If the integers 1-9 are randomly distributed into three sets of 3 integers, what is the probability that at least one set contains only odd integers?

The problem https://www.cut-the-knot.org/Probability/ThreeGroupsOfThree.shtml has been stated here,

If the integers 1-9 are randomly distributed into three sets of 3 integers, what is the probability that at least one set contains only odd integers?

My approach is that as there are 1, 3 , 5, 7 , 9 = 5 odd numbers, selecting 3 numbers from here would be 5C3 and total number of groups would be 9C3*6C3 so the probability is 5C3/

The number of ways of forming three labeled sets of three numbers from the set $\{1, 2, 3, 4, 5, 6, 7, 8, 9\}$ is $$\binom{9}{3}\binom{6}{3}\binom{3}{3}$$ since we must select three of the nine numbers to be in the first set, three of the remaining six numbers to be in the second set, and all three of the remaining three numbers to be in the third set.

However, if the sets are not labeled, the order in which the sets are selected does not matter. Hence, the number of ways of forming three sets of three numbers from the set $\{1, 2, 3, 4, 5, 6, 7, 8, 9\}$ is $$\frac{1}{3!}\binom{9}{3}\binom{6}{3}\binom{3}{3}$$

Since there are five odd numbers in the set $\{1, 2, 3, 4, 5, 6, 7, 8, 9\}$, there can be at most one set with three odd numbers since $2 \cdot 3 = 6 > 5$. If three odd numbers are selected to be in one of the sets, then the remaining two odd numbers are placed in one set or two different sets.

Three odd numbers in one set and the other two odd numbers in another set: There are $\binom{5}{3}$ ways to select the three odd numbers that appear in the same set and $\binom{4}{1}$ to select which even number will be placed in the set with the remaining two odd numbers. The remaining three even numbers form the third set. Hence, there are $$\binom{5}{3}\binom{4}{1}$$ such cases.

Three odd numbers in one set and the other two odd numbers are placed in two different sets: There are $\binom{5}{3}$ ways to select the three odd numbers that appear in the same set. There are $\binom{4}{2}$ ways to select which two of the four even numbers will be placed in the same set as the smaller of the two remaining odd numbers. The remaining three numbers form the third set. Hence, there are $$\binom{5}{3}\binom{4}{2}$$ such cases.

Thus, the probability that one of the three sets of three numbers will contain only odd integers is $$\frac{\dbinom{5}{3}\dbinom{4}{1} + \dbinom{5}{3}\dbinom{4}{2}}{\dfrac{1}{3!}\dbinom{9}{3}\dbinom{6}{3}\dbinom{3}{3}}$$