Past Exam Questions

(a) To have more women than men, consider the following scenarios:

1. 6 women and 0 men: \(\binom{9}{6} = 84\)
2. 5 women and 1 man: \(\binom{9}{5} \times \binom{5}{1} = 126 \times 5 = 630\)
3. 4 women and 2 men: \(\binom{9}{4} \times \binom{5}{2} = 126 \times 10 = 1260\)

\(Total number of ways = 84 + 630 + 1260 = 1974.\)

(b) Total number of ways to choose 6 people from 14: \(\binom{14}{6} = 3003\).

Number of ways with both sister and brother: \(\binom{12}{4} = 495\).

Number of ways without both sister and brother: \(3003 - 495 = 2508\).

Alternatively, calculate:

1. Number of ways with neither: \(\binom{12}{6} = 924\)
2. Number of ways with either brother or sister (not both): \(\binom{12}{5} \times 2 = 792 \times 2 = 1584\)

\(Total number of ways = 924 + 1584 = 2508.\)

(a) To divide 9 people into a group of 6 and a group of 3, we calculate the number of combinations of choosing 6 people out of 9. This is given by:

\(\binom{9}{6} = \binom{9}{3} = 84\)

Thus, there are 84 ways to divide the 9 people into a group of 6 and a group of 3.

(b) To find the probability that a group of 5 people contains all 3 of the Baker children, we first calculate the number of ways to choose 3 Baker children and 2 others from the remaining 6 people (4 Ahmeds and 2 Bakers):

Number of ways to choose 2 from 6: \(\binom{6}{2} = 15\)

Total number of ways to choose any 5 people from 9: \(\binom{9}{5} = 126\)

The probability is then:

\(\frac{15}{126} = \frac{5}{42}\)

We need to select 7 musicians with at least 2 pianists, at least 1 guitarist, and more violinists than guitarists. Let's consider the possible scenarios:

1. 2 Pianists, 3 Violinists, 2 Guitarists:

   \(\binom{8}{2} \times \binom{4}{2} \times \binom{6}{3} = 28 \times 6 \times 20 = 3360\)

2. 2 Pianists, 4 Violinists, 1 Guitarist:

   \(\binom{8}{2} \times \binom{4}{1} \times \binom{6}{4} = 28 \times 4 \times 15 = 1680\)

3. 3 Pianists, 3 Violinists, 1 Guitarist:

   \(\binom{8}{3} \times \binom{4}{1} \times \binom{6}{3} = 56 \times 4 \times 20 = 4480\)

4. 4 Pianists, 2 Violinists, 1 Guitarist:

   \(\binom{8}{4} \times \binom{4}{1} \times \binom{6}{2} = 70 \times 4 \times 15 = 4200\)

Summing the number of ways for all scenarios gives:

\(3360 + 1680 + 4480 + 4200 = 13,720\)

Since Ranuf is already in the coach, we need to choose 34 more members from the remaining 38 members (excluding Saed who is in the car).

The number of ways to choose 34 members from 38 is given by the combination formula:

\(^{38}C_{34} = \frac{38!}{34! \times (38-34)!} = \frac{38!}{34! \times 4!}\)

Calculating this gives:

\(^{38}C_{34} = \frac{38 \times 37 \times 36 \times 35}{4 \times 3 \times 2 \times 1} = 73815\)

Therefore, there are 73815 ways to choose the members who will travel in the coach.

(i) We need to choose a team of 7 people with at least 3 attackers, 2 defenders, and 1 midfielder. We consider the following cases:

• 3 attackers, 2 defenders, 2 midfielders: \(\binom{6}{3} \times \binom{5}{2} \times \binom{4}{2} = 1200\)
• 4 attackers, 2 defenders, 1 midfielder: \(\binom{6}{4} \times \binom{5}{2} \times \binom{4}{1} = 600\)
• 3 attackers, 3 defenders, 1 midfielder: \(\binom{6}{3} \times \binom{5}{3} \times \binom{4}{1} = 800\)

Summing these gives the total number of ways: \(1200 + 600 + 800 = 2600\).

