IGCSE, Edexcel, IB and AP Science Courses | Dr Dabour Hub Academy

Overview

Recall the counting principle, permutations and combinations, and use them in the binomial theorem.
Expand $(x + a)^{n}$ for a positive integer $n$ and use Pascal’s triangle and the general term.
Find the middle term(s), the term containing $x^{k}$ , the term free of $x$ , and the ratio between consecutive terms.
Find the greatest term of an expansion and use the expansion to prove relations between combinations.

🌳 Concept tree

Lesson 1Counting Principle, Permutations (nPr) and Combinations (nCr)

├── Recallcounting principle, permutations and combinations

│ ├── Counting principle

│ ├── Permutations (nPr)

│ └── Combinations (nCr)

├── 1-1 (a) Pascal's triangle and the binomial expansion

│ ├── Patterns in the expansions

│ ├── Pascal's triangle

│ ├── Sum of a row and the number of subsets

│ ├── The binomial theorem

│ └── Special cases

├── 1-1 (b) The general term

│ ├── Rule

│ ├── Term counted from the end

│ ├── Sum and difference of two expansions

│ ├── Recognising an expansion in disguise

│ └── Finding unknowns from coefficients

├── 1-1 (c) The middle term

│ ├── Rule

│ └── Middle term of a sum of two expansions

├── 1-2 The term containing $x^{k}$ and the term free of $x$

│ ├── Method

│ └── Equal middle terms

├── 1-3 (a) The ratio between two consecutive terms

│ └── Ratio of terms

└── 1-3 (b) The greatest term

├── Method

└── Using the binomial expansion to prove relations

Lesson 1: Counting Principle, Permutations (nPr) and Combinations (nCr) · 1 / 16

The counting principle

Counting principle questions ask how many outcomes a whole process has
without listing them. There are two basic rules.

Comparison aspect	Main idea	Keywords
Multiplication rule (AND)	if one choice can be made in m ways and a second choice can then be made in n ways the two choices together can be made in m × n ways	choicemadewayssecond
Addition rule (OR)	if the choices cannot happen together and one of them can be made in m ways the other in n ways then one of them can be made in m + n ways	choicescannothappentogether
Example.	A student has 4 shirts and 3 pairs of trousers. The number of outfits is 4 × 3 = 12. If instead the student travels either by one of 3 buses or by one of 2 trains the number of travel choices is 3 + 2 = 5	studentshirtspairstrousers

Factorial notation

For a whole number n ≥ 1, the factorial is n! = n × (n − 1) × … × 2 × 1.
By definition, 0! = 1.
For example, 5! = 5 × 4 × 3 × 2 × 1 = 120.

Lesson 1: Counting Principle, Permutations (nPr) and Combinations (nCr) · 2 / 16

Permutations (nPr)

A permutation is an arrangement in which the order matters.
The number of ways to arrange r objects chosen from n different objects is written nPr:
nPr = n! ÷ (n − r)! = n × (n − 1) × … × (n − r + 1)
The product has exactly r factors.
For example, 7P3 = 7 × 6 × 5 = 210, and 5P5 = 5! = 120.
When r = n, every object is used and the formula gives n!.

Seats, rankings, passwords, codes and arrangements on a shelf, where ABC and BCA are different results.
Choosing a president and a secretary from a group, because the two posts are different.

Lesson 1: Counting Principle, Permutations (nPr) and Combinations (nCr) · 3 / 16

Combinations (nCr)

A combination is a selection in which the order does not matter. The number of ways to choose r objects from n is written ⁿC_r:
nCr = n! ÷ (r! × (n − r)!) = nPr ÷ r!
Since each group of r objects can be ordered in r! ways, the number of combinations is the number of permutations divided by r!.
For example, 6C2 = 6! ÷ (2! × 4!) = 15.

ⁿC₀ = ⁿC_n = 1, and ⁿC₁ = n.
ⁿC_r = ⁿC_n−r. For example, 9C7 = 9C2 = 36.
Pascal rule nCr + nCr+1 = n+1Cr+1. For example, 5C2 + 5C3 = 10 + 10 = 20 = 6C3.

Permutations or combinations?

Ask whether swapping two chosen objects creates a new result.
If it does, use permutations.
If it does not, as in a committee, use combinations.

Lesson 1: Counting Principle, Permutations (nPr) and Combinations (nCr) · 4 / 16

Worked examples on counting, permutations and combinations

Comparison aspect	Main idea	Keywords
Example 1 (permutation with no repetition).	How many 3-letter codes can be made from 26 letters if no letter may be repeated?	manylettercodesmade
Solution	The order matters so use 26P3 = 26 × 25 × 24 = 15 600	ordermatters26P3
Example 2 (combination).	A committee of 3 is chosen from 8 people. How many committees are possible?	committeechosenpeoplemany
Solution	The order of the members does not matter so use 8C3 = (8 × 7 × 6) ÷ (3 × 2 × 1) = 56	ordermembersdoesmatter
Example 3 (combinations with the multiplication rule).	A team of 4 is chosen from 4 boys and 5 girls with exactly 2 boys and 2 girls. Find the number of teams	teamchosenboysgirls
Solution	Choose the boys and the girls separately then multiply. The number of teams is 4C2 × 5C2 = 6 × 10 = 60	Chooseboysgirlsseparately

