Combinatorics and the Binomial Theorem

In earlier counting problems, we often arranged distinct objects. For example, if 5 different letters are arranged in a row, there are \(5! = 120\) possible orders.

But what happens if some of the letters are the same? Then many of those arrangements look identical, even if we pretend the repeated letters are different while counting. Permutations with repetition corrects this overcounting.

This idea is important in combinatorics because real counting problems often involve repeated letters, repeated digits, or groups of identical objects.

Main idea: if there are \(n\) objects in total, and some are repeated, then the number of distinct arrangements is

Here:

• \(n\) = total number of objects
• \(n_1, n_2, n_3, \dots\) = the numbers of identical objects in each group

We divide by factorials because swapping identical objects does not create a new visible arrangement.

For example, if a word has 2 identical A's, then every arrangement has been counted \(2!\) times if we treat those A's as different. So we divide by \(2!\) to fix the overcount.

Let us build this idea carefully.

Step 1: Start with all objects treated as different.

If there are \(n\) objects, the number of arrangements is \(n!\).

Step 2: Notice the overcounting.

If some objects are identical, swapping them does not change the arrangement. So \(n!\) counts the same arrangement many times.

Step 3: Divide by the repeated groups.

If one letter appears 3 times, divide by \(3!\). If another letter appears 2 times, divide by \(2!\), and so on.

This gives the formula again:

This is sometimes called the number of permutations of a multiset.

When should you use this formula?

• When order matters
• When all objects are being arranged
• When some objects are identical

When should you not use it?

• If order does not matter, the problem is about combinations, not permutations.
• If all objects are different, just use \(n!\).
• If you are only arranging part of the objects, you may need a different method.

Worked Example 1: Arrange the letters in BOOK

How many distinct arrangements can be made using all the letters of the word BOOK?

The letters are B, O, O, K.

• Total letters: \(4\)
• The letter O is repeated \(2\) times

If all 4 letters were different, there would be \(4!\) arrangements.

But the two O's are identical, so each distinct arrangement has been counted \(2!\) times.

Answer: There are \(12\) distinct arrangements of BOOK.

Why divide by \(2!\)? If we label the O's as \(O_1\) and \(O_2\), then B\(O_1\)\(O_2\)K and B\(O_2\)\(O_1\)K look exactly the same as BOOK. So we must divide out this duplicate counting.

Worked Example 2: Arrange the letters in BANANA

Find the number of distinct arrangements of the letters in BANANA.

First count each letter:

• B appears 1 time
• A appears 3 times
• N appears 2 times

There are \(6\) letters in total, so begin with \(6!\).

Now divide by the repeated letters:

• Divide by \(3!\) for the 3 A's
• Divide by \(2!\) for the 2 N's

Answer: There are \(60\) distinct arrangements of BANANA.

This example shows that you divide once for each group of identical objects.

Worked Example 3: How many 7-letter arrangements can be formed from the letters of BALLOON?

The word BALLOON has 7 letters:

• B = 1
• A = 1
• L = 2
• O = 2
• N = 1

Total letters: \(7\)

Repeated groups: two L's and two O's

Answer: There are \(1260\) distinct arrangements.

Notice that letters appearing only once do not affect the denominator, since \(1! = 1\).

Worked Example 4: Arrangements of digits with repetition

How many distinct 5-digit arrangements can be formed using the digits 1, 1, 2, 2, 2?

We are arranging all 5 digits.

• Total digits: \(5\)
• Two 1's
• Three 2's

Use the formula:

Answer: There are \(10\) distinct arrangements.

This works exactly the same way as arranging letters in a word.

A useful way to think about the formula

Suppose we want to arrange the letters in AAB.

If the A's were different, say \(A_1, A_2, B\), there would be

arrangements.

But for each visible arrangement, the two A's can switch places in \(2! = 2\) ways without changing what we see.

So the true number is

The distinct arrangements are AAB, ABA, and BAA.

This simple example shows exactly why division is needed.

Common mistakes to avoid

• Forgetting to divide for repeated objects. If letters repeat, using just \(n!\) will be too large.
• Dividing by the repeated number instead of its factorial. For example, 3 repeated A's means divide by \(3!\), not by 3.
• Missing one repeated group. In BANANA, you must divide by both \(3!\) and \(2!\).
• Using this formula when order does not matter. This formula is only for arrangements.

