Past Exam Questions

(i) To find the number of different teams, calculate the combinations for each group:

Choose 5 batsmen from 7: \(\binom{7}{5}\)

Choose 4 bowlers from 5: \(\binom{5}{4}\)

Choose 1 all-rounder from 2: \(\binom{2}{1}\)

Choose 1 wicket-keeper from 2: \(\binom{2}{1}\)

The total number of combinations is:

\(\binom{7}{5} \times \binom{5}{4} \times \binom{2}{1} \times \binom{2}{1} = 21 \times 5 \times 2 \times 2 = 420\)

(ii) If one particular batsman refuses to be in the team when one particular bowler is in the team, calculate the number of teams where both are in the team and subtract from the total:

Choose 4 batsmen from the remaining 6: \(\binom{6}{4}\)

Choose 3 bowlers from the remaining 4: \(\binom{4}{3}\)

Choose 1 all-rounder from 2: \(\binom{2}{1}\)

Choose 1 wicket-keeper from 2: \(\binom{2}{1}\)

The number of teams with both in the team is:

\(\binom{6}{4} \times \binom{4}{3} \times 2 \times 2 = 15 \times 4 \times 2 \times 2 = 240\)

Subtract from the total number of teams:

\(420 - 240 = 180\)

We need to choose a team of 6 people with at least 1 swimmer, 1 athlete, and 1 cyclist, and more athletes than cyclists. Let's consider the possible distributions:

1. 1 swimmer, 3 athletes, 2 cyclists:

   \(egin{align*} 5 \times \binom{7}{3} \times \binom{4}{2} &= 5 \times 35 \times 6 = 1050 ext{ ways} div>\)

2. 1 swimmer, 4 athletes, 1 cyclist:

   \(egin{align*} 5 \times \binom{7}{4} \times 4 &= 5 \times 35 \times 4 = 700 ext{ ways} div>\)

3. 2 swimmers, 3 athletes, 1 cyclist:

   \(egin{align*} \binom{5}{2} \times \binom{7}{3} \times 4 &= 10 \times 35 \times 4 = 1400 ext{ ways} div>\)

4. 3 swimmers, 2 athletes, 1 cyclist:

   \(egin{align*} \binom{5}{3} \times \binom{7}{2} \times 4 &= 10 \times 21 \times 4 = 840 ext{ ways} div>\)

Adding all these possibilities gives the total number of ways:

\(egin{align*} 1050 + 700 + 1400 + 840 = 3990 ext{ ways} div>\)

We need to find the number of ways to select 7 flowers with at least 2 of each type (roses, tulips, daffodils) from the available stock: 6 roses, 5 tulips, and 4 daffodils.

Consider the possible distributions of flowers:

1. 2 roses, 2 tulips, 3 daffodils: \(\binom{6}{2} \times \binom{5}{2} \times \binom{4}{3} = 600\)
2. 2 roses, 3 tulips, 2 daffodils: \(\binom{6}{2} \times \binom{5}{3} \times \binom{4}{2} = 900\)
3. 3 roses, 2 tulips, 2 daffodils: \(\binom{6}{3} \times \binom{5}{2} \times \binom{4}{2} = 1200\)

Summing these possibilities gives the total number of different bunches:

\(600 + 900 + 1200 = 2700\)

(iii) To have more males than females in a group of 5, we can have the following combinations:

1. 3 males and 2 females: The number of ways to choose 3 males from 5 is \(\binom{5}{3}\), and the number of ways to choose 2 females from 10 is \(\binom{10}{2}\). Thus, the total number of ways is \(\binom{5}{3} \times \binom{10}{2} = 450\).
2. 4 males and 1 female: The number of ways to choose 4 males from 5 is \(\binom{5}{4}\), and the number of ways to choose 1 female from 10 is \(\binom{10}{1}\). Thus, the total number of ways is \(\binom{5}{4} \times \binom{10}{1} = 50\).
3. 5 males and 0 females: The number of ways to choose 5 males from 5 is \(\binom{5}{5}\), and the number of ways to choose 0 females from 10 is \(\binom{10}{0}\). Thus, the total number of ways is \(\binom{5}{5} \times \binom{10}{0} = 1\).