Comparison aspect	Main idea	Keywords
Example 4 (arrangements with repeated letters).	How many different arrangements are there of the letters of the word LEVEL?	manydifferentwordLEVEL
Solution	There are 5 letters with L repeated twice and E repeated twice. Divide by the arrangements of the repeats: 5! ÷ (2! × 2!) = 120 ÷ 4 = 30	lettersrepeatedtwiceDivide
Example 5 (a block arrangement).	Six people sit in a row and two particular people must sit next to each other. How many seatings are possible?	peopleparticularmustnext
Solution	Treat the two people as one block. Then there are 5 units which can be arranged in 5! ways. Inside the block the two people can swap places in 2! ways. The total is 5! × 2! = 240	TreatpeopleblockThen
Example 6 (at least a condition).	A group of 5 is chosen from 6 men and 4 women with at least 2 women. Find the number of groups	groupchosenwomenFind

Solution→All groups of 5 number 10C5 = 252. Subtract the groups with fewer than 2 women. With no women, there are 6C5 = 6 groups. With exactly 1 woman, there are 4C1 × 6C4 = 4 × 15 = 60 groups. So the answer is 252 − 6 − 60 = 186.

Lesson 1: Counting Principle, Permutations (nPr) and Combinations (nCr) · 5 / 16

Recall: counting principle, permutations and combinations

Counting principle

Multiplication rule (AND)if one task can be done in

m

ways and, after it, a second task in

n

ways, the two tasks together can be done in

m \times n

ways.

Addition rule (OR)if two tasks cannot happen together and they can be done in

m

and

n

ways, then one of them can be done in

m + n

ways.

Factorial

n! = n (n - 1) (n - 2) \dots 3 \times 2 \times 1

, with

0! = 1

.

Permutation an arrangement of $r$ objects chosen from $n$ different objects, where the order matters.
$^{n} P_{r} = \frac{n!}{(n - r)!} = n (n - 1) (n - 2) \dots (n - r + 1)$
$^{n} P_{n} = n!$ and $^{n} P_{1} = n$ .
Arranging $n$ different objects around a circle (no fixed reference point) gives $(n - 1)!$ ways.

Comparison aspect	Main idea	Keywords
Combination	a selection of $r$ objects out of $n$ different objects, where the order does not matter. $^{n} C_{r} = \frac{n!}{r! (n - r)!} = \frac{^{n} P_{r}}{r!}$	selectionobjectsdifferentorder
•	$^{n} C_{r} =^{n} C_{n - r}$ , and $^{n} C_{0} =^{n} C_{n} = 1$ , $^{n} C_{1} = n$ .
•	If $^{n} C_{x} =^{n} C_{y}$ , then either $x = y$ or $x + y = n$ .	theneither
Pascal’s rule	$^{n} C_{r} +^{n} C_{r - 1} =^{n + 1} C_{r}$ .
Ratio of consecutive combinations	nCrnCr−1=n−r+1r	nCrnCr

Point of comparison	Permutations $^{n} P_{r}$	Combinations $^{n} C_{r}$
Order	Matters	Does not matter
Formula	$\frac{n!}{(n - r)!}$	$\frac{n!}{r! (n - r)!}$
Typical use	Forming numbers, passwords, rankings	Choosing committees, cards, subsets
Size	$^{n} P_{r} \geq^{n} C_{r}$	$^{n} C_{r} \leq^{n} P_{r}$

✍️ Worked examples

Example 1:
How many different 3-digit numbers can be formed from the digits 1, 2, 3, 4, 5 without repetition?
In how many ways can a committee of 3 be chosen from 7 people?
Solution:
Numbers the order matters, so 5P3=5×4×3=60.
Committee the order does not matter, so 7C3=7×6×53×2×1=35.
Example 2 Solve $^{n} P_{2} = 42$ and $^{n} C_{2} = 28$ .
Solution:
$^{n} P_{2} = n (n - 1) = 42$ , so $n^{2} - n - 42 = 0$ , $(n - 7) (n + 6) = 0$ , hence $n = 7$ (reject $- 6$ ).
$^{n} C_{2} = \frac{n (n - 1)}{2} = 28$ , so $n (n - 1) = 56$ , hence $n = 8$ .

Lesson 1: Counting Principle, Permutations (nPr) and Combinations (nCr) · 6 / 16

1-1 (a) Pascal’s triangle and the binomial expansion

Patterns in the expansions

$(x + a)^{2} = x^{2} + 2 x a + a^{2}$ and $(x + a)^{3} = x^{3} + 3 x^{2} a + 3 x a^{2} + a^{3}$ .
$(x + a)^{4} = x^{4} + 4 x^{3} a + 6 x^{2} a^{2} + 4 x a^{3} + a^{4}$ .
The expansion of $(x + a)^{n}$ has $n + 1$ terms.
The powers of $x$ go down and the powers of $a$ go up, and in every term the two powers add to $n$ .

Pascal’s triangle

Pascal’s triangle a triangle of numbers whose row $n$ gives the coefficients of $(a + b)^{n}$ .
Each row starts and ends with 1, and every other number is the sum of the two numbers above it.
Written with combinations, row $n$ is $^{n} C_{0},^{n} C_{1}, \dots,^{n} C_{n}$ .
Rows 1; 1 1; 1 2 1; 1 3 3 1; 1 4 6 4 1; 1 5 10 10 5 1.

