00:01
Okay, so we want to know how many nine -bit strings are there that for a, start with the sub -string 101.
00:09
So in the first three spaces, we have these strings 101.
00:13
So in the six final spaces, we just have to put either zero or one.
00:20
So therefore, we'll just have two to the six options for the other six spaces.
00:27
So in that case we have 64 different strings there.
00:34
For b, we want to know how many 9 -bit strings have a weight of 5.
00:39
So if they have a weight of 5, that mean they contain exactly 5 ones.
00:53
So therefore, added in 9 spaces, we need to choose 5 to put a 1 into.
01:00
So therefore we have 9 and choose 5, which should be 126.
01:11
Now for c, we want to choose 5.
01:14
Have a weight of five and start with the substring of one zero one so we have the one zero one right then we have six spaces left so two of those values have already been taken so with the six other spaces we have to just find two more or three more spaces to put a one so in that case we would just have six choose three right so in that case just twenty now for d, we want to start with 101 or end with one one one.
02:12
So we have actually two different scenarios.
02:24
So we know with the possibilities for having 101 starting with, we have a total of 64.
02:36
Now we need to figure out the number of strings we have to end with one one.
02:43
So therefore two to nine spaces that have been taken.
02:45
So therefore for the other seven spaces we have a choice of one or zero so that would just be two to the seventh so that i'll give us or would that two to the seven and then we got to think about all the scenarios where we have i think of all the scenarios where we have one zero one starting off and then it also ends with one one well in this case we have just four numbers in the middle that need to be taken up or four spaces in the middle need to be taken up so we just have two to the fourth here.
03:34
So in this case, we have 64 plus two to the seventh minus two to the fourth...