8 friends are in line for a movie. 6 are female. How many different ways can they arrange themselves if: a) A female is first in line. b) There are 3 consecutive females.

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jeew-m | College Teacher | (Level 1) Educator Emeritus

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  • There are 6 female.
  • Out of 6 we can select one female to the first place and it is given by 6C1.
  • Now when we select one for the first place we have another 7 to fill the rest
  • This is given as 7!
  • So this can be done in 6C1*7! ways or 30240 ways.



  • There is one general way that three consecutive female cannot be seated.
  • That arrangement is two female one male then two female one male and two female.


  • Here 1 denotes female and 0 denotes male.
  • The six female can be changed in 6! ways
  • Two males can be selected in 2! ways
  • So this general arrangment can be done in 2!*6!= 1440 ways.
  • All other methods have three consecutive female.
  • All 8 can be arranged in 8!=40320 ways
  • From this 40320 ; 1440 doesn't have three consecutive female.
  • Therefore 40320-1440 = 38880 ways have 3 consecutive female.


nCr = n!/(r!*(n-r)!)

n! = 1*2*3*.......(n-1)*n

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