Sum of a row and the number of subsets

A set of $n$ elements has $2^{n}$ subsets.
Therefore the sum of the numbers in row $n$ is
$^{n} C_{0} +^{n} C_{1} +^{n} C_{2} + \dots +^{n} C_{n} = 2^{n}$

The binomial theorem

For x,a∈R and a positive integer n (x+a)n=xn+nC1 xn−1a+nC2 xn−2a2+⋯+nCn−1 x an−1+an
The number of terms is $n + 1$ .
Powers of $x$ descend and powers of $a$ ascend, and the sum of the powers in every term is $n$ .
In $^{n} C_{r}$ the value of $r$ equals the order of the term minus one.

✍️ Worked examples

Example 3 Expand $(2 x + 3 y)^{4}$ .
Solution:
$(2 x + 3 y)^{4} = (2 x)^{4} +^{4} C_{1} (2 x)^{3} (3 y) +^{4} C_{2} (2 x)^{2} (3 y)^{2} +^{4} C_{3} (2 x) (3 y)^{3} + (3 y)^{4}$ .
$= 16 x^{4} + 4 (8 x^{3}) (3 y) + 6 (4 x^{2}) (9 y^{2}) + 4 (2 x) (27 y^{3}) + 81 y^{4}$ .
$= 16 x^{4} + 96 x^{3} y + 216 x^{2} y^{2} + 216 x y^{3} + 81 y^{4}$ .

Special cases

Putting x=1 and a=x (1+x)n=1+nC1 x+nC2 x2+⋯+xn
Putting a=−x gives alternating signs (1−x)n=1−nC1 x+nC2 x2−⋯+(−1)nxn
Putting $x = 1$ in $(1 + x)^{n}$ gives the sum of the coefficients, $2^{n}$ .
Putting $x = 1$ in $(1 - x)^{n}$ gives $^{n} C_{0} -^{n} C_{1} +^{n} C_{2} - \dots = 0$ .

✍️ Worked examples

Example 4 Write the expansion of $(1 + x)^{8}$ and use it to find $^{8} C_{0} +^{8} C_{1} +^{8} C_{2} + \dots +^{8} C_{8}$ .
Solution:
$(1 + x)^{8} = 1 +^{8} C_{1} x +^{8} C_{2} x^{2} + \dots + x^{8}$ .
Put $x = 1$ on both sides.
$2^{8} =^{8} C_{0} +^{8} C_{1} +^{8} C_{2} + \dots +^{8} C_{8}$ .
The sum is $256$ .

Lesson 1: Counting Principle, Permutations (nPr) and Combinations (nCr) · 7 / 16

1-1 (b) The general term

Rule

The general term is the term of order $r + 1$ in the expansion of $(x + y)^{n}$ , with $0 \leq r \leq n$ :
$T_{r + 1} =^{n} C_{r} x^{n - r} y^{r}$
The coefficient of $T_{r + 1}$ is $^{n} C_{r} \times (coefficient of the first term)^{n - r} \times (coefficient of the second term)^{r}$ , including signs.

✍️ Worked examples

Example 5 In the expansion of ${(x^{3} + \frac{2}{x})}^{8}$ in descending powers of $x$ , find the sixth term and its coefficient.
Solution:
Sixth term means $r + 1 = 6$ , so $r = 5$ .
$T_{6} =^{8} C_{5} (x^{3})^{3} {(\frac{2}{x})}^{5}$ .
$= 56 \times x^{9} \times 32 x^{- 5} = 1792 x^{4}$ .
The coefficient is $1792$ .

The term of order $m$ from the end in an expansion with $n + 1$ terms is the term of order $n + 1 - m + 1 = n - m + 2$ from the beginning.

✍️ Worked examples

Example 6 Find the fourth term from the end in the expansion of ${(2 x - \frac{1}{x})}^{7}$ .
Solution:
The expansion has $8$ terms, so the order from the beginning is $8 - 4 + 1 = 5$ .
$T_{5} =^{7} C_{4} (2 x)^{3} {(- \frac{1}{x})}^{4}$ .
$= 35 \times 8 x^{3} \times x^{- 4} = 280 x^{- 1}$ .

The expansions of $(x + a)^{n}$ and $(x - a)^{n}$ differ only in the signs of the terms of even order, so:
$(x + a)^{n} + (x - a)^{n} = 2 (T_{1} + T_{3} + T_{5} + \dots)$
$(x + a)^{n} - (x - a)^{n} = 2 (T_{2} + T_{4} + T_{6} + \dots)$
where $T_{1}, T_{2}, \dots$ are the terms of $(x + a)^{n}$ .

✍️ Worked examples

Simplify $(x + 2)^{6} + (x - 2)^{6}$ .
Solution:
The sum equals $2 (T_{1} + T_{3} + T_{5} + T_{7})$ .
$= 2 (x^{6} +^{6} C_{2} x^{4} (2)^{2} +^{6} C_{4} x^{2} (2)^{4} + 2^{6})$ .
$= 2 (x^{6} + 60 x^{4} + 240 x^{2} + 64)$ .
$= 2 x^{6} + 120 x^{4} + 480 x^{2} + 128$ .

