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.
jeew-m | Certified Educator
- 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