Question

If you have 16 different games with 2 teams each game. How many combinations would you need to pick every game correctly?

Answer 1

16C2 ?

Answer 2

\[\Large C(16,2) = \frac{{16!}}{{(16 - 2)! \cdot 2!}} = \frac{{16!}}{{14! \cdot 2!}} = 120\]

Answer 3

Assuming you're familiar with the factorial (!), but just in case: n! = the product of every whole integer from n to 1 inclusive in decrements of 1 (or increments of -1, however you want to term it). So 5! = 5 * 4 * 3 * 2 * 1.
The combination formula is:
\[\Large C(n,k) = \frac{{P(n,k)}}{{k!}} = \frac{{n!}}{{(n - k)! \cdot k!}}\]

Answer 4

Sorry, skipped mention of the permutation formula P(n,k):
\[\Large \begin{array}{l}
C(n,k) = \frac{{P(n,k)}}{{k!}}\\
P(n,k) = \frac{{n!}}{{(n - k)!}}\\
C(n,k) = \frac{{n!}}{{(n - k)! \cdot k!}}
\end{array}\]

Answer 5

\(\binom{16}{2} =8\times 15=120 \)

Answer 6

yeah 16c2 ie 120