4 Women & 6 men have to be seated in a row given that no two women can sit together. How many different arrangements are there.

• aRRANGEMENT of 6 men is done in 6!ways.and it leaves 7 places as
  _M_M_M_M_M_M_ , where4 womens can be arranged in 7p4 ways.
  So arrangements are 6!*7P4.
• 10 years agoHelpfull: Yes(61) No(3)
• 6!x7x6x5x4
  
• 10 years agoHelpfull: Yes(11) No(2)
• total no of arrangements = 10!
  two women always sit together = 7!*4!
  two women never sit together = 10! - (7!*4!) Ans
• 10 years agoHelpfull: Yes(7) No(3)
• 6!*7C4*4!
  first arrange all the mens.after their arrangement we will have 7 choice to arrange the women....choose 4 out of 4 then arrange them.
• 10 years agoHelpfull: Yes(4) No(0)
• when 6 men are put in a row then there will be 7 gaps created.4 woman will have to be put in 7C4 ways.
• 10 years agoHelpfull: Yes(3) No(6)
• 6!*7P3= 6!*7*6*5*4
• 10 years agoHelpfull: Yes(3) No(1)
• 6 men can be arranged in 6! ways
  _M_M_M_M_M_M_ now there is 7 empty spaces
  4 places can be selected in 7c4 ways and further can be arranged in 4! ways
  so final ans is 6!*7c4*4!
• 10 years agoHelpfull: Yes(2) No(0)
• 7c4*4!*6!
  4 out of 7 positions can be selected in 7c4 ways and 4 women can be arranged there in 4! ways and 6 men can be arranged in 6! ways
• 10 years agoHelpfull: Yes(2) No(0)
• No of vacant spaces is 7 so 7C4 now women can also arrange them so 4! and men can also arrange them so 6!.
  hence ans is 6!*4!*7C4
• 10 years agoHelpfull: Yes(2) No(0)
• 6!*7P4=6!*[7!/(7 - 4)!]=6!*7*6*5*4
• 10 years agoHelpfull: Yes(1) No(1)
• total-2 women seated together
  10!-9!*2!
• 10 years agoHelpfull: Yes(0) No(1)
• 6!*7p4 is ans.
• 10 years agoHelpfull: Yes(0) No(1)
• 6 men arranged in 6! ways and
  in 7 gaps 4 women are arranged in 7x6x5x4 ways
  so the answer is 6!x7x6x5x4
• 10 years agoHelpfull: Yes(0) No(1)
• 6 man seats by 6!
  7 space
  6!*7C4*4!=6!*7*6*5*4
• 10 years agoHelpfull: Yes(0) No(1)