Yes it is correct.
How many bit strings of length six either begin with two 0s or end with three 0s?
This tells me that there are only two option available to begin with, 0 or 1.
For it to start with 2 0's, it would be 2^4=16
For it to end with 3 0's, it would be 2^3=8
Subtract 2^1 for the two instances that would have two 0's in the front and three 0's in the back.
So: 16+8-2=22
Is that correct?
Plato,
Would you mind helping me out on this one?
Suppose a department contains 5 men and 7 women. How many ways are there to form a committee with four members if there needs to be at least one man and at least one woman?
Would I set it up similar to this:
C(5,1)*(7,1)= (5!/1!4!)*(7!/1!6!)
I saw a similar problem:
C(9,3)*(11,4)= (9!/3!6!)*(11!/4!7!)=84*330=27720
What I dont get about it is, how did they get the 84*330?? That's the step I need to figure out my problem.