In symmetric key cryptographic system, both parties have access to key.
So, the first person has $\text{N-1 keys}$ with other $\text{N-1}$ people,
second one has another $\text{N-2}$ with $\text{N-2}$ people ( $1$ we already
considered ) and so on till $1.$
So, Total number of keys required $= N-1 + N-2 +\ldots + 1$
$=\dfrac{N(N-1)}{2}$
C choice.
Had we been using Public key cryptography we needed just $2N$ keys in the system.
Reference: https://en.wikipedia.org/wiki/Symmetric-key_algorithm