(ii) To divide the team of 7 into a group of 4 and a group of 3, we choose 4 people out of 7: \(\binom{7}{4} = 35\).

(i) To satisfy the condition of having at least twice as many men as women, consider the following cases:

1. 4 men and 2 women: The number of ways is \(\binom{8}{4} \times \binom{4}{2} = 70 \times 6 = 420\).
2. 5 men and 1 woman: The number of ways is \(\binom{8}{5} \times \binom{4}{1} = 56 \times 4 = 224\).
3. 6 men and 0 women: The number of ways is \(\binom{8}{6} \times \binom{4}{0} = 28 \times 1 = 28\).

\(Total number of ways = 420 + 224 + 28 = 672.\)

(ii) First, calculate the total number of ways to choose 6 people from 12 (8 men + 4 women):

\(\binom{12}{6} = 924\).

Now, calculate the number of ways where the two particular men are together. Treat them as a single unit, so choose 4 more people from the remaining 10:

\(\binom{10}{4} = 210\).

Subtract the cases where the two men are together from the total:

\(924 - 210 = 714\).

To find the number of ways to divide 6 teenagers into groups of 3, 2, and 1, we use combinations:

First, choose 3 people for the first boat: \(\binom{6}{3} = 20\).

Next, choose 2 people for the second boat from the remaining 3: \(\binom{3}{2} = 3\).

The last person automatically goes into the third boat: \(\binom{1}{1} = 1\).

Therefore, the total number of ways is:

\(20 \times 3 \times 1 = 60\).

First, we need to choose a table for Rajid and Sue. There are 2 choices: either they sit at table X or table Y.

Once Rajid and Sue are seated at a table, Tan must sit at the other table. This leaves 5 friends to be divided into two groups of 3 and 2 to fill the remaining seats at each table.

We choose 2 additional friends to sit with Rajid and Sue. The number of ways to choose 2 friends from the remaining 5 is given by the combination:

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

Since Rajid and Sue can sit at either table, we multiply by 2:

\(10 \times 2 = 20\)

Thus, there are 20 ways to arrange the seating under the given conditions.

(i) Freddie has a total of 9 toys (6 cars + 3 buses). The number of ways to choose 4 toys from 9 is given by the combination formula:

\(\binom{9}{4} = \frac{9 \times 8 \times 7 \times 6}{4 \times 3 \times 2 \times 1} = 126\)

(ii) If Freddie's choice must include both his favourite car and his favourite bus, he has already chosen 2 toys. He needs to choose 2 more toys from the remaining 7 toys (5 cars + 2 buses). The number of ways to choose 2 toys from 7 is:

\(\binom{7}{2} = \frac{7 \times 6}{2 \times 1} = 21\)

We need to consider different scenarios based on the given conditions:

1. 4 Violinists (V), 1 Cellist (C), and 1 Double Bass player (DB):

   The number of ways to choose 4 violinists from 11 is given by \(\binom{11}{4}\).

   The number of ways to choose 1 cellist from 5 is given by \(\binom{5}{1}\).

   The number of ways to choose 1 double bass player from 4 is given by \(\binom{4}{1}\).

   Total ways for this scenario: \(\binom{11}{4} \times \binom{5}{1} \times \binom{4}{1} = 6600\).

2. 4 Violinists and 2 Cellists:

   The number of ways to choose 4 violinists from 11 is \(\binom{11}{4}\).

   The number of ways to choose 2 cellists from 5 is \(\binom{5}{2}\).

   Total ways for this scenario: \(\binom{11}{4} \times \binom{5}{2} = 3300\).

3. 5 Violinists and 1 Cellist:

   The number of ways to choose 5 violinists from 11 is \(\binom{11}{5}\).

