Kenneth Rosen Edition 6th Exercise 5.3 Question 35 (Page No. 362)

Question

How many bit strings contain exactly eight 0s and 10 1s if every 0 must be immediately followed by a 1 ?

Comments

is the answer 9 * 9 = 81?

Answer 1

you can think this way

you have a string of length 10 and you have two letters to use : eight "01"s and two "1"s.

ie. You have ten spaces, eight filled by "01" and two filled by "1"

In other words, there are 10 locations in the string and you're choosing 2 of them to be special,

so the answer is  = ¹⁰C₂ = 45

Answer 2

Exactly eight 0's and ten 1's, if every 0 is immediately followed by a 1.

We have eight pairs of 01 and two 1's. So basically we have ten objects eight '01' and two 1's.

we have to arrange these ten objects, select 8 positions out of 10 and put '01' and in the remaining two positions put two 1's but note that two 1's can be arranged only in 1 way.

Using product rule, C(10,8)*C(2,2)= 45*1 = 45 total strings.

Comments

There's a mistake bro. _0_1_0....

Here selecting positions:  _011_0... and _0_110... will result in the same string.

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yes Rishabh, there is a mistake in the solution.

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I updated my solution, which is correct now.

Answer 3

After arranging $\color{red}{01}$ ,$8$ times in a line we have created $9$ gaps. In this gaps, we need to fill two $1$'s Now,

1. All $9$ gaps are distinct and $1$'s are obviously indistinguishable.

Therefore we can have $\begin{align*} \binom{9+2-1}{2} = \binom{10}{2} = 45 \\ \end{align*}$ arrangement of $1$'s. And finally $45$ overall bit strings. 

Answer 4

First place 1's. After placing 10 1's we are left with 11 gaps for filling 0's such that no two zeros are consecutive.

But every zero must be followed by one. So, we can't use the last gap.

Now we have 10 gaps to fill with 8 zeros which can be done in $\binom{10}{8} = \binom{10}{2} = 45$ ways