Permutations, Combinations and the Binomial Theorem

Counting problems are the ones candidates most often get wrong while being certain they are right, and the reason is almost always the same: the arrangement was counted when only the selection mattered, or the other way round. So the first question to ask of any counting problem is does the order matter? — permutation if yes, combination if no. The second is whether repetition is allowed. Answer those two before writing anything and most of this chapter becomes routine. The binomial theorem is then a direct application of combinations, which is why the two topics belong together.

Permutations and Combinations

Which formula for which situation
Situation	Count
Arrange r of n distinct objects	ⁿPᵣ = n!/(n − r)!
Select r of n distinct objects	ⁿCᵣ = n!/(r!(n − r)!)
Arrange r of n with repetition allowed	nʳ
Arrange n objects with p and q alike	n!/(p! q!)
Arrange n distinct objects in a circle	(n − 1)!, or (n−1)!/2 if reflections coincide
Distribute r identical objects into n boxes	^r+n−1C_n−1

🎯 Why a circular arrangement gives (n − 1)! and not n!

Around a circle there is no first seat, so rotating everybody one place to the left produces an arrangement that is not genuinely different. Each linear arrangement therefore has n rotations that all describe the same circular seating, and dividing n! by n gives (n − 1)!. Two refinements matter. If the circle can be flipped over — a garland of flowers, or a necklace, where clockwise and anticlockwise look the same — divide by 2 again to get (n − 1)!/2. But if the positions are distinguishable, as around a round table with numbered chairs, the rotations are different after all and the answer returns to n!. So the question always turns on whether the seats can be told apart.

🧠 Gaps for "no two together", and grouping for "always together"

Two constraints cover most arrangement questions, and each has a standard device. For "always together", tie the required objects into one block, arrange the resulting units, then multiply by the internal arrangements of the block — so the number of ways to seat 5 people with two specific ones adjacent is 4! × 2!. For "no two together", arrange the unrestricted objects first and then slot the restricted ones into the gaps between them: with 5 boys and 3 girls where no two girls are adjacent, arrange the boys in 5! ways, which creates 6 gaps, and choose 3 of those in ⁶C₃ ways with 3! orderings. Recognising which of the two devices applies is usually the whole difficulty of the question.

Three identities are worth having instantly: ⁿCᵣ = ⁿC_n−r, which halves the work whenever r is large; ⁿCᵣ + ⁿC_r−1 = ⁿ⁺¹Cᵣ, Pascal's rule, which is why the triangle works; and the total number of subsets, ΣⁿCᵣ = 2ⁿ. Note also that ⁿCᵣ is largest at the middle value of r, so the biggest term in a row of Pascal's triangle is the central one.

The Binomial Theorem

The expansion (x + a)ⁿ = Σ ⁿCᵣ x^n−r aʳ has n + 1 terms, and the general term is T_r+1 = ⁿCᵣ x^n−r aʳ. Almost every question is answered by writing that general term, imposing whatever condition is asked — a particular power of x, independence from x, a specific coefficient — and solving for r. Writing the general term first is the single most reliable move in this section.

Standard binomial questions and how to attack them
Asked for	Method
Coefficient of a given power of x	Set the exponent of x in T_r+1 equal to it and solve for r
Term independent of x	Set the exponent of x to zero
Middle term	n even gives one middle term; n odd gives two
Number of terms in (x + a)ⁿ	n + 1
Sum of the coefficients	Put x = 1 and a = 1, giving 2ⁿ
Sum of coefficients with alternating signs	Put x = 1, a = −1, giving 0 — so odd and even sums are equal

⚠️ The greatest coefficient and the greatest term are different questions

The greatest coefficient in (1 + x)ⁿ is the central binomial coefficient, ⁿC_n/2 for even n, and it does not depend on x at all. The greatest term in a numerical expansion does depend on x, because the powers of x can grow faster or slower than the coefficients shrink; you find it by computing the ratio T_r+1/T_r and locating where it crosses 1. So "find the greatest coefficient in (1 + 2x)¹⁰" and "find the greatest term in (1 + 2x)¹⁰ when x = 3" have different answers and different methods. Reading which of the two is being asked, before starting, saves an entire wasted calculation.

For a negative or fractional index the expansion (1 + x)ⁿ becomes an infinite series and is valid only when |x| < 1, which is a condition worth stating explicitly whenever you use it. The most useful special cases are (1 − x)⁻¹ = 1 + x + x² + … and (1 − x)⁻² = 1 + 2x + 3x² + …, and both appear in problems on summing series. Note that here the binomial coefficients are no longer ⁿCᵣ in the factorial sense, since n is not a positive integer.