Adding these, the total number of ways is \(450 + 50 + 1 = 501\).

(iv) Mr and Mrs Blake are already chosen, so we need to choose 3 more performers. We need more males than females, so the combinations are:

1. 3 males and 0 females: Choose 3 males from the remaining 4 males. The number of ways is \(\binom{4}{3} \times \binom{9}{0} = 4\).
2. 2 males and 1 female: Choose 2 males from the remaining 4 males and 1 female from 9 females. The number of ways is \(\binom{4}{2} \times \binom{9}{1} = 54\).

Adding these, the total number of ways is \(4 + 54 = 58\).

To solve this problem, we need to consider the different combinations of students that include at least one Chinese and one European student. We will use combinations to calculate the number of ways to choose students.

Let C, E, and A represent the number of Chinese, European, and American students chosen, respectively. We have the following cases:

1. C = 1, E = 1, A = 2: \(7 \times 6 \times \binom{4}{2} = 252\)

2. C = 1, E = 2, A = 1: \(7 \times \binom{6}{2} \times 4 = 420\)

3. C = 1, E = 3, A = 0: \(7 \times \binom{6}{3} \times 1 = 140\)

4. C = 2, E = 1, A = 1: \(\binom{7}{2} \times 6 \times 4 = 504\)

5. C = 2, E = 2, A = 0: \(\binom{7}{2} \times \binom{6}{2} \times 1 = 315\)

6. C = 3, E = 1, A = 0: \(\binom{7}{3} \times 6 \times 1 = 210\)

Adding these cases together gives the total number of ways:

\(252 + 420 + 140 + 504 + 315 + 210 = 1841\)

Since Jon and Sarah must travel in the same taxi, we consider them as a single unit. This leaves us with 6 other friends to arrange.

First, choose 2 more friends to join Jon and Sarah in the same taxi. This can be done in \\(^6C_2\\) ways.

The remaining 4 friends automatically go into the other taxi. Thus, the number of ways to choose the groups is \\(^6C_2\\).

However, Jon and Sarah can be in either taxi P or taxi Q, so we multiply by 2 to account for both possibilities.

Therefore, the total number of ways is \\(^6C_2 \\times 2 = 15 \\times 2 = 30\\).

The total number of people is 16 (7 men + 9 women). We need to find the probability that all 6 people chosen are women.

The number of ways to choose 6 women from 9 is given by the combination formula: \(\binom{9}{6}\).

The total number of ways to choose 6 people from 16 is given by: \(\binom{16}{6}\).

The probability that there are no men on the committee is:

\(P(\text{no men}) = \frac{\binom{9}{6}}{\binom{16}{6}}\)

Calculating these values:

\(\binom{9}{6} = 84\)

\(\binom{16}{6} = 8008\)

Thus, \(P(\text{no men}) = \frac{84}{8008} = \frac{21}{2002} = \frac{3}{286} \approx 0.0105\)

Alternatively, using the probability method:

\(P(\text{no men}) = \frac{9}{16} \times \frac{8}{15} \times \frac{7}{14} \times \frac{6}{13} \times \frac{5}{12} \times \frac{4}{11} = 0.0105\)

To solve this problem, we need to consider all possible combinations where Cherry chooses at least one of each type of ornament. The types are wooden animals (W), sea-shells (S), and pottery ducks (D).

We can break down the combinations as follows:

1. 1 W, 1 S, 3 D: The number of ways to choose 1 wooden animal is 6, 1 sea-shell is 4, and 3 pottery ducks is \(\binom{3}{3} = 1\). Total = \(6 \times 4 \times 1 = 24\).
2. 1 W, 3 S, 1 D: The number of ways to choose 1 wooden animal is 6, 3 sea-shells is \(\binom{4}{3} = 4\), and 1 pottery duck is 3. Total = \(6 \times 4 \times 3 = 72\).
3. 3 W, 1 S, 1 D: The number of ways to choose 3 wooden animals is \(\binom{6}{3} = 20\), 1 sea-shell is 4, and 1 pottery duck is 3. Total = \(20 \times 4 \times 3 = 240\).
4. 1 W, 2 S, 2 D: The number of ways to choose 1 wooden animal is 6, 2 sea-shells is \(\binom{4}{2} = 6\), and 2 pottery ducks is \(\binom{3}{2} = 3\). Total = \(6 \times 6 \times 3 = 108\).
5. 2 W, 1 S, 2 D: The number of ways to choose 2 wooden animals is \(\binom{6}{2} = 15\), 1 sea-shell is 4, and 2 pottery ducks is \(\binom{3}{2} = 3\). Total = \(15 \times 4 \times 3 = 180\).
6. 2 W, 2 S, 1 D: The number of ways to choose 2 wooden animals is \(\binom{6}{2} = 15\), 2 sea-shells is \(\binom{4}{2} = 6\), and 1 pottery duck is 3. Total = \(15 \times 6 \times 3 = 270\).

