Question

How many bit strings of length 15 have bits 1, 2, and 3 equal to 101, or have bits 12, 13, 14, and 15 equal to 1001 or have bits 3, 4, 5, and 6 equal to 1010? (Let’s count bits left to right so 101000000000000 has 101 for bits 1,2,and 3.)

Answer 1

careful-
you have 3 constraints given. first and last are overlapping !

Answer 2

1, 2, and 3 equal to 101 : 2^12 ways
12, 13, 14, and 15 equal to 1001 : 2^11 ways
3, 4, 5, and 6 equal to 1010 : 2^11 ways

so far its simple i suppose, eh ?

Answer 3

So would I then add 2^11 + 2^1

Answer 4

may i knw why you would do that... .

Answer 5

i want to know how many different possibilities for strings there are. I dont want to count the same ones more than once. H
ow do I know which ones not to count?

Answer 6

does my first reply make sense ?

Answer 7

2nd one... where i listed down, number of strings possible for each given constraint separately

Answer 8

so my solution would be to add those numbers together?

Answer 9

and get 16777216

Answer 10

thats half of the solution,
we need to subtract overlapping ones.

if you see, first and last constraints have "3" in common right ?

Answer 11

okay....

Answer 12

so, im still not sure if you're comfortable with-
how i worked out first part.

does that make sense to you fully ?

Answer 13

I am now very confused. I am sorry.

Answer 14

its okay. let me explain first part ok

Answer 15

I appreciate it