Partitioning a set of $n$ elements into $2$ parts, where one part has $3$ elements and the other has all the remaining elements?

Question

How many ways are there to partition the set of $n$ into $2$ parts -- one part with $3$ elements and the other part with all the other elements?

I think the answer is $C(, 3),$ but I'm not certain? Would love any tips.

Answer 1

Let $ n\ge 4$.

Your question is equivalent to :

How many subsets of three elements can we choose in a set containing $ n $ elements. Each one of these subsets defines a partition. the complementary will contain $ n-3 $ elements. If $ n-3=3 $ or $ n=6 $, there will be $$\frac 12 \binom{6}{3}=10\text{ partitions}$$

otherwise, there are $$\binom{n}{3}=\frac{n!}{3!(n-3)!} \text { possibilities}$$