Adding all these combinations gives the total number of selections:

\(24 + 72 + 240 + 108 + 180 + 270 = 894\).

First, calculate the total number of ways to select 9 people from 14 without any restrictions:

\(\binom{14}{9} = 2002\).

Next, calculate the number of ways where the two particular people (say T and M) are both included in the group. If T and M are both in the group, we need to select 7 more people from the remaining 12:

\(\binom{12}{7} = 792\).

Subtract the restricted cases from the total:

\(2002 - 792 = 1210\).

Alternatively, consider the cases where neither or only one of T or M is in the group:

• Neither T nor M in the group: \(\binom{12}{9} = 220\).
• Only T in the group: \(\binom{12}{8} = 495\).
• Only M in the group: \(\binom{12}{8} = 495\).

Summing these cases gives:

\(220 + 495 + 495 = 1210\).

(a) To form a committee of 6 people with 3 men and 3 women from 6 men and 8 women, we use combinations:

The number of ways to choose 3 men from 6 is given by \(^6C_3\).

The number of ways to choose 3 women from 8 is given by \(^8C_3\).

Thus, the total number of ways is \(^6C_3 \times ^8C_3 = 20 \times 56 = 1120\).

(b) We need to form a committee of 6 people with no more than 2 brothers. We consider the following cases:

1. 0 brothers: Choose 0 brothers from 3 and 6 others from the remaining 11 people (3 non-brothers + 8 women): \(^3C_0 \times ^{11}C_6 = 1 \times 462 = 462\).
2. 1 brother: Choose 1 brother from 3 and 5 others from the remaining 11 people: \(^3C_1 \times ^{11}C_5 = 3 \times 462 = 1386\).
3. 2 brothers: Choose 2 brothers from 3 and 4 others from the remaining 11 people: \(^3C_2 \times ^{11}C_4 = 3 \times 330 = 990\).

Adding these cases gives the total number of ways: \(462 + 1386 + 990 = 2838\).

Let T, S, and G represent train, social network, and games apps respectively. We need to select 5 apps such that there is at least one of each type.

Consider the following cases:

1. 1 train, 1 social, 3 games: The number of ways is given by:

   \(3 \times 6 \times \binom{14}{3} = 6552\)

2. 1 train, 3 social, 1 game: The number of ways is given by:

   \(3 \times \binom{6}{3} \times 14 = 840\)

3. 3 train, 1 social, 1 game: The number of ways is given by:

   \(\binom{3}{3} \times 6 \times 14 = 84\)

4. 2 train, 2 social, 1 game: The number of ways is given by:

   \(\binom{3}{2} \times \binom{6}{2} \times 14 = 630\)

5. 2 train, 1 social, 2 games: The number of ways is given by:

   \(\binom{3}{2} \times 6 \times \binom{14}{2} = 1638\)

6. 1 train, 2 social, 2 games: The number of ways is given by:

   \(3 \times \binom{6}{2} \times \binom{14}{2} = 4095\)

Summing all these possibilities gives the total number of ways:

\(6552 + 840 + 84 + 630 + 1638 + 4095 = 13839\)

To solve this problem, we need to distribute 9 different games among Wainah (W), Jingyi (J), and Hebe (H) such that each receives an odd number of games. The possible distributions of games are (1, 1, 7), (1, 7, 1), (7, 1, 1), (1, 3, 5), (1, 5, 3), (3, 1, 5), (3, 5, 1), (5, 1, 3), (5, 3, 1), and (3, 3, 3).

