Four boys and three girls stand in queue for an interview. The probability that they stand in alternate positions is?

• Total number of possible arrangements for 4 boys and 3 girls in a queue = 7!
  When they occupy alternate position the arrangement would be like:
  B G B G B G B
  
  Thus, total number of possible arrangements for boys = (4×3×2)
  Total number of possible arrangements for girls =(3×2)
  
  Required probability =
  (4×3×2×3×2)
  7!
  = 1/35
• 9 years agoHelpfull: Yes(0) No(1)
• (4*3*3*2*2)=144
• 8 years agoHelpfull: Yes(0) No(1)
• Boys=4!
  Girls=3!
  Total for both boys and girls=4!+3!=7!
  required propability=4!*3! / 7!
  =4*3*2*1*3*2*1 / 7*6*5*4*3*2*1
  =1/35
• 6 years agoHelpfull: Yes(0) No(0)