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Question

How many total arrangements of wins and ties are possible?___


Solution

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 × 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

TeamWinLoseTieA(SA/NZ),P(SA or NZ)ISA(A or NZ),P(A or NZ),IISA,NZ,PANZ(A or SA)(A or SA),IPPA,SA,INZ

Case II: I ties with NZ,A ties with P

TeamWinLoseTieASA,NZIPSANZ,PA,IINZ,PNZNZPA,SAIPI,SA,NZA

Case III: I ties with P,A ties with NZ.

TeamWinLoseTieASA,PINZSANZ,PA,IIA,SA,NZPNZPSA,IAPA,SA,NZI

Choice (c). Total number of arrangements
Case I  = 2
Case II = 1
Case III = 1
            --------
               4
            --------

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