We calculate the number of ways for each distribution:

1. For (1, 1, 7):
   egin{align*} 9C_1 imes 8C_1 imes 1C_1 &= 72 ext{(This is multiplied by 3 for permutations of (1, 1, 7))} ext{Total} &= 72 imes 3 = 216 ext{ways} ext{(for each permutation)} ext{Total for all permutations} &= 216
2. For (1, 3, 5):
   egin{align*} 9C_1 imes 8C_3 imes 5C_1 &= 504 ext{(This is multiplied by 6 for permutations of (1, 3, 5))} ext{Total} &= 504 imes 6 = 3024 ext{ways} ext{(for each permutation)} ext{Total for all permutations} &= 3024
3. For (3, 3, 3):
   egin{align*} 9C_3 imes 6C_3 imes 3C_3 &= 1680 ext{(This is multiplied by 1 as all are equal)} ext{Total} &= 1680 ext{ways} ext{(for each permutation)} ext{Total for all permutations} &= 1680

Adding all these possibilities gives the total number of ways:

egin{align*} 216 + 3024 + 1680 &= 4920 ext{ways} ext{(total)} ext{Thus, the total number of ways is } 4920. ext{This matches the mark scheme.}

Since the club president and the club treasurer must be included, we need to choose the remaining 43 members from the other 48 members (50 total members minus the 2 already chosen).

The number of ways to choose 43 members from 48 is given by the combination formula:

\(\binom{48}{43} = \binom{48}{5}\)

Calculating \(\binom{48}{5}\):

\(\binom{48}{5} = \frac{48 \times 47 \times 46 \times 45 \times 44}{5 \times 4 \times 3 \times 2 \times 1} = 1,712,304\)

Therefore, there are 1,712,304 ways to choose the 45 members.

(i) To have more women than men, the possible combinations are:

• 4 women and 2 men: \(\binom{8}{4} \times \binom{5}{2} = 700\)
• 5 women and 1 man: \(\binom{8}{5} \times \binom{5}{1} = 280\)
• 6 women and 0 men: \(\binom{8}{6} \times \binom{5}{0} = 28\)

\(Total ways = 700 + 280 + 28 = 1008\)

(ii) For 3 men and 3 women, but two particular men refuse to be together:

• Choose 3 men from 5, and 3 women from 8 without restriction: \(\binom{5}{3} \times \binom{8}{3} = 560\)
• Subtract the cases where the two particular men are together:
• Choose 2 men (including the two particular men) and 3 women: \(\binom{3}{2} \times \binom{8}{3} = 168\)

\(Total ways = 560 - 168 = 392\)

To find the number of possible selections that include at least one member from each year-group, we consider the different combinations of selecting members from each year.

1. Select 1 member from Year 1, 2 members from Year 2, and 2 members from Year 3:

\(7 \times 1 \times 1 = 7\)

2. Select 2 members from Year 1, 2 members from Year 2, and 1 member from Year 3:

\(\binom{7}{2} \times \binom{2}{2} \times \binom{2}{1} = 42\)

3. Select 2 members from Year 1, 1 member from Year 2, and 2 members from Year 3:

\(\binom{7}{2} \times \binom{2}{1} \times \binom{2}{1} = 42\)

4. Select 3 members from Year 1, 1 member from Year 2, and 1 member from Year 3:

\(\binom{7}{3} \times \binom{2}{1} \times \binom{2}{1} = 140\)

Summing these possibilities gives the total number of selections:

\(7 + 42 + 42 + 140 = 231\)

We need to calculate the number of ways to select 6 bicycles with the given constraints.

Consider the following cases:

1. Select 3 mountain bicycles, 1 racing bicycle, and 2 ordinary bicycles:
   The number of ways is given by:
   \(\binom{7}{3} \times \binom{5}{1} \times \binom{8}{2} = 35 \times 5 \times 28 = 4900\).
2. Select 3 mountain bicycles, 2 racing bicycles, and 1 ordinary bicycle:
   The number of ways is given by:
   \(\binom{7}{3} \times \binom{5}{2} \times \binom{8}{1} = 35 \times 10 \times 8 = 2800\).
3. Select 2 mountain bicycles, 2 racing bicycles, and 2 ordinary bicycles:
   The number of ways is given by:
   \(\binom{7}{2} \times \binom{5}{2} \times \binom{8}{2} = 21 \times 10 \times 28 = 5880\).