An expression such as $(3 + x)^{n} -^{n} C_{1} (3 + x)^{n - 1} (1 - 2 x) + \dots$ is the expansion of ${[(3 + x) - (1 - 2 x)]}^{n}$ , which simplifies before you pick a term.

Write the coefficients as combinations
then divide one equation by the other to cancel the unknown.

✍️ Worked examples

Example 8 If $(1 + c x)^{n} = 1 + 20 x + a_{2} x^{2} + a_{3} x^{3} + \dots$ and $16 a_{2} = 3 a_{3}$ , find $n$ and $c$ (where $c \neq 0$ ).
Solution:
Coefficient of x nC1 c=20, so nc=20 (1).
$a_{2} =^{n} C_{2} c^{2}$ and $a_{3} =^{n} C_{3} c^{3}$ .
$16^{n} C_{2} c^{2} = 3^{n} C_{3} c^{3}$ , so $16^{n} C_{2} = 3^{n} C_{3} c$ .
Use nC3nC2=n−23 16=(n−2) c (2).
From (2) nc−2c=16; with (1): 20−2c=16, so c=2.
Then $n = \frac{20}{2} = 10$ .

Lesson 1: Counting Principle, Permutations (nPr) and Combinations (nCr) · 8 / 16

1-1 (c) The middle term

Rule

The expansion of $(x + a)^{n}$ has $n + 1$ terms.
If $n$ is even, the number of terms is odd and there is one middle term of order $\frac{n}{2} + 1$ .
If $n$ is odd, the number of terms is even and there are two middle terms of orders $\frac{n + 1}{2}$ and $\frac{n + 3}{2}$ .

✍️ Worked examples

Example 9 Find the middle term in the expansion of ${(2 x + \frac{1}{x})}^{6}$ .
Solution:
$n = 6$ is even, so the middle term has order $\frac{6}{2} + 1 = 4$ .
$T_{4} =^{6} C_{3} (2 x)^{3} {(\frac{1}{x})}^{3}$ .
$= 20 \times 8 x^{3} \times x^{- 3} = 160$ .
Example 10 Find the two middle terms in the expansion of $(2 x + y)^{7}$ .
Solution:
$n = 7$ is odd, so the orders are $\frac{7 + 1}{2} = 4$ and $\frac{7 + 3}{2} = 5$ .
$T_{4} =^{7} C_{3} (2 x)^{4} y^{3} = 35 \times 16 x^{4} y^{3} = 560 x^{4} y^{3}$ .
$T_{5} =^{7} C_{4} (2 x)^{3} y^{4} = 35 \times 8 x^{3} y^{4} = 280 x^{3} y^{4}$ .

If $(x + a)^{n} + (x - a)^{n}$ has $n + 1$ terms, only the odd-order terms survive (doubled), so the middle term of the sum is $2 T_{middle}$ when its order is odd.

Lesson 1: Counting Principle, Permutations (nPr) and Combinations (nCr) · 9 / 16

1-2 The term containing xk and the term free of x

Method

Writing the whole expansion to find one term is slow, so use the general term.
Step 1 write Tr+1 in terms of r.
Step 2 add the powers of x in Tr+1 (an expression in r) and put the sum equal to k.
Step 3 solve for r. If r is a natural number (and 0≤r≤n), the required term is Tr+1.
If $r$ is not a natural number, there is no term containing $x^{k}$ .
For the term free of $x$ , put the sum of the powers of $x$ equal to $0$ .

✍️ Worked examples

Example 11 In the expansion of ${(x^{2} + \frac{2}{x})}^{9}$ find the coefficient of $x^{6}$ , the term free of $x$ , and show that there is no term containing $x^{7}$ .
Solution:
$T_{r + 1} =^{9} C_{r} (x^{2})^{9 - r} {(\frac{2}{x})}^{r} =^{9} C_{r} 2^{r} x^{18 - 3 r}$ .
Coefficient of x6 18−3r=6, so r=4 and the term is T5 with coefficient 9C4×24=126×16=2016.
Free of x 18−3r=0, so r=6 and the term is T7=9C6×26=84×64=5376.
For x7 18−3r=7 gives r=113, which is not a natural number.
So the expansion has no term containing $x^{7}$ .
Example 12 If $n$ is a positive integer, show that the expansion of ${(x^{5} + \frac{1}{x^{2}})}^{n}$ has a term free of $x$ only when $n$ is a multiple of 7, and find that term for $n = 7$ .
Solution:
$T_{r + 1} =^{n} C_{r} (x^{5})^{n - r} {(\frac{1}{x^{2}})}^{r} =^{n} C_{r} x^{5 n - 7 r}$ .
Free of x 5n−7r=0, so r=5n7.
$r$ is a whole number only when $n$ is a multiple of 7.
For n=7 r=5, so the term is T6=7C5=21.

Equal middle terms

To find the value of a quantity that makes two middle terms equal
write both middle terms with the general term and equate them.

✍️ Worked examples

Example 13 In the expansion of ${(2 + \frac{x}{3})}^{9}$ find $x$ if the two middle terms are equal.
Solution:
$n = 9$ is odd, so the middle terms are $T_{5}$ and $T_{6}$ .
$T_{5} =^{9} C_{4} (2)^{5} {(\frac{x}{3})}^{4}$ and $T_{6} =^{9} C_{5} (2)^{4} {(\frac{x}{3})}^{5}$ .
C4=9C5, so equate 25(x3)4=24(x3)5.
Divide by 24(x3)4 2=x3, so x=6.

