A theater group has eight members, of which four are females. How many ways are there to assign the roles of a play that involve one female lead, one male lead, and three different objects that can be played by either gender?




I'm supposed to use permutations/combinations for this question, for which I'm sort of stuck...

Would it work to choose?

4C1 + 4C1 + 6C3

^  choosing the female lead

             ^ choosing the male lead

                        ^ The other three random peoples


Thank you!



I ended up getting this wrong anyway haha

So my guesses were that you could add 4c1+4c1+6c3 or 4c1*4c1*6c3, and that was incorrect.


The correct solution says: 

There are 4 ways to select the female lead and 4 ways to select the male lead. Afterward, there are 6 members who can play the first inanimate object, 5 who can play the second, and 4 for the last.

Therefore, the answer is 4*4*6*5*4=1920


Which, I'll admit, has my brain spinning in a circle chasing butterflies.


I'm not sure why my solution completely flopped, could somebody explain? So sorry for the complicated question ​

 Jun 18, 2021
edited by TheOddOne  Jun 18, 2021

Ok I will try and explain.


Your answer assumes that the 3 objects are identical.  But they are not, they are different


So if you took your answer of  4*4*6C3  and then ordered the last 3 you would have to multiply by 3! (or 6)


4*4*6C3*6 = 1920 


Now the two answers are the same  laugh

 Jun 18, 2021

Thank you for helping!


A question though: would it also work if you split up the 6C3 into, say

6C1, 5C1, 4C1?


So the result would be:






I'm not sure if this is an one in a million accidentally correct answer.


Thank you for your help once more =^-^=

TheOddOne  Jun 18, 2021

No it is fine.  They are exactly the same.

nC1 always equals n


Think about it, there are n ways to choose an item from n.

Melody  Jun 18, 2021

Thank you so much! :DDD

TheOddOne  Jun 18, 2021

55 Online Users