Chapter 5: Combinatorics

Question 1

A restaurant offers 4 starters, 5 main courses, and 3 desserts. In how many ways can a 3-course meal be ordered?

Choose one starter, one main course and one dessert. By the Fundamental Principle of Counting, the number of choices is 4 × 5 × 3 = 60.

Question 2

There are 5 doors to enter a hall and 3 different doors to exit. In how many ways can a person enter and exit the hall?

There are 5 choices for entering and 3 choices for leaving. Therefore, total ways = 5 × 3 = 15.

Question 3

A bicycle lock has 3 dials, each with digits 0 to 9. How many different lock combinations are possible if a digit can be repeated?

Each dial has 10 choices. Since repetition is allowed, total combinations = 10 × 10 × 10 = 10³ = 1000.

Question 4

How many numbers between 2000 and 3000 can be formed from the digits 2, 3, 4, 5, 6, 7 when repetition of digits is not allowed?

The thousands digit must be 2, so it has 1 choice. The remaining three positions are filled from the other 5 digits without repetition: 5 × 4 × 3.

Total = 1 × 5 × 4 × 3 = 60.

Question 1

Find the HCF and LCM of 6! and 5!.

6! = 6 × 5!.

Hence 5! divides 6!. Therefore, HCF = 5! = 120 and LCM = 6! = 720.

Question 2

Find n if n!/(n − 3)! = 60.

n!/(n − 3)! = n(n − 1)(n − 2) = 60.

Since 5 × 4 × 3 = 60, n = 5.

Question 3

Find x, if 1/5! + 1/6! = x/7!.

Multiply by 7!:

x = 7!/5! + 7!/6! = (7 × 6) + 7 = 42 + 7 = 49.

Question 4

Find the value(s) of x, x ≥ 0:

(a) x! = 120. Since 5! = 120, x = 5.

(b) (x!)² = 576. Thus x! = 24, so x = 4.

(c) (x!)² − 25x! + 24 = 0. Put y = x!. Then y² − 25y + 24 = 0 = (y − 24)(y − 1). Thus y = 24 or 1, giving x = 4 or x = 1.

(d) ((x + 2)! − (x + 1)!)/x! = 49. Factor (x + 1)!: numerator = (x + 1)![(x + 2) − 1] = (x + 1)!(x + 1). Dividing by x! gives (x + 1)² = 49. Since x ≥ 0, x + 1 = 7, so x = 6.

Question 1

In how many ways can 4 distinct cars be parked in 6 empty spaces?

Arrange 4 distinct cars in 6 available positions: ⁶P₄ = 6 × 5 × 4 × 3 = 360.

Question 2

How many 3-letter words (with or without meaning) can be formed using the letters of the word “LOGIC”?

There are 5 distinct letters and order matters. Hence ⁵P₃ = 5 × 4 × 3 = 60.

Question 3

A library has 5 distinct science books and 3 distinct math books. In how many ways can they be arranged on a shelf if the math books must occupy the first 3 positions?

The 3 distinct mathematics books can be arranged in the first three positions in 3! ways. The remaining 5 science books occupy the remaining positions in 5! ways.

Total = 3! × 5! = 6 × 120 = 720.

Question 4

In how many ways can 5 boys and 2 girls be seated in a row of 7 chairs if the 2 girls must always sit together?

Treat the two girls as one block. Then there are 6 units to arrange: 5 boys and 1 girls-block, in 6! ways. The two girls can be arranged within the block in 2! ways.

Total = 6! × 2! = 1440.

Question 5

How many 4-digit numbers can be formed using the digits 2, 4, 6, 8, 9 without repetition if the number must be strictly greater than 6000?

The thousands digit can be 6, 8 or 9: 3 choices. The remaining three places can be filled from the remaining four digits in 4 × 3 × 2 ways.

Total = 3 × 4 × 3 × 2 = 72.

Question 6

How many words can be formed from DAUGHTER so that (i) all vowels occur together, (ii) all vowels do not occur together?

DAUGHTER has 8 distinct letters and 3 vowels A, U, E.

(i) All vowels together: Treat AUE as one block. We then arrange 6 units, giving 6! ways, and arrange the 3 vowels within the block in 3! ways.

Thus total = 6! × 3! = 4320.

(ii) All vowels do not occur together: Total arrangements = 8! = 40320. Therefore, required = 40320 − 4320 = 36000.

Question 1

In how many ways can 3 students be chosen from a class of 12 to represent the school?

Selection order does not matter. Therefore, ¹²C₃ = (12 × 11 × 10)/(3 × 2 × 1) = 220.

Question 2

How many triangles can be formed from 12 points in a plane, of which 5 are collinear?

Total ways to choose 3 points from 12 = ¹²C₃ = 220. The 5 collinear points produce ⁵C₃ = 10 selections, none of which forms a triangle.

Therefore, number of triangles = 220 − 10 = 210.

Question 3

An examination paper contains 12 questions divided into two parts, A and B. Part A contains 7 questions and Part B contains 5 questions. A candidate must attempt 7 questions, selecting at least 3 from each part. In how many ways can the candidate select the questions?

Possible selections are (3 from A, 4 from B) or (4 from A, 3 from B).

Total = ⁷C₃⁵C₄ + ⁷C₄⁵C₃ = 35 × 5 + 35 × 10 = 525.

Question 4

How many diagonals does a polygon with 10 sides have?

Choose any two vertices: ¹⁰C₂ = 45. Of these, 10 are sides. Hence diagonals = 45 − 10 = 35.

Question 5

If you invite 15 of your friends to a party and all shake hands exactly once, how many handshakes occur?

Each handshake corresponds to a pair of people. Thus the number is ¹⁵C₂ = 15 × 14 / 2 = 105.

Question 6

A committee of 3 persons is to be constituted from a group of 2 men and 3 women. In how many ways can this be done? How many committees consist of 1 man and 2 women?

Total committees = ⁵C₃ = 10.

For 1 man and 2 women: ²C₁ × ³C₂ = 2 × 3 = 6.