Combinations

What we see: a Data Sufficiency probability question. A group of 5 people has \(n\) men, and we need one exact value of \(n\).

Step 1. Set up what we know.
What we see: 5 people, \(n\) men, so \(5 - n\) women. A team of 2 is chosen.
What it means: there are \(\binom{5}{2} = 10\) possible teams, a number we will need for Statement 2.
What we do and why: each statement only has to pin down one value of \(n\).

Step 2. Statement 1 alone.
What we see: men = women + 1.
What it means: \(n = (5 - n) + 1\), so \(2n = 6\) and \(n = 3\).
What we do and why: one value. Sufficient.

Step 3. Statement 2 alone.
What we see: the chance of at least one man on the team is \(\tfrac{9}{10}\).
What it means: "at least one man" is everything except "no men." So the chance of an all-women team is \(1 - \tfrac{9}{10} = \tfrac{1}{10}\). Out of 10 teams, exactly 1 is all women.
What we do and why: with \(w\) women there are \(\binom{w}{2}\) all-women teams. \(\binom{w}{2} = 1\) only when \(w = 2\) (2 women make exactly one pair; 3 women make 3 pairs). So \(w = 2\) and \(n = 3\). Sufficient.

The trap: C. Statement 2 looks like it needs heavy algebra, and people expect a quadratic to give two answers. It does: \(\tfrac{w(w-1)}{2} = 1\) gives \(w = 2\) or \(w = -1\), but a count of people cannot be negative, so only one value is left.
Faster way: for "at least one," work with the opposite. "No men" here means 1 team out of 10, and only 2 women make a single pair.
Takeaway: P(at least one) = 1 - P(none). The "none" case is almost always one small count you can find in your head.

Answer: D