Lesson 1: Counting Principle, Permutations (nPr) and Combinations (nCr) · 10 / 16

1-3 (a) The ratio between two consecutive terms

Ratio of terms

Comparison aspect	Main idea	Keywords
For consecutive terms Tr and Tr+1 of (x+a)n	Tr+1Tr=n−r+1r×ax
•	For the coefficients (when the terms are written with a common variable), the factor $\frac{a}{x}$ is replaced by the ratio of the coefficients of the second and first terms of the binomial.	coefficientstermswrittencommon
For (1+x)n	coefficient of Tr+1coefficient of Tr=n−r+1r	coefficient1coefficient
Proof of the ratio	nCrnCr−1=n!r!(n−r)!×(r−1)!(n−r+1)!n!=n−r+1r	nCrnCr

✍️ Worked examples

Example 14 In the expansion of $(x + y)^{12}$ find $\frac{T_{4}}{T_{3}}$ and the ratio $\frac{T_{9}}{T_{7}}$ .
Solution:
For T4T3 take r=3 12−3+13⋅yx=103⋅yx.
$\frac{T_{9}}{T_{7}} = \frac{T_{9}}{T_{8}} \times \frac{T_{8}}{T_{7}}$ .
T9T8 has r=8 12−8+18⋅yx=58⋅yx.
T8T7 has r=7 12−7+17⋅yx=67⋅yx.
$\frac{T_{9}}{T_{7}} = \frac{5}{8} \times \frac{6}{7} \cdot \frac{y^{2}}{x^{2}} = \frac{15}{28} \cdot \frac{y^{2}}{x^{2}}$ .
Example 15 In the expansion of $(x + y)^{8}$ , if $2 T_{5} = T_{4} + T_{6}$ , find $\frac{x}{y}$ .
Solution:
Divide the equation by T5 2=T4T5+T6T5.
$\frac{T_{5}}{T_{4}} = \frac{8 - 4 + 1}{4} \cdot \frac{y}{x} = \frac{5 y}{4 x}$ , so $\frac{T_{4}}{T_{5}} = \frac{4 x}{5 y}$ .
$\frac{T_{6}}{T_{5}} = \frac{8 - 5 + 1}{5} \cdot \frac{y}{x} = \frac{4 y}{5 x}$ .
2=4x5y+4y5x; multiply by 5xy 10xy=4x2+4y2.
$2 x^{2} - 5 x y + 2 y^{2} = 0$ , so $(2 x - y) (x - 2 y) = 0$ .
Hence $y = 2 x$ or $x = 2 y$ , i.e. $\frac{x}{y} = \frac{1}{2}$ or $\frac{x}{y} = 2$ .
Example 16 The coefficients of three consecutive terms of $(1 + x)^{n}$ in ascending powers are $35, 21, 7$ . Find $n$ and the orders of the terms.
Solution:
Let the terms be $T_{r}, T_{r + 1}, T_{r + 2}$ .
$\frac{21}{35} = \frac{n - r + 1}{r}$ , so $3 r = 5 n - 5 r + 5$ and $5 n - 8 r = - 5$ (1).
$\frac{7}{21} = \frac{n - r}{r + 1}$ , so $r + 1 = 3 n - 3 r$ and $3 n - 4 r = 1$ (2).
Multiply (2) by 2 6n−8r=2; subtract (1): n=7.
From (2) 21−4r=1, so r=5.
The terms are $T_{5}, T_{6}, T_{7}$ , and indeed $^{7} C_{4} = 35$ , $^{7} C_{5} = 21$ , $^{7} C_{6} = 7$ .

Lesson 1: Counting Principle, Permutations (nPr) and Combinations (nCr) · 11 / 16

1-3 (b) The greatest term

Method

Compare $T_{r + 1}$ with $T_{r}$ using the ratio $\frac{T_{r + 1}}{T_{r}} = \frac{n - r + 1}{r} \cdot \frac{a}{x}$ at the given numerical values.
Solve $\frac{T_{r + 1}}{T_{r}} > 1$ to find the values of $r$ for which the terms are still increasing.
The greatest term is the last one for which the ratio is greater than (or equal to) 1; if the ratio equals 1 for some $r$ , two terms are equal and both are greatest.

✍️ Worked examples

Example 17 Find the greatest term in the expansion of $(x + y)^{10}$ when $x = 2$ and $y = 3$ .
Solution:
$\frac{T_{r + 1}}{T_{r}} = \frac{11 - r}{r} \cdot \frac{3}{2}$ .
The ratio is $\geq 1$ when $3 (11 - r) \geq 2 r$ , i.e. $33 \geq 5 r$ , i.e. $r \leq 6.6$ .
So $T_{7} \geq T_{6} \geq \dots \geq T_{1}$ in the sense that the terms increase up to $T_{7}$ (use $r = 1, \dots, 6$ ).
For $r \geq 7$ the ratio is below 1, so $T_{8} < T_{7}$ and the terms decrease after $T_{7}$ .
The greatest term is $T_{7} =^{10} C_{6} (2)^{4} (3)^{6} = 210 \times 16 \times 729 = 2 449 440$ .