Quick problem-solving steps

1. Count the total number of objects, \(n\).
2. Count how many times each repeated object appears.
3. Write \(n!\) in the numerator.
4. Divide by the factorial of each repetition number.
5. Simplify carefully.

Practice Check

Try these on your own:

• How many distinct arrangements of the word MUMMY are there?
• How many distinct arrangements of the letters in SUCCESS are there?
• How many distinct arrangements can be made from the digits 3, 3, 3, 5, 5, 8?

Answers:

• MUMMY: \(\frac{5!}{2!2!} = 30\)
• SUCCESS: \(\frac{7!}{3!2!} = 420\)
• Digits \(3,3,3,5,5,8\): \(\frac{6!}{3!2!} = 60\)

Summary

Permutations with repetition are used when arranging objects where some are identical. Start with the total number of arrangements, \(n!\), and divide by the factorial of each repeated group.

This prevents overcounting and gives the number of distinct arrangements. The key question to ask is: Are some of the objects the same? If yes, then you probably need permutations with repetition.

In ordinary linear permutations, changing the position of every object usually creates a new arrangement. But in a circular arrangement, some arrangements that look different in a line are actually the same when placed on a circle, because one can be rotated to match the other.

For example, if 4 students sit around a round table, the order ABCD is the same as BCDA, CDAB, and DABC, because each is just a rotation of the same seating pattern.

This is why circular permutations use a different counting rule from linear permutations.

Key Idea: For arranging n distinct objects in a circle, the number of different circular arrangements is

This formula works because rotations are considered identical.

Let us understand why.

If we arrange n distinct objects in a line, there are

possible arrangements.

But in a circle, each unique circular arrangement can be rotated into n equivalent linear forms. So we divide by n:

That is the reason the number of circular permutations of n distinct objects is

A helpful way to think about this is to fix one object in position. Since rotating the whole circle does not create a new arrangement, we can keep one object still and arrange the remaining objects around it.

Then we only need to arrange the remaining n - 1 objects, giving

Important Note: This rule applies when only rotation is ignored. In most round-table seating questions, this is the standard meaning.

Now let us go step by step through examples.

Worked Example 1: 5 students around a round table

How many different ways can 5 distinct students sit around a round table?

Since the students are distinct and seated in a circle, we use the circular permutation formula:

Answer: There are 24 different circular arrangements.

Why not \(5!\)? Because many of those linear arrangements are just rotations of each other and so count as the same circular seating.

Worked Example 2: 7 people standing in a circle

In how many ways can 7 different people stand in a circle?

Again, use the circular permutation formula:

Answer: There are 720 ways.

It does not matter whether they are sitting or standing. If the arrangement is around a circle and rotations are considered the same, the formula is still

Worked Example 3: 6 friends with 2 specific friends sitting next to each other

How many circular arrangements are possible for 6 friends around a table if A and B must sit next to each other?

When two people must stay together, treat them as one group first.

So instead of 6 separate objects, we now have:

• the group \((A,B)\)
• the other 4 friends

That gives a total of 5 objects to arrange in a circle.

The number of circular arrangements of 5 objects is

But inside the group, A and B can switch places in

ways: AB or BA.

So the total number of arrangements is

Answer: There are 48 circular arrangements.

Worked Example 4: 8 people with 3 specific people together

How many ways can 8 people sit around a circular table if 3 specific people must sit together?

Treat the 3 specific people as one block.

Then the objects to arrange are:

• 1 block of 3 people
• 5 other individual people

So there are 6 objects in total around the circle.

The number of circular arrangements of these 6 objects is

Now arrange the 3 people within their block. This can be done in

ways.

So the total number of arrangements is

Answer: There are 720 circular arrangements.

Steps for Solving Circular Permutation Problems

1. Check that the arrangement is circular. If objects are arranged around a circle, rotations usually count as the same.
2. Use \((n-1)!\) for n distinct objects.
3. If some objects must stay together, group them as one block.
4. Arrange the block and remaining objects in a circle.
5. Multiply by the number of ways to arrange objects inside the block.

