NDA Maths · Permutation & Combination

Combinations & Selections

Selecting a group where order doesn't matter — straight nCr selections, subsets, and the constrained selections (at-least / at-most, compulsory members, complementary counting).

All Permutation & Combination notes NDA Maths strategy

Why this matters

Selection problems hinge on spotting that order is irrelevant. The high-value trick is complementary counting — 'at least one' is total minus none — which beats summing cases.

Concept 1 of 2

Selections and subsets

Intuition

A combination counts groups, not orders. Choosing

r

from

n

^nC_r

; the number of subsets of an

n

-set is

2^n

. Fixing compulsory members just reduces the pool and the number to choose.

Definition

Choose $r$ of $n$ (order irrelevant): $^nC_r$ . Subsets of an $n$ -element set: $2^n$ . Compulsory members: if $k$ must be included, choose the rest: $^{n-k}C_{r-k}$ . Probabilities use $\dfrac{\text{favourable }C}{\text{total }C}$ .

Worked example

In how many ways can a committee of 3 be chosen from 8 people?

Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 1Permutation & CombinationEASY

In how many ways can a team of 5 players be selected from 8 players so as not to include a particular player?

[Q44 · Apr · 2021]

Use $^nC_r=\,^nC_{n-r}$ — and count the empty set

Symmetry

^nC_r=\,^nC_{n-r}

means choosing

r

to *keep* equals choosing

n-r

to *leave out* — so compute the easier one (

^{50}C_{48}=\,^{50}C_2

). Separately, the number of subsets of an

n

-set is

2^n

, which includes the empty set;

2^n-1

is the count of *non-empty* subsets only.

Drill 5 more on selections and subsets

Concept 2 of 2

Constrained selection: at-least, at-most, cases

Intuition

When a selection must satisfy 'at least one of X', it's almost always faster to count the complement (total minus the forbidden 'none') than to sum cases. Genuinely multi-part constraints split into disjoint cases that add.

Definition

At-least-one: $\text{total}-\text{none}$ . **At-most $k$ :** sum $^nC_0+\cdots+^nC_k$ . Cases: when the constraint forces distinct sub-situations (e.g. 'choose 3 from 4 women + 3 men with a balance rule'), count each disjoint case and add. Watch for over-/under-counting at the boundaries.

Worked example

From 6 programmers and 4 typists, choose 5 with at least one typist. How many ways?

Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 2Permutation & CombinationHARD

A man has 7 relatives (4 women and 3 men). His wife also has 7 relatives (3 women and 4 men). In how many ways can they invite 3 women and 3 men so that 3 of them are the man's relatives and 3 are his wife's relatives?

[Q5 · Apr · 2024]

'At least one' = total − none (don't sum cases)

For 'at least one X', count the complement:

\text{total}-(\text{none of X})

. Summing the cases 'exactly 1, exactly 2, …' is slower and easy to miscount. E.g. at least one typist when choosing 5 from 6 programmers + 4 typists is

^{10}C_5-\,^6C_5=252-6=246

, not a sum of four separate terms.

Drill 4 more on constrained selection: at-least, at-most, cases

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Watch out for (2)

Use $^nC_r=\,^nC_{n-r}$ — and count the empty set
→ Selections and subsets
'At least one' = total − none (don't sum cases)
→ Constrained selection: at-least, at-most, cases

Drill every past-year question on this subtopic

11 questions from the bank — paginated, with cart and Word-export support.

Drill the 11 questions

Drill every past-year question on this subtopic

Related notes