There is a row of 50 children of which 2 of them are girls.They are standing in an initial order such that a set of 6 boys, who are armature astronomers, stand next to each other in a certain order. Now the positions of boys and girls are shuffled such that ,the order of girls remain the same the amature astronomers are together in the same order, but the position and order of remaining boys are changed.How many distinct arrangements of the students can be done (including the initial order) by preserving the order of girls and keeping the amature astronomers together .
(a) 44 ! (b) 44 ! * 946 (c) 43 ! * 946 (d) 43 !

• 6 boys considered as 1 unit and 2 girls considered as 1 unit(orders are same)
  Remaining boys =50-6-2=42
  Each boy can be arranged in 1 way,so 42 boys in 42! ways
  so,42!+1!+1!=44!
  ans)44!
• 9 years agoHelpfull: Yes(7) No(2)
• The ans is 44!
  
  the questions says that the order of the girls and the astronomer boys are intact as the initial.
  
  So,we have [2girls]+[6astronomer]+other 42 guys
  
  {the bracket is used for girls and astronomers because they have to be intact in same order and there is no arrangement among them and therefore the 2 girls and 6 astronomer boys are taken as a single entity}
  
  so basically what we are doing is arranging 42+1+1 people.
  That is why the ans should be 44!
• 9 years agoHelpfull: Yes(2) No(0)
• GOURAV, as if we considered 6 boys together as 1 unit that can be arranged in 1! way
  and similarly 2 girls together as 1 unit that can be in 1! way..
• 9 years agoHelpfull: Yes(1) No(0)
• The Answer is 43 !.
  Two girls can not change their positions &The amateur astronomers are together needs to be in the same order so they can be in any position. So 50 -2-6 +1 = 43 !
• 9 years agoHelpfull: Yes(1) No(0)
• @RAYON DAS why after 42! , we added two times 1!
• 9 years agoHelpfull: Yes(0) No(1)
• 44! because out of 50, 2 of them are girls and 6 boys are taking as single entity.
  SO, 42+[2]+[6]=50;No arrangements are there in themselves
  ANSWER IS:( 42+1+1)!=44!
• 9 years agoHelpfull: Yes(0) No(0)