How many bit strings of length 10 have exactly 3 zeros? I'm unsure how to calculate it except by some brutish method: If the length was 4, then: 1000 0100 0010 0001 So (4) in total. If the length was 5, then: 1( result from 4) + 0 (1100) 0 (1010) 0 (1001) 0 (0110) 0 (0101) 0 (0011) So we get something like 4 + 3 + 2 + 1 = (10) in total. Does this go on for 6, 7, etc.? I don't want to do a proof by induction. Lol. By that logic, 10 would be 9+8+...+1 = 45. But that seems a little too low. I don't think that is right. I think I need to do another step (6) and check out the values and maybe then do some induction, but I'm lazy.
Love the avatar. It's just 10 choose 3, which is 10!/7!3! Which is = 120 Please Register or Log in to view the hidden image! In general, the way to choose x things from y choices is = Y!/X!(Y-X)! In this case, you have to choose 3 spots for 0's amongst 10 spots altogether. Please Register or Log in to view the hidden image! Oh, in case you don't know what a factorial is.... 7! = 7*6*5*4*3*2*1 = 5040 So 10!/7!3! is 362880/5040*6 = 120 Please Register or Log in to view the hidden image! Adam
Damn I feel stupid now. I did this in high school, but for some reason I always fail something like Occam's Razor. I feel ... stupid. Please Register or Log in to view the hidden image!
That's how I do it too! And it's quick in your head Please Register or Log in to view the hidden image! I mean I did it EXACTLY like that in my head, right down to the 10*3*4 Please Register or Log in to view the hidden image! Adam