Out of 4 people, why does ways to choose a 2person committee overcount by 2 the ways to divide the 4 into 2 teams of 2?

Please see the sentence alongside my red line below. Why does part (a) overcount part (b) by a factor of c?

Scilicet, why aren't the answers to parts (a) and (b) the same? Whenever you choose a 2person committee #1, the remaining unchosen 2 members automatically can form the 2person committee #2.

Thus where's the overcounting?
Blitzstein. Introduction to Probability (2019 2 ed). p. 15.
1 answer
The situations are different. In (a), two distinct groups of two are formed, “committee” and “not committee”. But in (b), we instead form two nondistinct “team” groups of two each.
In fairness, the nondistinctness of the teams in (b) is not explicit in the question. You could certainly argue that labelling the teams in distinct ways (A vs B, red vs blue, shirts vs bibs) is a valid assumption, and in that case there are six ways to do it.
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