9. How many bit strings contain exactly five 0s and fourteen 1s if
every 0 must be immediately followed by two 1?

Hi, math01-ga:

The requirement that every 0 must be immediately followed by two 1's
makes this problem quite simple.

First let's elaborate on the solution suggested by bokonon-ga.  We can
think of these arrangements as permutations of five blocks of the form
011 and four more singleton 1 bits.

That is, we permute the nine items (five 011's and four 1's), which
would give 9! arrangements, and divide by the number of ways in which
the identical blocks and singleton bits can be interchanged without
altering the result:

9!/( 5! 4!)

Second it might also be noticed that this is the same as the formula
for combinations of 9 things taken 4 at a time (equivalently, taken 5
at a time).  This is connected with another approach to thinking about
the problem that you may find useful.

Consider the placement of the various nine items in an arrangement. 
Essentially we need only choose which four of the nine "slots" are to
contain the singleton 1 bits (or equivalently, which five of the nine
slots are to hold the 011 blocks).  Nine choose four is another way of
saying:

C(9,4) = 9!/( 4! 5! ) = (9*8*7*6)/(1*2*3*4) = 9*2*7 = 126

regards, mathtalk-ga

First you can see that asking how many strings are there containing 5
'
011's and 4 '1's is the same problem.  To count the number of ways to
order a number of several sets of undistiguishable items, n1 one type,
n2 of the next, ... , nx of the last, use the formula

(sum of n1...nx)! / (n1!n2!...nx!)

or for here

9! / 4!5!

which is 126.