Each box contains 10 balls numbered from 1 to 10.
n1,n2,n3 are numbers on the balls drawn from the box B1,B2 and B3 respectively such that n1<n2<n3.
i.e., all 3 numbers n1,n2,n3 must be different and can be arranged only in one way (increasing).
Now n1,n2,n3 can be selected in 10C3 ways.
Hence, total number of ways =10C3.1=10C3=(3!)(7!)(10!)=3×210×9×8=120.