There are 4 boys and 3 girls.they sit in arrow randomly.what is the probability that all girls are
together?

A. 1/14 B. 2/14 C. 5/14 D. 3/14

• Let us assume the 3 girls as a single unit.
  
  So we've now 4 boys +1 = 5 units which can arrange themselves in 5! = 120 ways. Now in these , 3 girls can arrange themselves in 3!= 6 ways.
  
  So finally we've the favourable cases= 120*6=720
  
  The sample space consists of all possible arrangements of 4 boys and 3 girls which is = (4+3)!= 7! = 5040
  
  Required probability = (favourable cases)/(sample-space)=720/5040 =72/504= 1/7=2/14
• 6 years agoHelpfull: Yes(31) No(0)
• 4 boys and 3 girls total 7 people
  So total way of arrangement these 7 people=7!
  so n(s)=7!
  now let us consider all three girls as single one.
  now 4 boys-1 girl(because we considered 3girls as 1) total 5 people
  So number of way of arrangement=5!
  again group of girls that we considered as 1 can be arrange by 3! so total favourable ways n(e)=5!*3!
  required probability=5!*3!/7!
  after solving ans would be = 1/7
• 6 years agoHelpfull: Yes(4) No(0)
• b.2/14
  
  the probability that 3 girls can sit together is 3!*5! (___1st boy___2nd boy___3rd boy___4th boy___)and the total arrangement is 7! so 3!*5!/7!=2/14
• 6 years agoHelpfull: Yes(2) No(5)
• Simple!!
  
  Taking 3 girls as a single unit (coz they sit together) total students are now 5 & the arrangements foe 5 will be 5!
  Now, 3 girls will have their own arrangements which will be 3!.
  Total students are 7.
  So, (3! * 5!) / 7! = 2/14
• 6 years agoHelpfull: Yes(2) No(0)
• 2/14 option B is correct
• 4 years agoHelpfull: Yes(0) No(0)