Number of binary strings of length $56$ vs number of permutations of English alphabet

Question

This is exercise $1.2$ in Nicholas Loehr's book "Combinatorics".

Which is larger: the number of binary strings of length $56$, or the number of permutations of the English alphabet ($26$ letters)?

Letting $D$ and $P$ denote these sets, respectively, it is obvious that $|D|=2^{56}$ and $|P|=26!$. I checked using a calculator that $26!$ is indeed larger, but I want to find a combinatorial argument for this.

So far, I have been unsuccessful. I have been trying to find an injection $D\hookrightarrow P$, though I'm not sure this is the best strategy.

I know that any binary string can be represented uniquely in the form $0^{\alpha_1}10^{\alpha_2}1\cdots 10^{\alpha_k}$ where $0^j$ denotes a string of $j$ consecutive zeroes. However, this doesn't immediately help to get a permutation of $26$ letters, since the $\alpha_i$ could be between $0$ and $56$, and they could repeat. But maybe there is something that can be done with the $\alpha_i$ to get an injection.

On the other hand, one can uniquely represent a permutation as a $26\times 26$ matrix with exactly one $1$ per row and column and zeroes elsewhere. But I don't see a way to get a map from these matrices that would surject onto $D$.

I also considered that $2^{56}$ is the number of subsets of a $56$-element set, but that didn't seem to lead anywhere so far yet either.

Any hints on how to continue any of the above ideas to get a combinatorial proof, or alternative methods would be appreciated.

Edit: I am aware that it can be shown, e.g., inductively, that $n!$ is larger than $2^n$ for $n$ sufficiently large (or that $n! > 2^{2n+4}$ in this case), but I am interested in a more combinatorial argument for this.

Answer 1

One way to do it:

De Polignac's Formula allows us to factor factorials easily. Here we get $$26!= 2^{23}\times 3^{10}\times 5^6\times 7^3\times 11^2\times 13^2\times 17\times 19\times 23$$

But it is easy to see that $3^{10}>2^{10}$ and $5^6>2^{12}$ and $7^3>2^6$ and $11^2>2^6$ and that's more than sufficient.

Answer 2

I don't think a combinatorial argument necessarily gives you very much here, and I don't think that's the point of the exercise. If I was solving it I would do that by making some numerical estimates on $26!$.

That said, there is a natural way to biject $P$ with the set of finite sequences $(x_0, \dotsc, x_{25})$ such that $x_i \in \{0, \dotsc, i\}$ for each $i$ (it's a nice exercise, see if you can work it out! You can read about one such bijection here. Of course all you need for this problem is an injection to $P$). You can inject $D$ into this set by interpreting appropriately sized "blocks" of a binary string as nonnegative integers (and this clearly won't be surjective). I think that this argument straightforwardly extends to prove that $2^{78} < 26!$ (if I've calculated correctly!), and you can make it go further by being a bit cleverer about it.

Really this argument is a round-about way to spell out the bound $26! > 1 \cdot 2 \cdot 2 \cdot 4 \cdot 4 \cdot 4 \cdot 4 \cdot 8 \cdot \dotsb$, but it does give an explicit injection.

Answer 3

Alternative approach (but Calculus rather than Combinatorics):

For $~n \in \Bbb{Z^+},~$ let $~G(n)~$ denote the geometric mean of $~\{1,2,\cdots,n\}.~$ That is, $~\displaystyle G(n) = (n!)^{1/n}.$

As $~\displaystyle n \to \infty, ~\frac{G(n)}{n} \to \frac{1}{e}.~$ Further, as $~n~$ increases, the fraction $~\displaystyle \frac{G(n)}{n}~$ is strictly decreasing (for $~n > 5~$) and approaches $~\dfrac{1}{e}~$ from above.

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Now, this is the part where I am somewhat cheating. I know from personal experience that $~\dfrac{G(100)}{100} > 0.4.$ Since the fraction $~\dfrac{G(n)}{n}~$ is strictly decreasing, this implies that $~G(26) > 10 \implies (26)! > 10^{26} \implies \log_{10} \,(26)! > 26.~$

Further, $\log_{10} \,2 \approx 0.301 \implies \log_{10} \,2^{(56)} = 0.301 \times 56 < 20.~$ This makes it game over.

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In fact, the difference in base 10 logarithms is so extreme, that even without cheating, you would have that $~G(26) > \dfrac{26}{e} > \dfrac{26}{2.72}.~$

Note that $~2.72 < 2.\overline{72} = 2 + \dfrac{8}{11} = \dfrac{30}{11},~$ and that $~\log_{10} \,3 \approx 0.477, ~\log_{10} \,11 \approx 1.041.$

This means that the only numerical values needed to work the problem are that $~e < 2.72~$ and a knowledge of the approximate base 10 logarithms of the first few primes.

For what it's worth, base 10 logarithms are derivable via this method.

Answer 4

Here is a way to use the inequality $26! > 8^{26-7}$ to get an explicit injection.

First, pad our $56$-bit sequence to length $57$ and add on a $0$; divide the resulting sequence into $19$ blocks of length $3$, which we can read in binary and interpret as $19$ numbers $x_1, x_2, \dots, x_{19}$ between $0$ and $7$. We will use $x_i$ to help us place the $i^{\text{th}}$ letter of the alphabet (one of the letters A through S).

Now write down the following permutation of the alphabet:

1. Begin by writing down, in alphabetical order, all the letters A through S whose corresponding $x_i$ is $0$. Then write T.
2. Next, write down, in alphabetical order, all the letters A through S whose corresponding $x_i$ is $1$. Then write U.
3. We can
4. Keep going
5. Like this
6. Until we
7. Write down, in alphabetical order, all the letters A through S whose corresponding $x_i$ is $6$. Then write Z.
8. End by writing down, in alphabetical order, all the letters A through S whose corresponding $x_i$ is $7$.