@Angkit
– – – 3 places , 3 odd numbers (1,3,5) , 3 even numbers (2,4,6)
Now fixed first place containing odd number (we can have exactly one odd), so for first place, we will have 3 choices (1,3,5), for the second place we will have 3 choices (2,4,6), for the third place we will have 3 choices (2,4,6). so 3*3*3 =3$^{3}$ ways.
Now fixed second place containing odd number (we can have exactly one odd), so for second place, we will have 3 choices (1,3,5), for the first place we will have 3 choices (2,4,6), for the third place we will have 3 choices (2,4,6). so 3*3*3 =3$^{3}$ ways.
Now fixed third place containing odd number (we can have exactly one odd), so for third place, we will have 3 choices (1,3,5), for the second place we will have 3 choices (2,4,6), for the first place we will have 3 choices (2,4,6). so 3*3*3 =3$^{3}$ ways
These are favourable outcomes, now total outcomes will be, we will have 6 choices (1,2,3,4,5,6) for each of the 3 places, so 6*6*6 = 6$^{3}$ ways
P = favourable outcomes / total outcomes =( 3*3$^{3}$) / 6$^{3}$ = 3/8