Norah invited 4 friends to a table tennis party. Each of the 5 people at the party played every other person exactly 1 time. The table below shows the number of games won by each player except Norah. There were no ties. How many games did Norah win?

$$\begin{array}{|c|c|} \hline \text{Player} & \text{Games won} \\ \hline \text{Collier} & 2 \\ \text{Evangeline} & 1 \\ \text{Gabe} & 1 \\ \text{Norah} & ? \\ \text{Maria} & 3 \\ \hline \end{array}$$

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We can find the number of games played. We are looking for the number of ways to form unique pairs out of 5 people. Order does not matter.

$${}_nC_r = \frac{n!}{(n-r)!r!}$$ $${}_5C_2 = \frac{5!}{(5-2)!2!}$$ $${}_5C_2 = \frac{5!}{3!2!}$$ $$= \frac{5\cdot 4\cdot 3\cdot 2 }{3\cdot 2 \cdot 2}$$ $$= 10$$

Norah must have won:

$$10-2-1-1-3 = \boxed{3}$$

We can figure out the total number of games through inspection.

Starting at Collier, Collier must have played 4 games to play with everyone else.

Evangeline will play 4 games with everyone else, but 1 will be a duplicate with Collier, so we have $4+3$ games.

Gabe plays 4 games, but 2 have already been mentioned, so the total games is $4+3+2$.

Continuing the pattern...

$$\text{Games played} = 4+3+2+1 = 10$$

Each game had a winner, so there are 10 games won.

$$2+1+1+?+3 = 10$$ $$? = \boxed{3}$$