Common Mistakes to Avoid

• Using \(n!\) instead of \((n-1)!\): This is the most common mistake. In a circle, rotations are not new arrangements.
• Forgetting to arrange the objects inside a group: If A and B are together, AB and BA are different unless stated otherwise.
• Not reading the condition carefully: “Around a table” usually means circular permutation.

Linear vs Circular Permutations

• In a line: number of arrangements of \(n\) distinct objects is \(n!\)
• In a circle: number of arrangements of \(n\) distinct objects is \((n-1)!\)

For example, with 4 objects:

This big difference happens because many linear orders become identical when placed on a circle.

Quick Check

• 5 objects in a circle: \((5-1)! = 24\)
• 6 objects in a circle: \((6-1)! = 120\)
• 9 objects in a circle: \((9-1)! = 8! = 40320\)

Summary

Circular permutations are used when arranging distinct objects around a circle. Since rotations represent the same arrangement, we do not use \(n!\). Instead, for \(n\) distinct objects, the number of arrangements is

A useful strategy is to fix one object and arrange the rest. If some objects must stay together, treat them as one block, arrange the blocks in a circle, and then arrange the objects within the block.

This idea appears often in combinatorics. For example, if you are choosing 3 students for a committee, the group {A, B, C} is the same committee as {C, A, B}. Since the order does not change the group, this is a combinations problem.

Combinations are different from permutations. In permutations, order matters. In combinations, order does not matter. Learning to tell the difference is one of the most important skills in counting.

When do we use combinations?

• Choosing a team
• Selecting committee members
• Picking questions to answer
• Choosing toppings or cards or books, when only the selection matters

When do we not use combinations?

• Assigning gold, silver, and bronze medals
• Arranging students in a line
• Creating passwords or codes
• Choosing class president, vice president, and secretary

In those cases, order or role matters, so they are not combinations.

The notation for combinations is:

\(\binom{n}{r}\), which is also written as \(nC_r\).

It means: the number of ways to choose \(r\) objects from \(n\) objects, without caring about order.

The formula is

where \(n!\) means factorial:

For example,

Also remember that \(0! = 1\).

Why does the combinations formula work?

If we first count arrangements of \(r\) chosen objects from \(n\), we get permutations. But each group of \(r\) objects can be arranged in \(r!\) different orders.

Since combinations do not care about order, we divide by \(r!\) to remove all the repeated orderings.

So combinations can be thought of as:

This is why the denominator contains \(r!\).

Important facts about combinations

• \(\binom{n}{0} = 1\) because there is exactly one way to choose nothing.
• \(\binom{n}{n} = 1\) because there is exactly one way to choose all \(n\) objects.
• \(\binom{n}{1} = n\) because there are \(n\) ways to choose one object.
• \(\binom{n}{r} = \binom{n}{n-r}\). Choosing \(r\) objects is the same as leaving out \(n-r\) objects.

That last fact is very useful. For example,

because choosing 3 items is equivalent to deciding which 7 items are not chosen.

How to solve a basic combinations problem

1. Decide whether order matters.
2. If order does not matter, use combinations.
3. Identify \(n\), the total number of available objects.
4. Identify \(r\), the number of objects being chosen.
5. Use \(\binom{n}{r} = \frac{n!}{r!(n-r)!}\).
6. Simplify carefully.

Worked Example 1: Basic selection

How many ways can a coach choose 3 players from 8 players?

Step 1: Order does not matter. A group of 3 players is the same no matter the order they are listed.

Step 2: Use combinations with \(n=8\) and \(r=3\).

Cancel the common \(5!\):

Answer: There are 56 ways.

Worked Example 2: Using the symmetry rule

How many ways can 2 books be chosen from 12 books?

We use

Cancel \(10!\):

Answer: There are 66 ways.

Notice that

So choosing 2 books gives the same count as choosing which 10 books to leave behind.

Worked Example 3: Combination with a required item

A committee of 4 students is chosen from 9 students. One particular student, Maya, must be on the committee. How many different committees are possible?

If Maya must be included, then 1 of the 4 spots is already filled.

