Combinatorics problem from a contest
The following solution confirms the answer of $120.7528$, but it may use more machinery than is appropriate for a contest.
An alternative way to view the problem is that we have ten bins and we toss a succession of balls into the bins, one at a time, with each bin chosen independently and with the equal probability. Let $T$ be the number of balls tossed when we first have $n$ balls in bin $n$ for all $n=1,2,3,\dots,10$. We want to find the expected value of $T$.
To this end, let $a_n$ be the number of sequences of bin selections satisfying the constraint that bin $n$ contains at least $n$ balls for all $n$. Let $f(x)$ be the exponential generating function (EGF) of $a_n$. The EGF of the sequence of balls falling in a single bin $n$, since the bin must contain at least $n$ balls, is $$\frac{1}{n!} x^n + \frac{1}{(n+1)!}x^{n+1} + \frac{1}{(n+2)!}x^{n+2} + \dots = e^x - \sum_{i=0}^n \frac{1}{i!} x^i$$ The sequence of all balls is the labelled product of the sequences for each of the bins, so $$f(x) = \prod_{n=0}^9 \left( e^x - \sum_{i=0}^n \frac{1}{i!} x^i \right)$$ This EGF counts the acceptable sequences of ball tosses. What we are really interested in is the associated probabilities; but given the EGF of the number of acceptable sequences, the EGF of the associated probabilities, in which each ball is tossed into bin $n$ with probability $1/10$, is simply $f(x/10)$. I.e., if $p_n$ is the probability that a sequence of $n$ tosses satisfies the fill requirement for all the bins, then the EGF of $p_n$ is $f(x/10)$.
If we would like to know the probability that a sequence does not satisfy the fill constraint, that probability is $q_n = 1-p_n$, which has the EGF $g(x) = e^x-f(x/10)$. Another way to look at $q_n$ is that it is the probability that $T >n $, where $T$ is the number of balls tossed when the fill requirement of the bins is first met. So $$E(T) = \sum_{n=0}^{\infty} P(T>n) = \sum_{n=0}^{\infty} q_n$$ We can use the following trick to extract this infinite sum from the EGF of $q_n$. Because $$\int_0^{\infty} x^n e^{-x} \;dx = n!$$ we have $$\sum_{n=0}^{\infty} q_n = \int_0^{\infty} g(x) \; e^{-x} \; dx $$
The formula for $g(x)$ is quite complicated, so the prospect of computing this integral by hand is daunting, but a numerical integration using Mathematica is easy, with the result $$E(T) = 120.75280$$