Using the binomial expansion to prove relations

Put suitable values of $x$ in $(1 + x)^{n}$ to prove results such as $^{n} C_{0} +^{n} C_{1} + \dots +^{n} C_{n} = 2^{n}$ and $^{n} C_{0} -^{n} C_{1} +^{n} C_{2} - \dots = 0$ .
Adding the two results gives the sum of the even-ranked combinations equal to the sum of the odd-ranked ones, each equal to $2^{n - 1}$ .

Lesson 1: Counting Principle, Permutations (nPr) and Combinations (nCr) · 12 / 16

Key Scientific Tables & Anatomical Diagrams Extracted from Official Curricula

💡 Core Diagrams & Reference Data→All data, reference comparison tables, and anatomical figures have been accurately digitized from official curriculum textbooks.

🔬 binomial_theorem_and_pascal_triangle

Scientific details and precise observational breakdown:
(a) moTeACing (bh) decreasing (c) alternating. (cl) finite

(21) The eequence (T,) = (| Win’ 4 1)) is

ja) NCTeASINg ih) decreasing (c) alternating. (d) constant.
2) Which of the following sequences (T,) is decreasing where n € T°?
iaviT y= n’) (by (Ty = in) (ce) (T= (n’) (I(T y= (n+ 1)
(23) Which of the following represents a general term of a decreasing sequence ?

(ayn (b)n? (c)n’ ~§ (dy ‘4

(24) The sequence (1 +467 11 +.) is

(a) decreasing. (b) alternating. (c) finite. (d) infinite.

(25) |.) The sequence (3 +5 +7 +96… +21) is

Lesson 1: Counting Principle, Permutations (nPr) and Combinations (nCr) · 13 / 16

Part 14

📖

🔍 In-Depth Explanations, Deductions & Multi-Book Insights:

Mathematics Thanaweya Amma (2026/2027)

Lesson 1 – Counting Principle, Permutations (nPr) and Combinations (nCr)
Combinatorics analyzes discrete counting arrangements where order matters (Permutations) versus selection groupings where order is irrelevant (Combinations).
We apply factorial properties and pascal identity in solving algebraic ministerial problems.

🌳 Hierarchy Classification & Concept Tree

Combinatorial Foundations

├── 1. Fundamental Counting Principle

│ ├── 'And' ruleOperations performed simultaneously → Multiply possibilities (m × n)

│ └── 'Or' ruleMutually exclusive alternative operations → Add possibilities (m + n)

├── 2. Permutations (nPr)

│ ├── Arrangements where order mattersnPr = n! / (n – r)!

│ └── Circular permutations(n – 1)! without fixed reference point

└── 3. Combinations (nCr)

├── Selections where order does NOT matternCr = n! / [r! · (n – r)!] = nPr / r!

├── Complement rulenCr = nC(n-r)

└── Addition identity (Pascal's rule)nCr + nC(r-1) = (n+1)Cr

📋 Master Teacher's Blackboard & Exam Secrets

RULE 1Ratio between consecutive combinations: nCr / nC(r-1) = (n – r + 1) / r.

RULE 2Pascal's Identity: nCr + nC(r-1) = (n+1)Cr.

RULE 3If nCx = nCy, then either x = y or x + y = n.

📊 Comparative AnalysisDistinction Between Permutations and Combinations

Criterion	Permutations (nPr)	Combinations (nCr)
Order Sensitivity	Order is CRUCIAL and strictly matters	Order is IRRELEVANT (set grouping)
Typical Real Application	Forming numbers from digits passwords race ranks	Selecting committees playing cards geometric lines
Mathematical Formula	n! / (n – r)!	n! / [r! · (n – r)!]
Relative Magnitude	nPr ≥ nCr for all r ≥ 1	nCr ≤ nPr

Lesson 1: Counting Principle, Permutations (nPr) and Combinations (nCr) · 14 / 16

Lesson 1 – Counting Principle, Permutations and Combinations and the Binomial Theorem

★ Comprehensive Study & Detailed Pedagogical Guide ★
Comprehensive Explanation Fundamental Principle of Counting, Permutations, Combinations, and the Binomial Theorem
🎯 Targeted Learning Outcomes & Pedagogical Objectives:
• Apply the Fundamental Principle of Counting (Multiplication and Addition rules).
• Distinguish conceptually and algebraically between Permutations (ⁿP_r) and Combinations (ⁿC_r).
• Expand (x + a)ⁿ using the Binomial Theorem and compute the general term T_r+1.
• Determine the middle term(s) and find the term independent of x.
📖 Core Terminology, Scientific Definitions & Glossaries:
🔹 Fundamental Counting Principle If an operation can be performed in m ways and a second independent operation in n ways, both operations in succession can be performed in m × n ways.