So we only need to choose the remaining 3 students from the other 8 students.

Answer: There are 56 possible committees.

This is a common type of restriction: if someone must be included, fix that person first, then choose the rest.

Worked Example 4: Combination with inclusion and exclusion

A class has 6 boys and 5 girls. How many ways can a team of 4 students be chosen if the team must contain exactly 2 boys and 2 girls?

This problem has a condition, so we count each part separately.

First choose 2 boys from 6:

Then choose 2 girls from 5:

Now multiply, because each choice of boys can be paired with each choice of girls:

Answer: There are 150 possible teams.

This idea is very important: when a selection must satisfy different parts, count each part and then multiply.

Combinations with “at least” conditions

Sometimes a problem says at least, at most, or exactly. These words tell you how to organize your counting.

• Exactly means one specific case.
• At least means add several possible cases.
• At most also means add several possible cases.

For example, suppose a group has 4 boys and 3 girls, and you choose 3 students. If the group must contain at least 2 girls, then there are two possible cases:

• 2 girls and 1 boy
• 3 girls and 0 boys

We would count both cases and add them.

Mini Example: How many groups of 3 can be chosen from 4 boys and 3 girls if there must be at least 2 girls?

Case 1: 2 girls and 1 boy

Case 2: 3 girls and 0 boys

Total:

Answer: There are 13 such groups.

Common mistakes to avoid

• Using permutations when order does not matter. Always ask: does changing the order create a new outcome?
• Mixing up \(n\) and \(r\). \(n\) is the total number of available objects, and \(r\) is the number chosen.
• Forgetting restrictions. Words like “must include,” “exactly,” and “at least” change the setup.
• Not multiplying separate choices. If you choose from two different groups, you often multiply.
• Not adding separate cases. If a problem has several valid cases, count each and add.

Connection to the Binomial Theorem

Combinations are closely related to the Binomial Theorem. In the expansion of \((a+b)^n\), the coefficients come from combinations.

For example,

The coefficients \(1, 4, 6, 4, 1\) are:

So combinations are not only about counting groups. They also help us expand powers of binomials.

Quick check: Is it a combination?

• Choose 5 students from 20 for a field trip leader group: Yes
• Arrange 5 students in 5 chairs: No
• Select 2 pizza toppings from 8 options: Yes
• Assign first, second, and third place: No

Summary

Use combinations when you are selecting items and order does not matter. The formula is

To solve problems with conditions, think carefully about what must be included, what must be excluded, and whether you need to multiply different selections or add separate cases. With practice, combinations become a powerful tool for counting and for understanding the Binomial Theorem.

In this lesson, you will learn how to build Pascal’s Triangle, notice its patterns, and use it in two important ways:

• to find combinations, written as \(\binom{n}{r}\)
• to find coefficients in binomial expansions

Once you understand the triangle, many counting and expansion problems become much faster.

1. Building Pascal’s Triangle

Pascal’s Triangle begins with a single \(1\) at the top. Each new row starts and ends with \(1\). Every number inside the triangle is found by adding the two numbers directly above it.

The first few rows are:

$$ \begin{aligned} &1\\ &1\qquad 1\\ &1\qquad 2\qquad 1\\ &1\qquad 3\qquad 3\qquad 1\\ &1\qquad 4\qquad 6\qquad 4\qquad 1\\ &1\qquad 5\qquad 10\qquad 10\qquad 5\qquad 1\\ &1\qquad 6\qquad 15\qquad 20\qquad 15\qquad 6\qquad 1 \end{aligned} $$

For example, in the row \(1, 4, 6, 4, 1\), the \(6\) comes from adding the two numbers above it:

$$3+3=6$$

And one of the \(4\)s comes from:

$$1+3=4$$

This is why Pascal’s Triangle is called recursive: each row is built from the row before it.

2. Important patterns in Pascal’s Triangle

Pascal’s Triangle has several useful patterns.

• Each row starts and ends with 1.
• The rows are symmetric. For example, \(1, 5, 10, 10, 5, 1\) reads the same from left to right and right to left.
• The numbers increase and then decrease across a row.
• Each row corresponds to a power of a binomial.

