Ans will be (C)
A pair of dice rolled 3 times, like (a,b) ,(c,d) , (e,f)
So, total 6 throws
Now if sum of a pair is 10
that pair of throws could be (5,5) , (4,6) , (6,4)
Now, it is given among 3 pair of throws only one pair will show 10
Now, that pair we can choose among any one of (5,5),(4,6),(6,4)= so 3C1 ways (say (a,b) =3C1)
Now for other two pairs ( say (c,d) , (e,f) ) we cannot choose among that previous three pairs
So, that can be (say (c,d))
For 1 in one dice ,other dice can be 1,2,3,4,5,6 = 6 choice..................i
For 2 " " " " " " " 1,2,3,4,5,6 = 6 choice.................ii
For 3 " " " " " " " 1,2,3,4,5,6 = 6 choice..................iii
For 4 " " " " " " " 1,2,3,4,5 = 5 choice.....................iv
For 5 " " " " " " " 1,2,3,4,6 = 5 choice......................v
For 6 " " " " " " " 1,2,3,5,6 = 5 choice.....................vi
So, adding from (i) to (vi) we get 33 choices
So, for 1 pair we have 33 choices
Finally we can say for 3 pairs we have
3C1* (3C1 *33*33) / (36 * 36 * 36 )
=121 / 576