-2
$\begingroup$

guys, I have been trying to do this question for long time. I got recurrence relation (and its initial conditions) for the number of bit strings of length n that contain two consecutive 1s. But con

  • 1
    Using symmetry... – 2017-02-01

2 Answers 2

1

Binary strings with odd length $n$ may only have

  • More $0$s than $1$s
  • More $1$s than $0$s

and if a string belongs to a subset, its complement belongs to the other one.
It follows that the number of strings lying in the first subset is just half of the total number of strings $$ \frac{1}{2}\cdot 2^{2n+1} = \color{red}{4^n}.$$ Symmetry is powerful, not only in combinatorics.

0

Hint:

If $n$ is odd, then any bit string contains either more $1$'s or more $0$'s. So, if $a_n$ is the number with more $1$'s, and $b_n$ is the number with more $0$'s, then $a_n+b_n$ is the number of ALL bitstrings of length $n$.

Now, can you see a one-to-one correspondence between those bitstrings with more $1$s and those with more $0$'s? This would establish that $a_n=b_n$, and make finishing the problem very simple.