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),I−ISA,NZ,P−ANZ(A or SA)(A or SA),IPP−A,SA,INZ Case II: I ties with NZ,A ties with P TeamWinLoseTieASA,NZIPSANZ,PA,I−INZ,P−NZNZPA,SAIP−I,SA,NZA Case III: I ties with P,A ties with NZ. TeamWinLoseTieASA,PINZSANZ,PA,I−IA,SA,NZ−PNZPSA,IAP−A,SA,NZI Choice (c). Total number of arrangements Case I  = 2 Case II = 1 Case III = 1             --------                4             --------

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