Jamb Mathematics - Lesson Notes on Permutation and Combination for UTME Candidate

Permutation and Combination | Jamb Mathematics

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As you prepare for your upcoming examination on permutation and combination, I encourage you to review fundamental principles, including factorial notation, the distinction between ordered and unordered selections, and the application of combinatorial formulas. Mastery of these concepts will enhance your ability to solve problems efficiently, particularly in scenarios involving arrangements, selections, and probability calculations. Please ensure that you engage with practice exercises to reinforce your understanding and application of these mathematical techniques.

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Are you preparing for your JAMB Mathematics exam and feeling a bit uncertain about how to approach the topic of Permutation and combination? Don’t worry—you’re in the right place! This lesson is here to break it down in a simple, clear, and engaging way, helping you build the strong foundation you need to succeed. Whether you're struggling with complex questions or just seeking a quick refresher, this guide will boost your understanding and confidence. Let’s tackle Permutation and combination together and move one step closer to achieving your exam success! Blissful learning.

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Calculation problems involving Linear and circular arrangements in permutation and combination

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Linear Arrangements

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Question 1:

In how many ways can 5 different books be arranged on a shelf?

Solution:

The number of ways to arrange $n$ distinct objects in a line is given by $n!$ .
Here, $n = 5$ , so the number of ways is:
$5! = 5 \times 4 \times 3 \times 2 \times 1 = 120$ .
Answer: 120 ways.

Question 2:

How many different ways can 7 people be arranged in a row?

Solution:

The number of ways to arrange $n$ people in a row is $n!$ .
For $n = 7$ :
$7! = 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 5040$ .
Answer: 5040 ways.

Question 3:

In how many ways can 4 red, 3 blue, and 2 green balls be arranged in a line?

Solution:

Since balls of the same color are identical, the formula for arranging $n$ objects with groups of identical objects is:
$\frac{n!}{r_1! \times r_2! \times r_3!}$ ,
where $r_1, r_2, r_3$ are the number of identical objects of each type.
Here, $n = 4 + 3 + 2 = 9$ and $r_1 = 4$ , $r_2 = 3$ , $r_3 = 2$ . $\frac{9!}{4! \times 3! \times 2!} = \frac{362880}{24 \times 6 \times 2} = \frac{362880}{288} = 1260$ .
Answer: 1260 ways.

Question 4:

In how many ways can the letters in the word "STATISTICS" be arranged?

Solution:

Total letters = 10.
Identical letters: S (3), T (3), I (2), C (1), A (1).
Formula:
$\frac{10!}{3! \times 3! \times 2! \times 1! \times 1!} = \frac{3628800}{6 \times 6 \times 2} = \frac{3628800}{72} = 50400$ .
Answer: 50,400 ways.

Question 5:

How many different ways can 8 people be seated in a row if 2 specific people must always sit together?

Solution:

Treat the 2 specific people as one unit.
This creates 7 units to arrange: $7!$ ways.
The 2 people within their unit can switch among themselves in 2! ways.
Total arrangements:
$7! \times 2! = 5040 \times 2 = 10080$ .
Answer: 10,080 ways.

Circular Arrangements

Question 6:

How many ways can 6 people be arranged in a circle?

Solution:

The formula for arranging $n$ distinct objects in a circle is $(n-1)!$ .
For $n = 6$ :
$(6-1)! = 5! = 5 \times 4 \times 3 \times 2 \times 1 = 120$ .
Answer: 120 ways.

Question 7:

How many ways can 8 people be seated around a circular table if there is no restriction?

Solution:

The formula for circular permutations: $(n-1)!$ .
For $n = 8$ :
$(8-1)! = 7! = 5040$ .
Answer: 5040 ways.

Question 8:

How many ways can 5 people be arranged in a circle if two of them must always sit together?

Solution:

Treat the 2 people as one unit.
This results in 4 units to arrange in a circle: $(4-1)! = 3! = 6$ .
The 2 people in their unit can switch among themselves in 2! ways.
Total arrangements:
$3! \times 2! = 6 \times 2 = 12$ .
Answer: 12 ways.

Question 9:

How many ways can 10 guests be seated at a circular table if two particular guests must not sit next to each other?

