Answer:
:
Let us take the first letter of the name of each team to represent its name. Now, it is given that
(i)A + I = 12
(ii)I + SA = 11
(iii)NZ + I = 10
(iv)I + P = 8
Adding the above four equations, we get :
A + SA + 4I + NZ + P = 41 ........(1)
The total number of games played is = 4 + 3 + 2 + 1 = 10
Each game is worth two points hence total points
= 10 $\times$ 2 = 20 or A + SA + I + NZ + P = 20 .......(2)
Comparing with equation (1) and (2) we get:
3I = 21 or I = 7 (3 wins, 1 tie)
A = 5, SA = 4, NZ = 3, P=1
(2 win, 1 tie) (2 wins) (1 win, 1 tie) (1 tie)
As there are two ties they must be between A, NZ, I and P, but not SA. Now we can have
the following possibilities of ties:
Case I: I ties with A the NZ ties with P
$\begin{array}{cccc}\text{Team}& \text{Win}& \text{Lose}& \text{Tie}\\ A& \left(SA/NZ\right),P& \left(SAorNZ\right)& I\\ SA& \left(AorNZ\right),P& \left(AorNZ\right),I& -\\ I& SA,NZ,P& -& A\\ NZ& \left(AorSA\right)& \left(AorSA\right),I& P\\ P& -& A,SA,I& NZ\end{array}$
Case II: I ties with NZ,A ties with P
$\begin{array}{cccc}\text{Team}& \text{Win}& \text{Lose}& \text{Tie}\\ A& SA,NZ& I& P\\ SA& NZ,P& A,I& -\\ I& NZ,P& -& NZ\\ NZ& P& A,SA& I\\ P& -& I,SA,NZ& A\end{array}$
Case III: I ties with P,A ties with NZ.
$\begin{array}{cccc}\text{Team}& \text{Win}& \text{Lose}& \text{Tie}\\ A& SA,P& I& NZ\\ SA& NZ,P& A,I& -\\ I& A,SA,NZ& -& P\\ NZ& P& SA,I& A\\ P& -& A,SA,NZ& I\end{array}$
Choice (c). Total number of arrangements
Case I = 2
Case II = 1
Case III = 1
--------
4
--------