Arrangements in a queue | Permutation and Combination

In how many ways can 20 boys and 18 girls make a queue such that no two girls are together?

2Comments
Sumit Verma @sumitverma
18 Apr 2017 12:50 pm

Number of ways boys can be arranged = 20!
Since no two girls are allowed to be together, so boys will stand in the following manner,
_B_B_B_B........... (At blank spaces, we can place a girl)
So out of 21 blank spaces we need to arrange 18 girls. So we can do this in 21P18 ways.
So total ways = 20! x 21P18 

shivani @shivani1234
18 Apr 2017 12:57 pm
  • This problem can be solved by putting boys first and then girls, like this :

_B1_B2_...._B20_  , in this we are left with 21 places to put girls,

so girls have got  21C18 choices and no.of ways of arranging girls is 18! and boys arranging is 20! ,

so ans becomes 21C18 *18! * 20!.

  • Now, if you think of putting girls first  and then boys than  you will be wrong as there might be case no boy sits b/w two girls, like this:

_G1B1G2B3G3_G4 ....G18_

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