If I have a deck with 5 suits of 14 cards each, what is the chance not drawing at least one card of each suit, after 16 draws with replacement? Using inclusion-exclusion rule: $$ \binom{5}{1}\left(\frac{4}{5}\right)^{16}-\binom{5}{2}\left(\frac{3}{5}\right)^{16}+\binom{5}{3}\left(\frac{2}{5}\right)^{16}-\binom{5}{4}\left(\frac{1}{5}\right)^{16}\approx0.1379 $$ There's a 4/5 chance to not draw a given suit with $\binom{5}{1}=5$ combinations of suits, 3/5 chance to not draw two given suits with $\binom{5}{2}=10$ combinations of two suits, and so on.
Using the same deck, what is the chance not drawing at least one card of each suit, after 16 draws without replacement? Again, using the inclusion-exclusion rule: $$ \frac{\binom{5}{1}\binom{56}{16}-\binom{5}{2}\binom{42}{16}+\binom{5}{3}\binom{28}{16}}{\binom{70}{16}}\approx0.08329 $$ There are 70 cards in the deck. If I only choose 4 suits ($\binom{5}{1}=5$ combinations), then there are 56 cards to pick from. If I only choose 3 suits ($\binom{5}{2}=10$ combinations), then there are 42 cards to pick from, etc. It's not possible to choose only 1 suit as there are 14 cards in a suit and 16 draws, so this term is not included.
It's surprising to me that there's this large of a difference between the two. I know that there is some self-correction in the non-replacement case; while the probability of choosing each suit is 1/5 to begin with, as soon as you draw a card, the probability of drawing its suit again decreases while that of the others' increases. I didn't think this self-correction would have as big of an impact as it seems to, so it makes me wonder if I calculated everything correctly.