Is Permutations and Combinations part of the EAPCET Mathematics syllabus?

Yes. Permutations and Combinations is taught in Intermediate second year · Paper IIA Mathematics, and the EAPCET Engineering syllabus is based on the Intermediate first- and second-year syllabus. Check the official notification each year for any changes.

How many Mathematics questions are there in EAPCET?

80 of the 160 questions in the Engineering paper are Mathematics, spread across 37 chapters.

What mistakes do students make in Permutations and Combinations?

Using n! for people around a table; circular arrangements of n distinct people number (n − 1)!. Dividing by 2 for seating arrangements; the extra division applies only when clockwise and anticlockwise orders are the same, as with garlands and necklaces. Forgetting to multiply by the internal arrangements of a block of objects that must stay together. Using ⁿPᵣ for committees or selections; when order does not matter, use ⁿCᵣ.

EAPCET Mathematics · Intermediate second year · Paper IIA

Permutations and Combinations

Permutations and Combinations is the mathematics of counting — arrangements where order matters and selections where it does not.

Subject: Mathematics
Taught in: Intermediate second year · Paper IIA
Mathematics in the paper: 80 of 160 questions

Key formulas and results

${}^{n}P_{r} = \dfrac{n!}{(n - r)!}$ and ${}^{n}C_{r} = \dfrac{n!}{r!\,(n - r)!}$
${}^{n}C_{r} = {}^{n}C_{n-r}$ and ${}^{n}C_{r} + {}^{n}C_{r-1} = {}^{n+1}C_{r}$
$n$ distinct objects in a circle: $(n - 1)!$ arrangements; a necklace or garland: $\dfrac{(n - 1)!}{2}$
$n$ objects with $p, q, r$ alike: $\dfrac{n!}{p!\,q!\,r!}$ arrangements
Selections of at least one from $n$ distinct objects: $2^{n} - 1$
$\dfrac{{}^{n}P_{r}}{{}^{n}P_{r-1}} = n - r + 1$ and ${}^{n}P_{r} = n \cdot {}^{n-1}P_{r-1}$
${}^{n}P_{r} + r \cdot {}^{n}P_{r-1} = {}^{n+1}P_{r}$
Arrangements of $n$ distinct things $r$ at a time with repetition allowed: $n^{r}$
Selections of any number (including none) from $n$ identical things: $n + 1$
Selections of at least one from $p$ alike, $q$ alike and $r$ alike things: $(p + 1)(q + 1)(r + 1) - 1$ ; with $n$ further distinct things: $(p + 1)(q + 1)(r + 1) \cdot 2^{n} - 1$
Number of diagonals of an $n$ -sided polygon: ${}^{n}C_{2} - n$
Derangements (no object in its own place): $D_{n} = n!\left(1 - \dfrac{1}{1!} + \dfrac{1}{2!} - \dfrac{1}{3!} + \dots + \dfrac{(-1)^{n}}{n!}\right)$ ; $D_{2} = 1$ , $D_{3} = 2$ , $D_{4} = 9$ , $D_{5} = 44$ , $D_{6} = 265$

What this chapter covers

Fundamental principle of counting
Permutations with and without repetition
Circular permutations
Permutations of objects that are alike
Combinations and their properties
Rank of a word in dictionary order

How it is usually asked

Count arrangements with restrictions (together, never together)
Find the rank of a word in dictionary order
Count committees or selections with conditions
Count circular seatings

Understanding Permutations and Combinations

Counting starts with the fundamental principle: if one task can be done in m ways and a second in n ways, both together can be done in mn ways. A permutation is an arrangement where order matters, and ⁿPᵣ = n!/(n − r)! counts arrangements of r objects chosen from n distinct ones. A combination is a selection where order does not matter, and ⁿCᵣ = n!/[r!(n − r)!].

Restrictions change the count in standard ways. Objects that must stay together are tied into one block, which is arranged with the rest and then internally. Objects that must never be together are counted as all arrangements minus those where they are together, or by placing them in gaps. Identical objects divide the count: n objects with p alike and q alike give n!/(p! q!) arrangements.

Arrangements in a circle have no starting point, so n distinct objects give (n − 1)! circular arrangements, and (n − 1)!/2 when clockwise and anticlockwise look the same, as with a necklace. EAPCET questions combine these ideas in seating, committee, word-arrangement and dictionary-rank problems, and the first decision is always whether order matters.

Worked example

EAPCET-style question

Six people sit around a round table. The number of arrangements in which two particular people never sit next to each other is

(1)48
(2)72
(3)96
(4)120

Solution

Total circular arrangements of 6 people $= (6 - 1)! = 120$ .
Arrangements with the two together: treat them as one unit, giving $(5 - 1)! = 24$ circular arrangements, times 2 for their internal order $= 48$ .
Never together $= 120 - 48 = 72$ .

Answer: 72

Shortcuts and results to remember

Gap method: to keep $m$ particular things apart among $n$ , arrange the other $n - m$ first, then place the $m$ in the gaps: $(n - m)! \times {}^{n-m+1}P_{m}$ .
Together: tie the $m$ things into one block: $(n - m + 1)! \times m!$ ; not all $m$ together: $n! - (n - m + 1)! \times m!$ .
Number of anagrams of a word = (total arrangements of its letters) − 1.
For ‘neither begins with X nor ends with Y’ counts, use inclusion–exclusion: total $- n(A) - n(B) + n(A \cap B)$ .

Common mistakes to avoid

Using n! for people around a table; circular arrangements of n distinct people number (n − 1)!.
Dividing by 2 for seating arrangements; the extra division applies only when clockwise and anticlockwise orders are the same, as with garlands and necklaces.
Forgetting to multiply by the internal arrangements of a block of objects that must stay together.
Using ⁿPᵣ for committees or selections; when order does not matter, use ⁿCᵣ.

How to prepare it

Decide first whether order matters — that single choice separates permutations from combinations.

For "at least" conditions, count everything and subtract the cases that break the rule.

Once the formulas feel familiar, test them the way the exam does: practise Permutations and Combinations questions chapter-wise, then attempt a full previous year paper to see whether it holds up under a 3-hour timer.

Related Mathematics chapters

See every chapter in the EAPCET Mathematics syllabus.

Practise Permutations and Combinations chapter-wise

Chapter-wise quizzes show exactly which topics in this chapter you still get wrong — before a full paper does.

Frequently asked questions

Is Permutations and Combinations part of the EAPCET Mathematics syllabus?: Yes. Permutations and Combinations is taught in Intermediate second year · Paper IIA Mathematics, and the EAPCET Engineering syllabus is based on the Intermediate first- and second-year syllabus. Check the official notification each year for any changes.
How should I prepare Permutations and Combinations for EAPCET?: Decide first whether order matters — that single choice separates permutations from combinations. For "at least" conditions, count everything and subtract the cases that break the rule.
How many Mathematics questions are there in EAPCET?: 80 of the 160 questions in the Engineering paper are Mathematics, spread across 37 chapters.
What mistakes do students make in Permutations and Combinations?: Using n! for people around a table; circular arrangements of n distinct people number (n − 1)!. Dividing by 2 for seating arrangements; the extra division applies only when clockwise and anticlockwise orders are the same, as with garlands and necklaces. Forgetting to multiply by the internal arrangements of a block of objects that must stay together. Using ⁿPᵣ for committees or selections; when order does not matter, use ⁿCᵣ.