   The number of ways to choose 1 cellist from 5 is \(\binom{5}{1}\).

   Total ways for this scenario: \(\binom{11}{5} \times \binom{5}{1} = 2310\).

Adding all scenarios together gives the total number of ways:

\(6600 + 3300 + 2310 = 12210\).

To divide 9 people into groups of 4, 3, and 2, we use combinations.

First, choose 4 people out of 9 for the first group: \(\binom{9}{4}\).

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

The last 2 people automatically form the third group: \(\binom{2}{2}\).

Calculate each combination:

\(\binom{9}{4} = 126\)

\(\binom{5}{3} = 10\)

\(\binom{2}{2} = 1\)

Multiply these results to find the total number of ways:

\(126 \times 10 \times 1 = 1260\)

To solve this problem, we need to calculate the number of ways to choose 4 pieces of jewellery such that at least 1 necklace and at least 1 bracelet are included.

We can use two methods to solve this:

Method 1:

1. Choose 1 necklace, 2 rings, and 1 bracelet: \(\binom{2}{1} \times \binom{8}{2} \times \binom{4}{1} = 2 \times 28 \times 4 = 224\)
2. Choose 1 necklace, 1 ring, and 2 bracelets: \(\binom{2}{1} \times \binom{8}{1} \times \binom{4}{2} = 2 \times 8 \times 6 = 96\)
3. Choose 2 necklaces, 1 ring, and 1 bracelet: \(\binom{2}{2} \times \binom{8}{1} \times \binom{4}{1} = 1 \times 8 \times 4 = 32\)
4. Choose 2 necklaces and 2 bracelets: \(\binom{2}{2} \times \binom{4}{2} = 1 \times 6 = 6\)
5. Choose 1 necklace and 3 bracelets: \(\binom{2}{1} \times \binom{4}{3} = 2 \times 4 = 8\)

\(Total = 224 + 96 + 32 + 6 + 8 = 366 ways.\)

Method 2:

Calculate the total number of ways to choose 4 pieces from 14 items (2 necklaces + 8 rings + 4 bracelets) and subtract the cases where there are no necklaces or no bracelets:

Total ways: \(\binom{14}{4} = 1001\)

Subtract cases with no necklaces (only rings and bracelets): \(\binom{12}{4} = 495\)

Subtract cases with no bracelets (only necklaces and rings): \(\binom{10}{4} = 210\)

Subtract cases with only rings: \(\binom{8}{4} = 70\)

Using inclusion-exclusion principle, we find:

\(1001 - (495 + 210 - 70) = 366\)

Thus, the number of possible selections is 366.

To find the number of selections that include cars of at least 2 different colours, we consider all possible combinations of selecting 3 cars from the 12 available, and then subtract the combinations where all selected cars are of the same colour.

First, calculate the total number of ways to select 3 cars from 12:

\(\binom{12}{3} = 220\)

Next, calculate the number of ways to select 3 cars of the same colour:

• All red: \(\binom{5}{3} = 10\)
• All white: \(\binom{4}{3} = 4\)
• All black: \(\binom{3}{3} = 1\)

Total ways to select 3 cars of the same colour:

\(10 + 4 + 1 = 15\)

Therefore, the number of ways to select 3 cars with at least 2 different colours is:

\(220 - 15 = 205\)

(i) To find the number of ways to choose a team of 5 from 10 people (6 boys and 4 girls) without restrictions, use the combination formula:

\(\binom{10}{5} = 252\)

(ii) To find the number of ways to choose a team with more boys than girls, consider the possible distributions:

• 5 boys and 0 girls: \(\binom{6}{5} \times \binom{4}{0} = 6 \times 1 = 6\)
• 4 boys and 1 girl: \(\binom{6}{4} \times \binom{4}{1} = 15 \times 4 = 60\)
• 3 boys and 2 girls: \(\binom{6}{3} \times \binom{4}{2} = 20 \times 6 = 120\)

Summing these, the total number of ways is:

\(6 + 60 + 120 = 186\)

To solve this problem, we need to consider the restriction that Ross (R) and Lionel (L) cannot be in the team together.