If we number the top row as row \(0\), then:

• row \(0\): \(1\)
• row \(1\): \(1, 1\)
• row \(2\): \(1, 2, 1\)
• row \(3\): \(1, 3, 3, 1\)

This row numbering is important because row \(n\) gives the coefficients of \((a+b)^n\).

3. Connection to combinations

The entries in Pascal’s Triangle are combinations. A combination counts how many ways we can choose objects when order does not matter.

The combination notation is:

$$\binom{n}{r}$$

This means “choose \(r\) objects from \(n\) objects.”

Each row of Pascal’s Triangle gives the values of \(\binom{n}{r}\) for a fixed \(n\).

For example, row \(5\) is:

$$1\qquad 5\qquad 10\qquad 10\qquad 5\qquad 1$$

So:

$$ \binom{5}{0}=1,\quad \binom{5}{1}=5,\quad \binom{5}{2}=10,\quad \binom{5}{3}=10,\quad \binom{5}{4}=5,\quad \binom{5}{5}=1 $$

This means, for example, that there are \(10\) ways to choose \(2\) items from \(5\), and also \(10\) ways to choose \(3\) items from \(5\).

The symmetry in the triangle matches the fact that:

$$\binom{n}{r}=\binom{n}{n-r}$$

Choosing \(2\) objects to keep is the same as choosing \(3\) objects to leave out when there are \(5\) objects total.

4. Pascal’s rule

The recursive rule in the triangle can be written using combinations:

$$\binom{n}{r}=\binom{n-1}{r-1}+\binom{n-1}{r}$$

This is called Pascal’s Rule.

It says that any interior number in the triangle is the sum of the two numbers above it.

For example:

$$\binom{6}{2}=\binom{5}{1}+\binom{5}{2}=5+10=15$$

This matches row \(6\):

$$1\qquad 6\qquad 15\qquad 20\qquad 15\qquad 6\qquad 1$$

5. Connection to the Binomial Theorem

One of the most important uses of Pascal’s Triangle is finding coefficients in expansions of \((a+b)^n\).

The coefficients come directly from row \(n\) of the triangle.

For example:

• row \(2\): \(1, 2, 1\), so \((a+b)^2=a^2+2ab+b^2\)
• row \(3\): \(1, 3, 3, 1\), so \((a+b)^3=a^3+3a^2b+3ab^2+b^3\)
• row \(4\): \(1, 4, 6, 4, 1\), so \((a+b)^4=a^4+4a^3b+6a^2b^2+4ab^3+b^4\)

Notice two things in a binomial expansion:

• the powers of the first term decrease
• the powers of the second term increase

In \((a+b)^4\), the powers are:

$$a^4b^0,\ a^3b^1,\ a^2b^2,\ a^1b^3,\ a^0b^4$$

The exponents in each term always add up to \(4\), the power of the binomial.

6. Worked Examples

Example 1: Generate the next row of Pascal’s Triangle

Suppose you are given row \(4\):

$$1\qquad 4\qquad 6\qquad 4\qquad 1$$

Find row \(5\).

Step 1: Start and end with \(1\).

So the new row will look like:

$$1\qquad \_\qquad \_\qquad \_\qquad \_\qquad 1$$

Step 2: Add neighboring numbers from the row above.

$$1+4=5$$

$$4+6=10$$

$$6+4=10$$

$$4+1=5$$

Answer:

$$1\qquad 5\qquad 10\qquad 10\qquad 5\qquad 1$$

Example 2: Use Pascal’s Triangle to find a combination

Find \(\binom{6}{2}\).

Step 1: Go to row \(6\) of Pascal’s Triangle.

Row \(6\) is:

$$1\qquad 6\qquad 15\qquad 20\qquad 15\qquad 6\qquad 1$$

Step 2: Count positions starting with \(\binom{6}{0}\).

So:

• first entry is \(\binom{6}{0}=1\)
• second entry is \(\binom{6}{1}=6\)
• third entry is \(\binom{6}{2}=15\)

Answer:

$$\binom{6}{2}=15$$

Example 3: Expand \((x+y)^4\) using Pascal’s Triangle

Step 1: Use row \(4\) of Pascal’s Triangle.

