So many melodies!

Since the melodies do not contain repeated notes, each melody of $n$ notes will be of the form $88\choose n$ $=\frac{88!}{n!(88-n)!}$ . Since melodies are sequences of notes, the order in which they are played matters, so we need to multiply the $88\choose n$ by $n!$ to count all possible permutations of melodies of $n$ notes.

Now, to count all possible, distinct melodies, we need to add up every melody of $n$ length with $n$ being a positive integer from 2 to 88 inclusive:

$\sum_{n=2}^{88}\dfrac{88!}{(88-n)!}$

We can rewrite this sum to reveal an astonishing fact:

$\sum_{n=2}^{88}\dfrac{88!}{(88-n)!}=88!\sum_{n=2}^{88}\dfrac{1}{(88-n)!}=88!\green{\sum_{n=0}^{86}\dfrac{1}{n!}}$

We get the taylor series of $e^x$ when $x=1$ up to the 87th term! The value of the green expression can be approximated to $e$ so the number of possible, distinct melodies in an 88 key grand piano is roughly:

$\sum_{n=2}^{88}\dfrac{88!}{(88-n)!}\approx\boxed{88!\cdot e}$

Finally, we need to compare this to the number of atoms in the universe. First we take the ln:

$\ln(10^{86})=86\ln(10)=86(\ln(5)+\ln(2))$ $\ln(88!\cdot e)=\ln(88!)+\ln(e)=\ln(88!)+1$

We know that $\ln(a!)=\displaystyle \sum_{j=1}^a\ln j$ . We can set upper and lower bounds for the value of this sum as follows:

$\int_{1}^{a}\ln(x+1)dx > \sum_{j=1}^{a}\ln j \geq \int_{1}^{a}\ln(x)dx , \hspace{0.1in} \text{for} \hspace{0.07in} a\geq1$

We evaluate the integrals:

$(a+1)\ln(a+1)-a > \sum_{j=1}^{a}\ln j \geq a\ln a-a+1$

In our case $a=88$ :

$(88+1)\ln(88+1)-88-((1+1)\ln(1+1)-1) > \sum_{j=1}^{88}\ln j > 88\ln(88)-88+1-(1\ln(1)-1+1)$ $89\ln(89)-2\ln(2)-89+1+2-1 > \ln(88!) > 88(\ln(11)+3\ln(2))-88+1$ $89(\ln(89)-1)-2(\ln(2)-1) > \ln(88!) > 88(\ln(11)+3\ln(2)-1)+1$

We observe that $\hspace{0.03in}\boxed{88(\ln(11)+3\ln(2)-1)+2>86(\ln(5)+\ln(2))}$ . Therefore, the number of possible, distinct melodies in an 88 key grand piano is bigger than the number of atoms in the universe.

Bonus

To find the minimum number of keys needed for this to happen we need to find the smallest $a$ such that $\ln(a!)+1 > 86\ln(10)$ . For this we evaluate the lower bound of the sum $\displaystyle\sum_{j=1}^{a}\ln j$ :

$a\ln(a)-a+2\geq 86\ln(10)$ $a\ln(a)-a+2 = 86\ln(10)$ $\implies\boxed{a\approx 62.5}$

So the minimum number of keys needed would be 63.

Now, if repeated notes are allowed, the total number of distinct possible melodies would be given by the following finite geometric series:

$S=\sum_{n=2}^{88}88^n=88\sum_{n=1}^{87}88^n$ $S=88\cdot\dfrac{88(88^{87}-1)}{88-1}$

We take the log of S:

$\log(S)=2\log(88)+\log(88^{87}-1)-\log(88-1)$ $\log(88^{87}-1)\approx\log(88^{87})$ $\implies\log(S)\approx2\log(88)+\log(88^{87})-\log(87)$ $\log(S)\approx89\log(88)-\log(87)\approx 171$ $S\approx 10^{171}$

This means that this sum is $10^{85}$ times bigger than the number of atoms in the universe. It is almost the number of atoms in the universe squared.