placed in 5p3 ways.
  total is 7! ways
  therefore, it is (4!*5p3)/7!= 2/7
• 10 years agoHelpfull: Yes(0) No(4)