Solution:

Total unrestricted circular arrangements: $(10-1)! = 9! = 362880$ .
Ways in which the two must sit together:
- Treat the two guests as one unit → reduces the arrangement count to $(9-1)! = 8! = 40320$ .
- The two guests can switch places in 2! ways.
- Ways with them together: $8! \times 2! = 40320 \times 2 = 80640$ .
Valid cases where they are NOT together:
$9! - 8! \times 2! = 362880 - 80640 = 282240$ .
Answer: 282,240 ways.

Question 10:

How many ways can 7 different colored beads be arranged in a necklace?

Solution:

For a necklace, we consider circular arrangements with reflection symmetry.
The formula is $\frac{(n-1)!}{2}$ .
For $n = 7$ :
$\frac{(7-1)!}{2} = \frac{6!}{2} = \frac{720}{2} = 360$ .
Answer: 360 ways.

Mixed Problems

Question 11:

A school has 10 students and needs to select 4 for a competition. How many ways can this be done?

Solution:

This is a combination problem:
$_{10}C_4 = \frac{10!}{4!(10-4)!} = \frac{10!}{4!6!} = \frac{10 \times 9 \times 8 \times 7}{4 \times 3 \times 2 \times 1} = 210$ .
Answer: 210 ways.

Question 12:

How many ways can 5 men and 3 women be seated in a row if women must sit together?

Solution:

Treat the 3 women as one unit.
This results in 6 units to arrange: $6! = 720$ .
The 3 women can switch places among themselves in 3! ways.
Total arrangements:
$6! \times 3! = 720 \times 6 = 4320$ .
Answer: 4,320 ways.

Question 13:

How many ways can a teacher sit 4 students on a bench if two of them refuse to sit next to each other?

Solution:

Total unrestricted arrangements: $4! = 24$ .
Arrangements where the two sit together:
- Treat them as one unit, so 3 units: $3! = 6$ .
- They can switch places in 2! ways.
- Together cases: $3! \times 2! = 6 \times 2 = 12$ .
Cases where they are not together:
$4! - 12 = 24 - 12 = 12$ .
Answer: 12 ways.

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Calculation problems involving Arrangements involving repeated objects.

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Basic Arrangements with Repeated Objects

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Question 1:

How many distinct ways can the letters in the word "APPLE" be arranged?

Solution:

The total number of letters = 5.
The letter P is repeated twice.
Formula for arrangements with repeated objects:
$\frac{n!}{r_1! \times r_2! \times ... \times r_k!}$
Applying the formula:
$\frac{5!}{2!} = \frac{120}{2} = 60$ Answer: 60 ways.

Question 2:

How many distinct permutations can be formed from the word "MISSISSIPPI"?

Solution:

Total letters = 11.
Repeated letters: M = 1, I = 4, S = 4, P = 2.
Applying the formula:
$\frac{11!}{4! \times 4! \times 2!} = \frac{39916800}{24 \times 24 \times 2}$ $= \frac{39916800}{1152} = 34650$ Answer: 34,650 ways.

Question 3:

How many different ways can the letters in "BALLOON" be arranged?

Solution:

Total letters = 7.
Repeated letters: L = 2, O = 2.
Applying the formula:
$\frac{7!}{2! \times 2!} = \frac{5040}{4} = 1260$ Answer: 1,260 ways.

Question 4:

In how many distinct ways can the letters in "TATTOO" be arranged?

Solution:

Total letters = 6.
Repeated letters: T = 3, O = 2.
Applying the formula:
$\frac{6!}{3! \times 2!} = \frac{720}{6 \times 2} = \frac{720}{12} = 60$ Answer: 60 ways.

Question 5:

How many different ways can the letters in "SUCCESS" be arranged?

Solution:

Total letters = 7.
Repeated letters: S = 3, C = 2.
Applying the formula:
$\frac{7!}{3! \times 2!} = \frac{5040}{6 \times 2} = \frac{5040}{12} = 420$ Answer: 420 ways.

Arrangements of People in a Row with Repetition

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Question 6:

In how many ways can 8 people, including 2 pairs of identical twins, be seated in a row?

Solution:

Total people = 8.
Two sets of identical twins, so 2 groups of 2.
Applying the formula:
$\frac{8!}{2! \times 2!} = \frac{40320}{4} = 10080$ Answer: 10,080 ways.

