Let us denote the mothers by M and children by C. M1 is mother of C1, M2->C2, M3->C3. Now, we need to ensure that M1, C1 are in order, M2, C2 in order and M3 , C3 are in order. Out of 6 places, choose 2 for M1,C1=$\binom{6}{2}$, then out of remaining 4 places choose 2 for M2,C2 then $\binom{4}{2}$ and rest 2 places can be occupied in 1 way by M3,C3.
$\therefore$ Total no of ways=$\binom{6}{2}*\binom{4}{2}=15*6=90$