In how many ways can we distribute $26$ shirts into $5$ washing machines such that at least one washing machine contains $6$ shirts or more? My attempt:
Choose a machine in $5$ ways. Then choose six shirts out of $26$ to put in the chosen machine which can be done in $\displaystyle{\binom{26}{6}}$ ways. Distribute the rest of the shirts into five machines in $5^{20}$ ways. In all there $\displaystyle{5\binom{26}{6}5^{20}}$ such possibilities.
Now, pick any two machines out of five in $\displaystyle{\binom52}$ ways. Then, first choose six shirts out of $26$ and for those choices pick another six out of $20$. That is, $\displaystyle{\binom{26}{6}\binom{20}{6}}$ choices. Next, distribute $14$ shirts into the five machines in $5^{14}$ ways. In all, there are $\displaystyle{\binom52\binom{26}{6}\binom{20}{6}5^{14}}$ such possibilities.
Continuing this way, our third and fourth cases should be  $\displaystyle{\binom53\binom{26}{6}\binom{20}{6}\binom{14}{6}5^8}$ and $\displaystyle{\binom54\binom{26}{6}\binom{20}{6}\binom{14}{6}\binom{8}{6}5^2}$ possibilities, respectively.
Finally, we have
$\displaystyle{5\binom{26}{6}5^{20} + \binom52\binom{26}{6}\binom{20}{6}5^{14} + \binom53\binom{26}{6}\binom{20}{6}\binom{14}{6}5^8 + \binom54\binom{26}{6}\binom{20}{6}\binom{14}{6}\binom{8}{6}5^2}$.

Please see (and comment) if the reasoning above is correct. Thanks.
Edit:
This problem in its entirety reads like this below (if that helps). I rephrased the problem cause counting the number of favorable outcomes is the difficult part of such problems:

What is the probability that, when $26$ shirts are distributed into $5$ washing machines, that at least one washing machine contains $6$ shirts or more?

 A: As mentioned, the strong pigeonhole principle says that if you have more than $kn$ objects to distribute into $n$ locations, at least one location will receive more than $k$ objects. Here $k=n=5$ and $26>5\times 5$, allowing us to be certain that there are more than $5$ shirts in some washing machine.
However to show this "the hard way", let's try sharing the shirts among washing machines. I'll assume that the shirts are indistinguishable and the machines are labelled. Then we can split the shirts between machines via the sticks-and-stones principle in $\binom{26+4}{4}$ ways, total $$\binom{30}{4} = 27405$$
Then using inclusion-exclusion, we can see how many of these ways will break the constraints of only 5 shirts per machine. Pre-allocating $6$ shirts to each machine in turn gives $5\binom{24}{4}$, then pre-allocating $6$ shirts to pairs of machines gives $\binom{5}{2}\binom{18}{4}$, then triplets of machines $\binom{5}{3}\binom{12}{4}$ and finally four machines at once with pre-broken constraint is $\binom{5}{4}\binom{6}{4}$. We can't break the constraint on all five machines at once.
So using inclusion-exclusion to remove double-counting etc., the count of constraint-breaking options is \begin{align}
\binom{5}{1}\binom{24}{4}-\binom{5}{2}\binom{18}{4}+\binom{5}{3}\binom{12}{4}&-\binom{5}{4}\binom{6}{4} \\
&= 5\cdot 10626 -10\cdot 3060 + 10\cdot 495 - 5\cdot 15\\
&= 53130-30600+4950-75 \\
&= 27405
\end{align}
Certainty.