Row \(4\) is:

$$1\qquad 4\qquad 6\qquad 4\qquad 1$$

Step 2: Write terms with powers of \(x\) decreasing from \(4\) to \(0\), and powers of \(y\) increasing from \(0\) to \(4\).

$$x^4,\ x^3y,\ x^2y^2,\ xy^3,\ y^4$$

Step 3: Attach the coefficients.

$$ (x+y)^4=x^4+4x^3y+6x^2y^2+4xy^3+y^4 $$

Answer:

$$ (x+y)^4=x^4+4x^3y+6x^2y^2+4xy^3+y^4 $$

Example 4: Expand \((2x+3)^3\) using Pascal’s Triangle

This example is a little harder because the terms are not just \(x\) and \(y\).

Step 1: Use row \(3\) of Pascal’s Triangle.

Row \(3\) is:

$$1\qquad 3\qquad 3\qquad 1$$

Step 2: Write the power pattern for \((2x+3)^3\):

$$ (2x)^3,\ (2x)^2(3),\ (2x)(3)^2,\ (3)^3 $$

Step 3: Add the coefficients from Pascal’s Triangle.

$$ (2x+3)^3=1(2x)^3+3(2x)^2(3)+3(2x)(3)^2+1(3)^3 $$

Step 4: Simplify each term.

$$1(2x)^3=8x^3$$

$$3(2x)^2(3)=3(4x^2)(3)=36x^2$$

$$3(2x)(3)^2=3(2x)(9)=54x$$

$$1(3)^3=27$$

Answer:

$$ (2x+3)^3=8x^3+36x^2+54x+27 $$

7. Common mistakes to avoid

• Using the wrong row. For \((a+b)^n\), use row \(n\), starting with the top as row \(0\).
• Forgetting the power pattern. The first term’s exponent goes down, while the second term’s exponent goes up.
• Ignoring coefficients when terms are more complicated. In \((2x+3)^3\), you must still multiply by the Pascal coefficients.
• Confusing combinations with permutations. Pascal’s Triangle gives combinations, where order does not matter.

8. Quick check for understanding

1. What comes after the row \(1, 3, 3, 1\)?
2. What is the value of \(\binom{5}{2}\)?
3. Which row gives the coefficients of \((a+b)^6\)?
4. Use Pascal’s Triangle to expand \((m+n)^2\).

Answers:

1. \(1, 4, 6, 4, 1\)
2. \(10\)
3. row \(6\)
4. \((m+n)^2=m^2+2mn+n^2\)

9. Summary

Pascal’s Triangle is built by starting and ending each row with \(1\), then adding pairs of numbers from the row above to fill the middle. The entries in row \(n\) are the combinations \(\binom{n}{r}\).

The triangle is also closely connected to the Binomial Theorem. Row \(n\) gives the coefficients of \((a+b)^n\), which makes polynomial expansions quicker and easier. If you can build the triangle and follow the power pattern, you can solve many counting and expansion problems with confidence.

For small powers, you may already know how to expand by repeated multiplication:

\((a+b)^2=(a+b)(a+b)=a^2+2ab+b^2\)

But what about \((a+b)^5\), \((x-2)^8\), or \((2x+3)^6\)? Expanding by multiplying again and again becomes slow and messy. The Binomial Theorem gives a systematic way to write every term directly.

This topic also connects algebra to combinatorics, because the coefficients in the expansion come from combinations such as \(\binom{n}{r}\).

1. What is a binomial?

A binomial is an expression with two terms, such as:

• \(a+b\)
• \(x-3\)
• \(2m+n\)

The Binomial Theorem tells us how to expand a binomial raised to a power.

2. The Binomial Theorem formula

For any positive integer \(n\),

This is often written more compactly as:

Here, \(\binom{n}{r}\) is called a combination or binomial coefficient.

3. Meaning of \(\binom{n}{r}\)

The notation \(\binom{n}{r}\) means the number of ways to choose \(r\) objects from \(n\) objects, without caring about order.

Its formula is:

For example:

These values become the coefficients in the expansion.

