Chart of Rounds for a Game I need to solve the following problem for actual use.


*

*10 people will be playing a game.

*They play the game 4 people at a time.

*Each time they play they garner points within the game.

*Each person needs to play against each other person at some time.

*Each person needs to play the same number of times.

*The winner is the person who has accumulated the most points at the end.


10 choose 4 is 210, so one solution is to have 210 rounds where every combination of 4 players plays the game. But this is an impractical number of rounds!
Is there a solution to this problem in less than 20 rounds? I suppose it would be okay if some rounds were played with only 3 people. How would I figure this out in the fairest way?
EDIT:
An additional useful constraint would be that no player plays twice in a row, if that's possible.
 A: Here is an ad hoc solution that I believe satisfies your requirements using 10 games, with each player being in four games.  A drawback is that some people play each other more than twice (the max is I and J who meet four times.  This could possibly be improved with more tinkering).
Game 1: ABCD
Game 2: AEFG
Game 3: AHIJ
Game 4: BEGH
Game 5: BFIJ
Game 6: DEIJ
Game 7: CDFH
Game 8: CGIJ
Game 9: CDEG
Game 10: ABFH

A B C D E F G H I J
* * * *
*       * * *
*             * * *
  *     *   * *
  *       *     * *
      * *       * *
    * *   *   *
    *       *   * *
    * * *   *
* *       *   *

Chart of number of meetups:

   A  B  C  D  E  F  G  H  I  J
A  x  2  1  1  1  2  1  2  1  1
B  2  x  1  1  1  2  1  2  1  1
C  1  1  x  3  1  1  2  1  1  1
D  1  1  3  x  2  1  1  1  1  1
E  1  1  1  2  x  1  3  1  1  1
F  2  2  1  1  1  x  1  2  1  1
G  1  1  2  1  3  1  x  1  1  1
H  2  2  1  1  1  2  1  x  1  1
I  1  1  1  1  1  1  1  1  x  4
J  1  1  1  1  1  1  1  1  4  x

Number of number of meetups:

 0:   0
 1:  68
 2:  16
 3:   4
 4:   2

A: As there are $10$ players, and each round involves $4$ players, and each player must play an equal number of times, the number of rounds must be a multiple of $5$, so $10$ rounds is the simplest solution.
Here is a solution in which $ABCD$ occurs twice:

Chart of number of meetups:

   A  B  C  D  E  F  G  H  I  J
A  x  2  2  2  1  1  1  1  1  1
B  2  x  2  2  1  1  1  1  1  1
C  2  2  x  2  1  1  1  1  1  1
D  2  2  2  x  1  1  1  1  1  1
E  1  1  1  1  x  1  1  2  2  2
F  1  1  1  1  1  x  4  1  1  1
G  1  1  1  1  1  4  x  1  1  1
H  1  1  1  1  2  1  1  x  2  2
I  1  1  1  1  2  1  1  2  x  2
J  1  1  1  1  2  1  1  2  2  x

Number of number of meetups:

 0:   0
 1:  64
 2:  24
 3:   0
 4:   2

A: Generally if we have $m$ players and $n$ players per game. Let total number of games must be at least $N$.Everyone plays everyone, which means,  everyone need to participate in $k_1 = ceil(\frac {m-1}{n-1})$ games. 
Let &Player 1& participate in $k_1$ first games and each one plays with him. Now we can not count him anymore. Now we have $m-1$ players each of who need to play $k_2=ceil(\frac {m-1-(n-1)}{n-1})= ceil(\frac {m-n}{n-1})$. After $k_2$ games we have one player less, and rest need to play $k_3=ceil(\frac {m-n-(n-1)}{n-1})= ceil(\frac {m-2n-1}{n-1})$ and so on until $k_s=0$. As you can see, numerator if fraction is a member of arithmetic progression: $a_1=m-1 a_p=a_1-(p-1)\cdot (n-1)$. $a_s<0$, so 
$m-1-s \cdot n +s+n-1<0$
$s \cdot (n-1) = m+n-2$
$s = \frac {m+n-2}{n-1}$
So $N=\sum \limits_{p=0}^{p \leqslant s} k_p = \frac { \sum \limits_{p=0}^{p \leqslant s} a_p }{n-1}+s$ ($s$ added because of $ceil$, which adds 1 game per round) 
which is
$N=\frac {\frac {a_1+a_s}{2} \cdot s}{n-1} + s = \frac {m-1}{2 \cdot (n-1)} \cdot \frac {m+n-2}{n-1} + \frac {m+n-2}{n-1} = \frac {m+n-2}{n-1} \cdot \frac {m-1+2n-2}{2 \cdot (n-1)} = \frac {(m+2 \cdot n-3) \cdot (m+n-2)}{2 \cdot (n-1)^2} $, if I got everything right of course. Then for $m=10, n=4, N=\frac {15 \cdot 12}{18}=10$
