+48 Problem:
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?
Thoughts:
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!
Edit:
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
+118673 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 
+48 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:
4*4*6C1*5C1*4C1
or
4*4*6*5*4?
I'm not sure if this is an one in a million accidentally correct answer.
Thank you for your help once more =^-^=
