So the problem lies in the fact that you're overcounting, Why? You're clearly calculating ${5 \choose 1}$ combinations of suits and ${5 \choose 3}$ combinations of ranks. Where it goes wrong is when you're multiplying by ${20 \choose 7}$, because it adds the option to draw 3 of the same suit at least another time (from another suit clearly).

Let
H = Hearts
D = Diamonds
C = Clubs
S = Spades
X = Cross

Which means that HHHDDDCXSS and DDDHHHCXSS are both being counted, which is wrong as they're different permutations, but the same combination (assuming they represent the same rank in this example). Where HHH is the 3 cards you select first and then the rest gets selected at ${20 \choose 7}$, at the second option the DDD gets selected first, then the rest by ${20 \choose 7}$.

I hope this is a simplistic and extensive enough explanation of why your answer is false and why your probability exceeds 1.

The only way to not reveal 3 cards of the same suit is if your hand consists of 2 cards of every suit.

Which can be calculated as follows: ${5 \choose 5} {5 \choose 2}^5$ You're obviously choosing all 5 suits, and for every of the 5 suits you choose 2 cards.

Then $1 - {5 \choose 5} {5 \choose 2}^5$ is the probability you will reveal 3 cards of the same suit. Which should be about 97% chance.

So the problem lies in the fact that you're overcounting, Why? You're clearly calculating ${5 \choose 1}$ combinations of suits and ${5 \choose 3}$ combinations of ranks. Where it goes wrong is when you're multiplying by ${20 \choose 7}$, because it adds the option to draw 3 of the same suit at least another time (from another suit clearly).

Let
H = Hearts
D = Diamonds
C = Clubs
S = Spades
X = Cross

Which means that HHHDDDCXSS and DDDHHHCXSS are both being counted, which is wrong as they're different permutations, but the same combination (assuming they represent the same rank in this example). Where HHH is the 3 cards you select first and then the rest gets selected at ${20 \choose 7}$, at the second option the DDD gets selected first, then the rest by ${20 \choose 7}$.

I hope this is a simplistic and extensive enough explanation of why your answer is false and why your probability exceeds 1.

The only way to not reveal 3 cards of the same suit is if your hand consists of 2 cards of every suit.

Which can be calculated as follows: ${5 \choose 5} {5 \choose 2}^5$ You're obviously choosing all 5 suits, and for every of the 5 suits you choose 2 cards.

Then $1 - \frac{{5 \choose 5} {5 \choose 2}^5}{25 \choose 10}$ is the probability you will reveal 3 cards of the same suit. Which should be about 97% chance.