Key takeaways

Ask first whether order matters — permutation if yes, combination if no — then whether repetition is allowed.
A circular arrangement gives (n − 1)!, halved again if the circle can be flipped, but n! if the seats are numbered.
"Always together" means group and multiply; "no two together" means arrange the rest and use the gaps.
ⁿCᵣ = ⁿC_n−r halves the arithmetic, and ⁿCᵣ + ⁿC_r−1 = ⁿ⁺¹Cᵣ is why Pascal's triangle works.
Write the general term T_r+1 = ⁿCᵣ x^n−r aʳ first — nearly every binomial question follows from it.
(x + a)ⁿ has n + 1 terms; n odd gives two middle terms, n even gives one.
The greatest coefficient is independent of x; the greatest term is not — check which is asked.
For a negative or fractional index the expansion is infinite and needs |x| < 1.

Practice questions (8)

Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.

The number of ways to seat 6 people around a round table is:
1. 720
2. 120
3. 60
4. 36
Show answer
Answer: B — 120
Around a circle there is no distinguished first seat, so the n rotations of any linear arrangement all describe the same seating; dividing 6! = 720 by 6 gives 120 = 5!. The answer 720 would be right only if the chairs were numbered, and 60 would be right for a garland that can also be flipped over — so the phrasing of the question decides which of the three applies.
The number of arrangements of the letters of the word ALLAHABAD is:
1. 9!
2. 9!/(4! 2!)
3. 9!/4!
4. 9!/(4! 2! 2!)
Show answer
Answer: B — 9!/(4! 2!)
The word has 9 letters with A appearing 4 times and L twice, the rest single, so the count is 9!/(4! 2!) = 7560. Dividing by 4! alone ignores the repeated L, and dividing by an extra 2! would assume a third repeated letter that is not there — so the first step in any such question is simply to tally each letter.
In how many ways can 5 boys and 3 girls be seated in a row so that no two girls are together?
1. 5! × 3!
2. 5! × ⁶P₃
3. 8!
4. 8! − 3!
Show answer
Answer: B — 5! × ⁶P₃
Seat the 5 boys first in 5! ways; that creates 6 gaps, including the two ends, and the 3 girls must occupy 3 distinct gaps, which can be done in ⁶P₃ ways since the girls are distinguishable. So the total is 5! × ⁶P₃ = 120 × 120. The option 5! × 3! would apply if the girls all had to sit in one particular block, which is the opposite constraint.
The number of terms in the expansion of (x + y)¹⁵ is:
1. 15
2. 16
3. 30
4. 32
Show answer
Answer: B — 16
The exponent of y runs from 0 to 15 inclusive, giving 16 terms — the count is always n + 1. The answer 15 comes from forgetting the r = 0 term, and 32 is 2ⁿ⁻¹⁰ style confusion with the sum of the coefficients, which is a different quantity: putting x = y = 1 gives 2¹⁵ as the coefficient sum, not the number of terms.
The expansion of (1 + x)ⁿ with n odd has how many middle terms?
1. One
2. Two
3. None
4. It depends on x
Show answer
Answer: B — Two
With n odd the expansion has n + 1 terms, an even count, so there is no single central position and the two terms numbered (n+1)/2 and (n+3)/2 share the middle. When n is even the term count n + 1 is odd and exactly one term sits in the middle. The number of terms, not x, decides this, which is why the last option cannot be right.
The sum of the coefficients in the expansion of (1 + x)¹⁰ is:
1. 10
2. 1024
3. 512
4. 100
Show answer
Answer: B — 1024
Substituting x = 1 makes every power of x equal to 1, so the expansion collapses to the sum of the coefficients, which is 2¹⁰ = 1024. The value 512 is 2⁹, and it is the sum of either the odd-indexed or the even-indexed coefficients alone — putting x = −1 shows those two sums are equal, so each is half the total.
The expansion of (1 + x)⁻² as an infinite series is valid when:
1. For all real x
2. |x| < 1
3. x > 0 only
4. x is an integer
Show answer
Answer: B — |x| < 1
For a negative or fractional index the binomial expansion is an infinite series and converges only for |x| < 1; outside that range the terms grow and the series is meaningless. This restriction does not arise for a positive integral index, where the expansion terminates after n + 1 terms and holds for every x — which is why the condition must be stated whenever the index is not a positive integer.
The number of ways to distribute 7 identical balls into 3 distinct boxes is:
1. 3⁷
2. ⁹C₂ = 36
3. 7!
4. 21
Show answer
Answer: B — ⁹C₂ = 36
Because the balls are identical only the counts per box matter, so the problem is to write 7 as an ordered sum of three non-negative integers; the standard bars-and-stars argument gives ⁷⁺³⁻¹C₃₋₁ = ⁹C₂ = 36. The answer 3⁷ would be right if the balls were distinguishable, since then each ball independently chooses a box — so noticing "identical" rather than "distinct" is the whole question.