Two dice thrown together. Each face of a die is marked with a different number from 1 to 6. The number on the faces of the die are marked in such a way that the sum of the numbers on any pair of opposite faces is 7. Two such dice are thrown. Assume that one can always see three faces of each die. What are the total number of ways in which a specified number is visible on both the dice?
I have no idea how to solve this. Help needed. 
P.S- This was asked by my friend who is preparing for CAT(Common Aptitude Test). The question has been copied exactly from his mock test sheet.
 A: The question asks about "a specified number is visible on both dice". This greatly simplifies the problem over the variant "a number is visible on both dice".
For, suppose the specified number is $1$.
What correspondence is there between the face the $1$ is on and the number thrown with a die? How many possibilities are there for a $1$ to be visible on one die? On two dice?
A: It's about probability I think. Well, let's take die 1. If we roll die 1, we have 6 probabilities for die 2. So, if we get 1 from die 1, we may get 1, 2, 3, 4, 5, 6 from die 2. if we get 2 from die 1, we still have the same. We may get 1, 2, 3, 4, 5, 6 from die 2. So we have 6 probabilities from each number of die 1. We may get 1, 2, 3, 4, 5, 6 from die 1 and each of them have 6 probabilities. Let's make a table for it. In this table, I showed that, we throw die 1 and what can we get from die 2 (the PROBABILITIES for die 2)

The same goes for 3, 4, 5, 6. So for each number from Die 1, we have 6 probabilities. This means; 6 (Numbers for die1) x 6 (Each one's amount of probabilities) = 36. We have 36 probabilities total. 
If we throw die 1,



So now, we threw both of them. It says sum of opposite faces is 7. The question is What is the possibility to get the same number from both dice. Here's another graph for it.

This graph is take from here; http://www.edcollins.com/backgammon/diceprob.htm
Actually, it describes the subject very well. You can visit it.
A: On one die, a $1$ can be seen together with one of the four pairs $23$, $35$, $54$, and $42$.
If we have thrown two indistinguishable dice and see a $1$ on both dice then the other two numbers can either be the same pair on both dice in four ways or two different pairs in ${4\choose2}=6$ ways. In all, there are 10 possibilities.
When it play a rôle which numbers show on top of the dice lying on the table the number of possibilities is larger, of course.
A: Regarding the question "What are the total number of ways in which a specified number is visible on both the dice?" I am assuming that a particular number (i.e. 1) if visible on the top face after the throw is distinguishable from the case where the 1 is visible on one of the side faces after the throw.
If this is the case then one die can be viewed in 8 ways---again taking 1 as the number we have 1 on top with one of the 4 pairs (23 , 35 , 54 , or 42)  showing on the side PLUS 1 on the side face with one of the 4 pairs (23 , 35 , 54 , or 42), giving a total of 8 ways for 1 die.  If the two dice are distinguishable then there are 64 way to view the number. If the 2 dice are indistinguishable then there are 28 ways that are indistinguishable leaving only 36 ways to view the number. I have this in a table format that makes the explanation easier to see but being new here I don't know how to insert it.
A: Lets fix a number '1'.
As the total on opposite sides adds to 7, '1' and 6'' must be on opposite sides.
And when a die is rolled, 1 is visible with the numbers on the adjacent sides and not with the numbers on the opposite side because it will be on the opposite side of it. (Go on roll a dice and check for yourself)'
So, on dice 1, the number visible with "1" are- (1,2,3)(1,4,5)(1,2,4)and(1,3,5)
But (1,3,4) is not a possibility because 3 and 4 are on the opposite(adds to 7) sides and will not show on the 3 faces viewed form a single angle. Similarly (1,5,2) is not.
We have 4 ways to view '1' in die 1 and which can pair along another 4 ways of die 2 to view 1, as the question specifies BOTH the dice should have the number( in this case 1).
Which gives us when rolled together: 
Die 1: (1,2,3) Die 2: (1,2,3)or (1,4,5)or (1,2,4)or (1,3,5) - 4 WAYS
Die 1: (1,4,5) Die 2: (1,2,3)or (1,4,5)or (1,2,4)or (1,3,5) - 4 WAYS
Die 1: (1,2,4) Die 2: (1,2,3)or (1,4,5)or (1,2,4)or (1,3,5) - 4 WAYS
Die 1: (1,3,5) Die 2: (1,2,3)or (1,4,5)or (1,2,4)or (1,3,5) - 4 WAYS
ADDS to 16 ways (ANSWER)
