How many keys in a Hybrid Cryptosystem? If symmetric cryptosystems have $C^2_n$ keys, and asymmetric systems have $2n$ keys, how many does a hybrid system have?  This was in an online quiz from school  


  Company A is considering to adopt a cryptosystem for secure communications among its employees. Currently it has $5$ employees. Assume that every employee has used the adopted cryptosystem to communicate with all other employees once.


3) c. If the company adopts hybrid cryptosystem, how many keys would be used in this case? Show your working.

My answer was 


  $C_n^2$ symmetric session key + $2n$ asymmetric keys
  
  $= C_5^2  + 2 \times 5$
  
  $= 20$ keys


This was marked wrong. Where did i go wrong here? As far as i understand, in a hybrid system they make use of asymmetric keys to encrypt the symmetric session key, which is used to encrypt/decrypt the document.
So first, for each employee there is a public-private-key pair, which is $2n$.

For each session one symmetric key is created. Your solution assumes the communication partners initially agree on a symmetric key they both use in the communication with each other. 

Considering a session as a communication in just one direction, instead of $n*(n-1)/2$ symmetric keys, there would be $n*(n-1)$ symmetric keys, making a total of $2n + n*(n-1) = 30$ keys

Edit for clarification:

The initial question states "each employee communicated with all other employees once"

This could mean two things:

$A$ sends a message encrypted with $symkey_1$ to $B$
$B$ responds with a message also encrypted with $symkey_1$
So $A$ communicated once with $B$ and $B$ communicated once with $A$

Another interpretation:

$A$ sends a message encrypted with $symkey_1$ to $B$
Two days later, $B$ sends a message to $A$ encrypted with $symkey_2$
So $A$ communicated once with $B$ and $B$ communicated once with $A$

The second interpretation will create $n*(n-1)$ symmetric keys, while the first one created only $n*(n-1)/2$ symmetric keys