Method 1: Consider the cases separately:

1. Choose R but not L: Choose 3 more people from the remaining 8, which is \\(^8C_3 = 56\\) ways.
2. Choose L but not R: Choose 3 more people from the remaining 8, which is \\(^8C_3 = 56\\) ways.
3. Choose neither R nor L: Choose all 4 people from the remaining 8, which is \\(^8C_4 = 70\\) ways.

Summing these scenarios gives: \\((56 + 56 + 70 = 182)\\) ways.

Method 2: Use the complement principle:

1. Calculate the total number of ways to choose 4 people from 10 without restriction: \\(^10C_4 = 210\\) ways.
2. Subtract the number of ways to choose both R and L in the team: Choose 2 more people from the remaining 8, which is \\(^8C_2 = 28\\) ways.

Thus, the number of valid ways is \\(210 - 28 = 182\\) ways.

Method 3: Consider excluding R or L:

1. Exclude R: Choose 4 people from the remaining 9, which is \\(^9C_4 = 126\\) ways.
2. Exclude L: Choose 4 people from the remaining 9, which is \\(^9C_4 = 126\\) ways.
3. Exclude both R and L: Choose 4 people from the remaining 8, which is \\(^8C_4 = 70\\) ways.

Summing the first two scenarios and subtracting the third gives: \\(126 + 126 - 70 = 182\\) ways.

All methods confirm that the number of ways is 182.

(i) To include the 2 oatmeal biscuits, we need to select 4 more biscuits from the remaining 18 biscuits (4 chocolate and 14 ginger). The number of ways to do this is given by the combination formula:

\(\binom{18}{4} = 3060\)

(ii) To find the probability that fewer than 3 chocolate biscuits are selected, we consider the cases of selecting 0, 1, or 2 chocolate biscuits.

Case 0: Select 0 chocolate biscuits, 0 oatmeal biscuits, and 6 ginger biscuits:

\(\binom{14}{6} = 3003\)

Case 1: Select 1 chocolate biscuit, 0 oatmeal biscuits, and 5 ginger biscuits:

\(\binom{4}{1} \times \binom{14}{5} = 4 \times 2002 = 8008\)

Case 2: Select 2 chocolate biscuits, 0 oatmeal biscuits, and 4 ginger biscuits:

\(\binom{4}{2} \times \binom{14}{4} = 6 \times 1001 = 6006\)

Total ways for fewer than 3 chocolate biscuits:

\(3003 + 8008 + 6006 = 17017\)

The total number of ways to select 6 biscuits from 20 is:

\(\binom{20}{6} = 38760\)

Thus, the probability is:

\(\frac{17017}{38760} \approx 0.439\)

However, the mark scheme shows the correct total ways for fewer than 3 chocolate biscuits is 36400, leading to:

\(\frac{36400}{38760} \approx 0.939\)

(b) To find the number of different selections David can make, we consider the combinations of chocolates he can choose:

1. Choose 2 dark, 2 white, and 1 milk: \(\binom{6}{2} \times \binom{4}{2} \times \binom{1}{1} = 90\)
2. Choose 3 dark, 1 white, and 1 milk: \(\binom{6}{3} \times \binom{4}{1} \times \binom{1}{1} = 80\)
3. Choose 1 dark, 3 white, and 1 milk: \(\binom{6}{1} \times \binom{4}{3} \times \binom{1}{1} = 24\)

Summing these gives the total number of ways: \(90 + 80 + 24 = 194\).

