imc

Let us label the players of the first team $A$, $B$, and $C$, and those of the second team, $X$, $Y$, and $Z$. $\text{1}$. One way of scheduling all six distinct rounds could be: Round 1: $AX$ $BY$ $CZ$ Round 2: $AX$ $BZ$ $CY$ Round 3: $AY$ $BX$ $CZ$ Round 4: $AY$ $BZ$ $CX$ Round 5: $AZ$ $BX$ $CY$ Round 6: $AZ$ $BY$ $CX$ The above mentioned schedule ensures that each player of one team plays twice with each player from another team. Now you can generate a completely new schedule by permutating those $6$ rounds and that can be done in $6!=720$ ways. $\text{2}$. One can also make the schedule in such a way that two rounds are repeated. (a) Round 1: $AX$ $BZ$ $CY$ Round 2: $AX$ $BZ$ $CY$ Round 3: $AY$ $BX$ $CZ$ Round 4: $AY$ $BX$ $CZ$ Round 5: $AZ$ $BY$ $CX$ Round 6: $AZ$ $BY$ $CX$ (b) Round 1: $AX$ $BY$ $CZ$ Round 2: $AX$ $BY$ $CZ$ Round 3: $AY$ $BZ$ $CX$ Round 4: $AY$ $BZ$ $CX$ Round 5: $AZ$ $BX$ $CY$ Round 6: $AZ$ $BX$ $CY$ As mentioned earlier any permutation of (a) and (b) will also give us a new schedule. For both (a) and (b) the number of permutations are $\frac{6!}{2!2!2!}$ = $90$ So the total number of schedules is $720+90+90$ =$\boxed{900}$.