Comparison aspect	Main idea	Keywords
🔹 Permutation (nPr)	An arrangement of r objects selected from n distinct objects where ORDER MATTERS: nPr = (n!) / ((n-r)!)	arrangementobjectsselecteddistinct
🔹 Combination (nCr)	A selection of r objects from n distinct objects where ORDER DOES NOT MATTER: nCr = (n!) / (r!(n-r)!) = nPr / (r!)	selectionobjectsdistinctORDER
🔹 Binomial Theorem	An algebraic expansion of the power of a sum of two terms: (x + a)n = Σr=0n nCr xn-r ar	algebraicexpansionpowerterms
🔹 Term Independent of x	A specific term in a binomial expansion where the overall exponent of the variable x reduces to zero (x0 = 1) representing a pure numerical constant	specificbinomialexpansionoverall
•	⚖️ Governing Scientific Laws Theorems & Mathematical Models	GoverningScientificLawsTheorems
★ Permutations and Combinations Formulas	[ Formula: nPr = (n!) / ((n-r)!) \| nCr = (n!) / (r!(n-r)!) \| nCr = nCn-r (Symmetry Law)nCr + nCr-1 = n+1Cr (Pascal’s Addition Law) \| nCr / (nCr-1) = (n – r + 1) / r (Ratio Law). ]	SymmetryPascalAdditionRatio

★ General Term in Binomial Expansion [ Formula: In (x + a)n: Tr+1 = nCr xn-r ar (Notice the lower index is r for term (r+1)). ]
★ Ratio Law for Successive Terms [ Formula: (Tr+1) / Tr = (n – r + 1) / r × (second term) / (first term) = (n – r + 1) / r · a / x. ]
★ Middle Term Rules [ Formula: If n is EVEN: Exactly ONE middle term T(n / 2 + 1).If n is ODD: Exactly TWO middle terms T((n+1) / 2) and T((n+3) / 2). ]
📌 Exhaustive In-Depth Pedagogical Breakdown:
◄ I. Foundational Combinatorics: Order vs Unordered Selection
Combinatorics governs the mathematics of finite discrete structures.
When selecting r items from n possibilities, the deciding question is: “Does sequence or order matter?”- Order Matters ⇒ Permutations (nPr): Arranging runners on a podium (1st, 2nd, 3rd), forming multi-digit passwords, or electing President, Secretary, and Treasurer.- Order Irrelevant ⇒ Combinations (nCr): Selecting a committee of 3 students from a class of 20, drawing a lottery hand of cards, or choosing 4 points on a circle to form a quadrilateral.

Pascal’s Identity ⁿC_r + ⁿC_r-1 = ⁿ⁺¹C_r mathematically proves that choosing a committee of r from n+1 people equals the sum of committees excluding a specific leader plus committees including that leader.
Figure (binomial_theorem_and_pascal_triangle.png): Figure (7): Pascal’s Triangle coefficients and binomial expansion distribution curves
◄ II. The Binomial Expansion Architecture and the (r+1)-th Term
The expansion of (x + a)n produces exactly (n + 1) terms.
The powers of the first term x descend steadily from n to 0, while powers of the second term a ascend from 0 to n.
The sum of the exponents of x and a in every individual term is perpetually equal to n.

To locate any term without writing out the full polynomial, mathematicians rely on the General Term: Tr+1 = nCr xn-r ar.
When terms involve fractional powers, such as (x2 + 1 / x)12, one simplifies the expression into xpower and equates that power to 0 to identify the term independent of x, or equates it to k to find the coefficient of xk.
◄ III. Ratio Rules and Greatest Term Identification
Calculating consecutive terms by direct evaluation is tedious.
The Ratio Property (Tr+1) / Tr = (n – r + 1) / r × a / x provides a swift recursive tool.
To locate the Greatest Term in magnitude for a given numerical substitution x = x0, one solves the simultaneous inequalities (Tr+1) / Tr ≥ 1 and Tr / (Tr-1) ≥ 1, bounding the optimal integer index r.

Comparison aspect	Main idea	Keywords
•	🧮 Exemplary Step-by-Step Worked Problems & Applied Exercises	ExemplaryStepWorkedProblems
Worked Example 1	Worked Problem 1 – Binomial General Term and Constant Evaluation	ProblemBinomialGeneralTerm
Problem Statement	Problem: In the expansion of (x2 – 2 / x)9 in descending powers of x:1) Find the term independent of x.2) Find the coefficient of x6.Detailed Solution:1) General term formula: Tr+1 = nCr (1st)n-r (2nd)r.Here n = 9 1st = x2 2nd = -2x-1:Tr+1 = 9Cr (x2)9-r (-2x-1)r = 9Cr (-2)r x18 – 2r x-r = 9Cr (-2)r x18 – 3r.For the term independent of x set the power of x equal to zero:18 – 3r = 0 ⇒ 3r = 18 ⇒ r = 6.Thus the term independent of x is T6+1 = T7:T7 = 9C6 (-2)6 x0 = (9 × 8 × 7) / (3 × 2 × 1) × 64 = 84 × 64 = 5376.2) For the coefficient of x6 set 18 – 3r = 6:3r = 12 ⇒ r = 4.Term is T5 = 9C4 (-2)4 x6 = (9 × 8 × 7 × 6) / (4 × 3 × 2 × 1) × 16 · x6 = 126 × 16 · x6 = 2016 x6.Coefficient of x6 = 2016	expansiondescendingpowersFind

