A card game called "Happy Family" is played with 28 cards, consisting of 7 sets of 4 cards. Each set consists of a father, a mother, a son and a daughter from the same family. The family names are Painter, Postman, Plumber, Butcher, Carpenter, Singer and Teacher. So for example, the complete set of Teacher family cards consists of father Teacher, mother Teacher, son Teacher and daughter Teacher. The objective of the game is to collect as many complete sets of family cards as possible. At the end of the game, all the cards are collected. Each player will either be left with no cards or complete sets of family cards. The winner is the one with the most number of complete sets of family cards. A and B play a game of "Happy Family".

(a)
At the end of the game, A has exactly 3 complete sets of family cards and one of the sets is the Teacher family cards. How many possible combinations of complete sets of family cards can B have?
(b)
If A is the winner at the end of the game, how many possible combinations of complete sets of family cards can A have?
(c)
Given that A has exactly 3 particular complete sets of family cards at the end of the game, and he arranges these 12 cards in a circle. How many different ways can these cards be arranged so that no two mothers are next to each other?

Permutation combination Arrangements Counting principles 中文

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Worked solution

9(a)

1. Determine the number of sets B must have and the available pool of sets to choose from.

Since there are 7 complete sets in total and A has exactly 3 sets, B must hold the remaining $7 - 3 = 4$ sets.
A has the Teacher set, so B cannot have it. Thus, B's 4 sets must be formed from the remaining $7 - 1 = 6$ sets.

2. Calculate the number of combinations for B.

Number of combinations = $^6C_4 = 15$

Answer

$15$

9(b)

1. Determine the number of complete sets A must have in order to win.

Since there are $7$ complete sets in total and A is the winner, A must have strictly more sets than B.
Therefore, A must have $4$ , $5$ , $6$ , or $7$ complete sets.

2. Calculate the total number of combinations of complete sets A can have.

Total possible combinations = $^7C_4 + {}^7C_5 + {}^7C_6 + {}^7C_7$
$= 35 + 21 + 7 + 1$
$= 64$

Answer

$64$

9(c)

1. Find the number of ways to arrange the non-mother cards in a circle.

There are 3 sets of cards, meaning there are 3 mothers and $12 - 3 = 9$ non-mother cards.

Number of ways to arrange the 9 distinct non-mother cards in a circle $= (9-1)! = 8! = 40\,320$ .

2. Find the number of ways to arrange the 3 mother cards such that no two are together.

Arranging the 9 non-mother cards creates exactly 9 spaces between them.
To ensure no two mothers are next to each other, we insert the 3 distinct mother cards into these 9 spaces.

Number of ways to choose spaces and arrange the mothers $= ^9P_3 = 9 \times 8 \times 7 = 504$ .

3. Calculate the total number of arrangements.

Total number of arrangements $= 8! \times ^9P_3 = 40\,320 \times 504 = 20\,321\,280$ .

Answer

$20\,321\,280$

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Worked solution

9(a)

9(b)

9(c)

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