How many subsets have cardinality $4$?
Since the set $S = \{1, 2, 3, 4, 5, 6\}$ has six elements, a four-element subset that contains the three-element subset $\{2, 3, 5\}$ must contain one element of the subset $\{1, 4, 6\}$. There are $$\binom{3}{1} = 3$$ ways of choosing one of those three elements. As TMM stated in the comments, the three resulting subsets are
$$\{1, 2, 3, 5\}, \{2, 3, 4, 5\}, \{2, 3, 5, 6\}$$
How many subsets of cardinality $4$ contain at least one odd number?
Since the set $S$ contains three odd numbers and three even numbers, there can be at most three even numbers in a subset of size $4$. Therefore, at least one element in any four-element subset of $S$ must be odd. Since there are
$$\binom{6}{4} = 15$$
four-element subsets of $S$, there are $15$ subsets of cardinality $4$ that contain at least one odd number.
How many subsets of cardinality $4$ contain exactly one even number?
If a subset of cardinality $4$ contains exactly one even number, then it must contain all three odd numbers and one of the three even numbers in set $S$. There are
$$\binom{3}{1} = 3$$
ways of choosing one of the three even numbers in set $S$ to be the only even number in the subset of cardinality $4$. The three subsets that contain exactly one even number are
$$\{1, 2, 3, 5\}, \{1, 3, 4, 5\}, \{1, 3, 5, 6\}$$
Best Answer
It can be reduced to the case when choosing $k-1$ elements from the set $\{2,3,...,n\}$.
So the answer is $\displaystyle\binom{n-1}{k-1}$.