Comparison aspect	Main idea	Keywords
•	💡 Master Teacher Insights & Critical Exam Cautions	MasterTeacherInsightsCritical
★ Exam Trap	If the second term has a negative sign (x – a)n do not forget (-1)r inside the general term! Odd r terms will be negative even r terms positive	secondtermnegativesign
★ Number of Terms	The expansion of (x + a)n always contains (n + 1) terms	expansionalwayscontains
★ Sum of Coefficients	To find the sum of all coefficients in (2x – 3y)5 simply substitute x = 1 y = 1 ⇒ (2(1) – 3(1))5 = (-1)5 = -1	findsimplysubstitute
•	🌍 Practical Real-World Industrial & Societal Connections	PracticalRealWorldIndustrial
✔ Cryptography & Cybersecurity	Combinatorics and binomial distributions form the cryptographic backbone of RSA public-key encryption and password entropy calculations	Combinatoricsbinomialdistributionsform

✔ Bioinformatics & Gene Sequencing→Combinatorial permutations determine the number of possible nucleotide base permutations in genetic DNA sequencing.

Lesson 1: Counting Principle, Permutations (nPr) and Combinations (nCr) · 15 / 16

Lesson summary

Comparison aspect	Main idea	Keywords
•	★ High-Yield Summary Mind Maps & Comparison Tables ★	HighYieldSummaryMind
Summary	Fundamental Principle of Counting Permutations Combinations and the Binomial Theorem	FundamentalPrincipleCountingPermutations
•	🗺️ Core Concept Map & Knowledge Architecture	CoreConceptKnowledgeArchitecture
✔ Counting	Multiplication (m × n) \| Addition (m + n)	MultiplicationAddition
✔ Permutations vs Combinations	nPr (order matters) vs nCr (order does not matter)	ordermattersdoesmatter
✔ Binomial Expansion	(x+a)n = Σ nCr xn-r ar \| Tr+1 = nCr xn-r ar
✔ Key Terms	Independent of x (power = 0) \| Middle term (n/2 + 1 if even)	IndependentpowerMiddleeven
•	📊 Classification & Comparative Summary Tables	ClassificationComparativeSummaryTables
Table	Comparison Between Permutations and Combinations	ComparisonBetweenPermutationsCombinations

Criterion	Permutations (ⁿP_r)	Combinations (ⁿC_r)
Order Relevance	Order is strictly significant (AB ≠ BA)	Order is completely irrelevant (AB = BA)
Formula	ⁿP_r = (n!) / ((n-r)!)	$^nC_r = (n!) / (r!(n-r)!) = ⁿP_r / (r!)
Relation	ⁿP_r gt; ⁿC_r (for r gt; 1)	ⁿC_r = ⁿP_r / (r!) ≤ ⁿP_r
Standard Scenarios	Passwords line arrangements race podium finishes	Committees card hands geometric polygons and lines

💡 Exam Flashcard Capsules & High-Yield Secrets:
★ ⁿC₀ = ⁿC_n = 1 and ⁿC₁ = n.
★ ⁿP_n = n! and ⁿP₀ = 1.
★ Ratio of consecutive terms (Tr+1) / Tr = (n-r+1) / r · 2nd / 1st.
★ If (x+a)ⁿ has an odd power n, it has two middle terms.

Lesson 1: Counting Principle, Permutations (nPr) and Combinations (nCr) · 16 / 16

Curriculum

Mathematics

Unit 1

Unit 2

Unit 3

Unit 4

Unit 5

Unit 6

Unit 7

Lesson 1: Counting Principle, Permutations (nPr) and Combinations (nCr)

🌳 Concept tree

│ ├── Counting principle

│ ├── Permutations (nPr)

│ └── Combinations (nCr)

│ ├── Patterns in the expansions

│ ├── Pascal's triangle

│ ├── Sum of a row and the number of subsets

│ ├── The binomial theorem

│ └── Special cases

├── 1-1 (b) The general term

│ ├── Rule

│ ├── Term counted from the end

│ ├── Sum and difference of two expansions

│ ├── Recognising an expansion in disguise

│ └── Finding unknowns from coefficients

├── 1-1 (c) The middle term

│ ├── Rule

│ └── Middle term of a sum of two expansions

│ ├── Method

│ └── Equal middle terms

│ └── Ratio of terms

└── 1-3 (b) The greatest term

├── Method

Factorial notation

Permutations or combinations?

Counting principle

✍️ Worked examples

Patterns in the expansions

Pascal’s triangle

Sum of a row and the number of subsets

The binomial theorem

✍️ Worked examples

Special cases

✍️ Worked examples

Rule

✍️ Worked examples

✍️ Worked examples

✍️ Worked examples

✍️ Worked examples

Rule

✍️ Worked examples

Method

✍️ Worked examples

Equal middle terms

✍️ Worked examples

Ratio of terms

✍️ Worked examples

Method

✍️ Worked examples

Using the binomial expansion to prove relations

🔬 binomial_theorem_and_pascal_triangle

(21) The eequence (T,) = (| Win’ 4 1)) is

(ayn (b)n? (c)n’ ~§ (dy ‘4

(24) The sequence (1 +467 11 +.) is

(25) |.) The sequence (3 +5 +7 +96… +21) is

🔍 In-Depth Explanations, Deductions & Multi-Book Insights:

Mathematics Thanaweya Amma (2026/2027)

🌳 Hierarchy Classification & Concept Tree

Combinatorial Foundations

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