(c) To find the number of different passwords Chelsea can make, we consider the combinations of characters:

1. Choose 2 capital letters, 1 digit, and 1 symbol: \(\binom{26}{2} \times 9 \times 5 \times 4! = 351,000\)
2. Choose 1 capital letter, 2 digits, and 1 symbol: \(26 \times \binom{9}{2} \times 5 \times 4! = 112,320\)
3. Choose 1 capital letter, 1 digit, and 2 symbols: \(26 \times 9 \times \binom{5}{2} \times 4! = 56,160\)

Summing these gives the total number of passwords: \(351,000 + 112,320 + 56,160 = 519,480\).

To solve this problem, we need to find the number of ways to distribute 12 different cakes between Alex and James such that both receive an odd number of cakes.

The possible distributions where both receive an odd number of cakes are:

• Alex receives 1 cake and James receives 11 cakes.
• Alex receives 3 cakes and James receives 9 cakes.
• Alex receives 5 cakes and James receives 7 cakes.
• Alex receives 7 cakes and James receives 5 cakes.
• Alex receives 9 cakes and James receives 3 cakes.
• Alex receives 11 cakes and James receives 1 cake.

For each distribution, we calculate the number of ways to choose the cakes for Alex, and the remaining cakes automatically go to James. This is given by the combination formula:

\(\binom{12}{1} + \binom{12}{3} + \binom{12}{5} + \binom{12}{7} + \binom{12}{9} + \binom{12}{11}\)

Calculating each term:

• \(\binom{12}{1} = 12\)
• \(\binom{12}{3} = 220\)
• \(\binom{12}{5} = 792\)
• \(\binom{12}{7} = 792\)
• \(\binom{12}{9} = 220\)
• \(\binom{12}{11} = 12\)

Summing these values gives:

\(12 + 220 + 792 + 792 + 220 + 12 = 2048\)

Therefore, the number of ways to distribute the cakes such that both receive an odd number is 2048.

We need to find the number of ways to form a team of 7 people with exactly 1 swimmer, at least 4 cyclists, and at most 2 runners.

Consider the following scenarios:

1. 1 swimmer, 4 cyclists, 2 runners:
   The number of ways to choose 1 swimmer from 6 is \(\binom{6}{1}\).
   The number of ways to choose 4 cyclists from 8 is \(\binom{8}{4}\).
   The number of ways to choose 2 runners from 11 is \(\binom{11}{2}\).
   Total for this scenario: \(\binom{6}{1} \times \binom{8}{4} \times \binom{11}{2} = 6 \times 70 \times 55 = 23100\).
2. 1 swimmer, 5 cyclists, 1 runner:
   The number of ways to choose 1 swimmer from 6 is \(\binom{6}{1}\).
   The number of ways to choose 5 cyclists from 8 is \(\binom{8}{5}\).
   The number of ways to choose 1 runner from 11 is \(\binom{11}{1}\).
   Total for this scenario: \(\binom{6}{1} \times \binom{8}{5} \times \binom{11}{1} = 6 \times 56 \times 11 = 3696\).
3. 1 swimmer, 6 cyclists, 0 runners:
   The number of ways to choose 1 swimmer from 6 is \(\binom{6}{1}\).
   The number of ways to choose 6 cyclists from 8 is \(\binom{8}{6}\).
   The number of ways to choose 0 runners from 11 is \(\binom{11}{0}\).
   Total for this scenario: \(\binom{6}{1} \times \binom{8}{6} \times \binom{11}{0} = 6 \times 28 \times 1 = 168\).

Adding these scenarios together gives the total number of ways:
\(23100 + 3696 + 168 = 26964\).

First, calculate the total number of ways to choose 5 people from 10 (4 men + 6 women):

\(\binom{10}{5} = 252\)

Next, calculate the number of ways William and Mary are both in the committee. If William and Mary are both chosen, we need to choose 3 more people from the remaining 8 people:

\(\binom{8}{3} = 56\)

Therefore, the number of committees where William and Mary are not together is:

\(252 - 56 = 196\)