Question 7:

In how many ways can 10 students, including 3 identical twins, be seated in a row?

Solution:

Total students = 10.
Identical twins: 3 groups of 2.
Applying the formula:
$\frac{10!}{2! \times 2! \times 2!} = \frac{3628800}{8} = 453600$ Answer: 453,600 ways.

Question 8:

How many different ways can a class of 12 students be arranged in a row if 4 of them are triplets?

Solution:

Total students = 12.
Identical triplets = 4.
Applying the formula:
$\frac{12!}{3! \times 3! \times 3! \times 3!} = \frac{479001600}{1296} = 369600$ Answer: 369,600 ways.

Arrangements in a Circle with Repeated Objects

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Question 9:

How many distinct circular arrangements can be made from "BANANA"?

Solution:

Total letters = 6.
Repeated letters: A = 3, N = 2.
Circular arrangement formula:
$\frac{(n-1)!}{r_1! \times r_2! \times ... \times r_k!}$
Applying the formula:
$\frac{(6-1)!}{3! \times 2!} = \frac{5!}{6 \times 2} = \frac{120}{12} = 10$ Answer: 10 ways.

Question 10:

How many ways can 7 identical red, 4 identical blue, and 3 identical green beads be arranged in a circular necklace?

Solution:

Total beads = 7 + 4 + 3 = 14.
Identical beads: Red = 7, Blue = 4, Green = 3.
Circular formula:
$\frac{(14-1)!}{7! \times 4! \times 3!} = \frac{13!}{7! \times 4! \times 3!}$
Computing:
$\frac{6227020800}{5040 \times 24 \times 6} = 480700$ Answer: 480,700 ways.

Mixed Problems

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Question 11:

How many distinct ways can the letters in "AAABBBCCC" be arranged?

Solution:

Total letters = 9.
Repeated: A = 3, B = 3, C = 3.
Applying the formula:
$\frac{9!}{3! \times 3! \times 3!} = \frac{362880}{6 \times 6 \times 6} = \frac{362880}{216} = 1680$ Answer: 1,680 ways.

Question 12:

How many ways can the letters in "EERIE" be arranged?

Solution:

Total letters = 5.
Repeated letters: E = 3, R = 1, I = 1.
Applying the formula:
$\frac{5!}{3!} = \frac{120}{6} = 20$ Answer: 20 ways.

Question 13:

How many ways can 10 people be seated in a row if 2 specific people must sit together?

Solution:

Treat them as one unit, reducing total to 9.
Arrange 9 units: 9! ways.
Within their unit, they can switch: 2! ways.
Total:
$9! \times 2! = 362880 \times 2 = 725760$ Answer: 725,760 ways.

Question 14:

How many different ways can the letters in "PEPPER" be arranged?

Solution:

Total letters = 6.
Repeated letters: P = 3, E = 2.
Applying the formula:
$\frac{6!}{3! \times 2!} = \frac{720}{6 \times 2} = \frac{720}{12} = 60$ Answer: 60 ways.

Question 15:

How many distinct ways can the letters in "COMMITTEE" be arranged?

Solution:

Total letters = 9.
Repeated letters: M = 2, T = 2, E = 2.
Applying the formula:
$\frac{9!}{2! \times 2! \times 2!} = \frac{362880}{8} = 45360$ Answer: 45,360 ways.

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- Jamb Mathematics- Lesson notes on Probability for utme Success

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Permutation and Combination | Jamb Mathematics

Calculation problems involving Linear and circular arrangements in permutation and combination

Linear Arrangements

Question 1:

Question 2:

Question 3:

Question 4:

Question 5:

Circular Arrangements

Question 6:

Question 7:

Question 8:

Question 9:

Question 10:

Mixed Problems

Question 11:

Question 12:

Question 13:

Calculation problems involving Arrangements involving repeated objects.

Basic Arrangements with Repeated Objects

Question 1:

Question 2:

Question 3:

Question 4:

Question 5:

Arrangements of People in a Row with Repetition

Question 6:

Question 7:

Question 8:

Arrangements in a Circle with Repeated Objects

Question 9:

Question 10:

Mixed Problems

Question 11:

Question 12:

Question 13:

Question 14:

Question 15:

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