Adding all these possibilities gives the total number of selections:
\(4900 + 2800 + 5880 = 13,580\).

We need to select 7 dogs with at least 1 spaniel, 2 retrievers, and 3 poodles. We consider the following cases:

1. 1 spaniel, 2 retrievers, 4 poodles:

   The number of ways to choose 1 spaniel from 10 is \(\binom{10}{1}\).

   The number of ways to choose 2 retrievers from 14 is \(\binom{14}{2}\).

   The number of ways to choose 4 poodles from 6 is \(\binom{6}{4}\).

   Total for this case: \(\binom{10}{1} \times \binom{14}{2} \times \binom{6}{4} = 13650\).

2. 1 spaniel, 3 retrievers, 3 poodles:

   The number of ways to choose 1 spaniel from 10 is \(\binom{10}{1}\).

   The number of ways to choose 3 retrievers from 14 is \(\binom{14}{3}\).

   The number of ways to choose 3 poodles from 6 is \(\binom{6}{3}\).

   Total for this case: \(\binom{10}{1} \times \binom{14}{3} \times \binom{6}{3} = 72800\).

3. 2 spaniels, 2 retrievers, 3 poodles:

   The number of ways to choose 2 spaniels from 10 is \(\binom{10}{2}\).

   The number of ways to choose 2 retrievers from 14 is \(\binom{14}{2}\).

   The number of ways to choose 3 poodles from 6 is \(\binom{6}{3}\).

   Total for this case: \(\binom{10}{2} \times \binom{14}{2} \times \binom{6}{3} = 81900\).

Adding all cases together gives the total number of selections:

\(13650 + 72800 + 81900 = 168350\).

Alternatively, rounding or different interpretations may lead to 168000.

We need to select 12 trees with at least 2 of each type: hibiscus (H), jacaranda (J), and oleander (O).

First, allocate 2 trees of each type: 2H, 2J, 2O. This uses up 6 trees, leaving 6 more to select.

Consider the following combinations for the remaining 6 trees:

1. 1H, 8J, 2O: Choose 1 more hibiscus, 8 jacarandas, and 2 oleanders. The number of ways is given by: \(\binom{4}{2} \times \binom{9}{8} \times \binom{2}{2} = 54\).
2. 3H, 7J, 2O: Choose 3 hibiscus, 7 jacarandas, and 2 oleanders. The number of ways is given by: \(\binom{4}{3} \times \binom{9}{7} \times \binom{2}{2} = 144\).
3. 4H, 6J, 2O: Choose 4 hibiscus, 6 jacarandas, and 2 oleanders. The number of ways is given by: \(\binom{4}{4} \times \binom{9}{6} \times \binom{2}{2} = 84\).

Summing these possibilities gives the total number of ways: \(54 + 144 + 84 = 282\).

To solve this problem, we need to divide 8 children into two groups of 3 and one group of 2.

First, choose 3 children out of 8 for the first group: \(\binom{8}{3}\).

Next, choose 3 children out of the remaining 5 for the second group: \(\binom{5}{3}\).

The remaining 2 children automatically form the last group: \(\binom{2}{2}\).

The total number of ways is given by multiplying these combinations:

\(\binom{8}{3} \times \binom{5}{3} \times \binom{2}{2} = 56 \times 10 \times 1 = 560\).

Consider two cases: twins in the team and twins out of the team.

Case 1: Twins in the team

If the twins are in the team, we need to choose 2 boys from 6 and 0 more girls from the remaining 5 girls.

The number of ways to choose 2 boys is given by:

\(\binom{6}{2} = 15\)

Case 2: Twins out of the team

If the twins are not in the team, we need to choose 2 girls from the remaining 5 girls and 2 boys from 6.

The number of ways to choose 2 girls is:

\(\binom{5}{2} = 10\)

The number of ways to choose 2 boys is:

\(\binom{6}{2} = 15\)

Thus, the total number of ways for this case is:

\(10 \times 15 = 150\)

Total number of ways

Adding both cases together, the total number of ways is:

\(15 + 150 = 165\)