4. Pattern in the expansion

In the expansion of \((a+b)^n\):

• the powers of \(a\) start at \(n\) and decrease by 1 each term,
• the powers of \(b\) start at 0 and increase by 1 each term,
• the coefficients are \(\binom{n}{0}, \binom{n}{1}, \binom{n}{2}, \dots, \binom{n}{n}\).

So every term follows a clear structure:

This is called the general term. It helps us find any specific term without writing the whole expansion.

5. Why combinations appear

When we expand \((a+b)^n\), we are multiplying \(n\) copies of \((a+b)\):

Each term in the final expansion comes from choosing either \(a\) or \(b\) from each bracket.

If a term contains \(b^r\), that means we chose \(b\) from exactly \(r\) of the \(n\) brackets, and \(a\) from the remaining \(n-r\) brackets. The number of ways to do that is \(\binom{n}{r}\). That is why the coefficient is \(\binom{n}{r}\).

6. First few expansions

It helps to notice some common results:

The coefficients \(1, 4, 6, 4, 1\) come from combinations:

7. Worked Example 1: Expand \((x+y)^4\)

Use the Binomial Theorem:

Now calculate the coefficients:

So:

Important check: the power of \(x\) goes down from 4 to 0, and the power of \(y\) goes up from 0 to 4.

8. Worked Example 2: Expand \((2x+3)^3\)

Here, \(a=2x\), \(b=3\), and \(n=3\).

Apply the formula:

Now simplify each term:

Notice: when the terms inside the bracket are not just \(x\) and \(y\), we must raise the entire term correctly. For example, \((2x)^3=8x^3\), not \(2x^3\).

9. Worked Example 3: Find the 4th term of \((x-2)^6\)

We do not need the full expansion. We use the general term:

For \((x-2)^6\), we have:

• \(a=x\)
• \(b=-2\)
• \(n=6\)

The 4th term means \(r+1=4\), so \(r=3\).

Substitute into the formula:

So the 4th term is:

Important idea: if the second term is negative, signs can alternate because of powers of a negative number.

10. Worked Example 4: Find the term containing \(x^2\) in \((3x+1)^5\)

We want a specific power of \(x\), so again we use the general term:

Simplify the variable part:

We want the term containing \(x^2\), so set:

Now substitute \(r=3\):

So the term containing \(x^2\) is:

11. How to find a particular term

When asked for a specific term, follow these steps:

1. Write the general term: \(T_{r+1}=\binom{n}{r}a^{n-r}b^r\).
2. Identify \(a\), \(b\), and \(n\).
3. If asked for the \(k\)-th term, set \(r+1=k\).
4. If asked for the term containing a certain power, match the exponent and solve for \(r\).
5. Substitute and simplify carefully.

12. Common mistakes to avoid

• Forgetting the coefficient: each term needs \(\binom{n}{r}\).
• Using the wrong term number: the general term uses \(r+1\), not \(r\). For example, the 4th term means \(r=3\).
• Incorrect powers: one exponent decreases while the other increases.
• Ignoring brackets: \((2x)^2=4x^2\), not \(2x^2\).
• Sign errors: in expressions like \((x-2)^n\), remember that \(b=-2\).

13. Quick connection to Pascal’s Triangle

The binomial coefficients can also be arranged in Pascal’s Triangle:

\(1\)

\(1\quad 1\)

\(1\quad 2\quad 1\)

\(1\quad 3\quad 3\quad 1\)

\(1\quad 4\quad 6\quad 4\quad 1\)

\(1\quad 5\quad 10\quad 10\quad 5\quad 1\)

Each row gives the coefficients of \((a+b)^n\).

For example, row 5 gives the coefficients of \((a+b)^5\):

So:

14. Why this matters

The Binomial Theorem is useful because it lets you:

• expand powers quickly and accurately,
• find a chosen term without full expansion,
• see how combinatorics appears in algebra,
• solve many exam questions more efficiently.

Brief Summary

The Binomial Theorem gives a formula for expanding \((a+b)^n\):

The coefficients are combinations, written as \(\binom{n}{r}\). The general term

helps you find any term directly. As you move through the expansion, the power of the first term decreases while the power of the second term increases. With practice, you can expand binomials and find specific terms confidently.