MHT-CET Maths · Permutations and Combinations

Fundamental Principle, nPr and nCr — Definitions and Identities

Multiply the choices of successive stages; nPr counts ordered selections and nCr unordered ones — and the identities nCr + nCr−1 = n+1Cr and nCr = nCn−r settle the equation-style stems.

All Permutations and Combinations notes MHT-CET Maths strategy

Why this matters

12 PYQs, one HARD — the cheapest page in the chapter and the one the other four assume. The recurring stems are a two-stage arrangement written as a product of nPr terms, a ratio of two binomial coefficients set equal to a number, a Pascal-rule inequality, and the domain of a function defined through nPr. Each is answered by one identity; the work is choosing which.

Concept 1 of 3: The Fundamental Principle: Multiply Stages, Add Alternatives

If a task is done in stages and the first stage has

m

ways and the second

n

ways (whatever happened first), the task has

mn

ways. If instead the task is done in one of two mutually exclusive ways, add.

Definition

Multiplication: stages in sequence — women choose $2$ of chairs $1$ – $4$ in ${}^4P_2$ ways, THEN men choose $3$ of the remaining $6$ in ${}^6P_3$ ways: ${}^4P_2 \times {}^6P_3$ .
Addition: alternatives that cannot both happen — a number ending in $25$ OR ending in $75$ .
The test for multiplication: does the count of the second stage depend on WHICH first-stage choice was made? If not, multiply. (The men's count is $6$ chairs whichever $2$ the women took.)
'Select and then arrange' is two stages: ${}^7C_3 \times {}^4C_2$ picks $3$ consonants and $2$ vowels, then $5!$ arranges the five letters chosen: $35 \times 6 \times 120 = 25{,}200$ .

Fundamental principle

\text{stages: } m \times n \qquad \text{alternatives: } m + n

Worked example

A password is one letter from

\{A, B, C\}

followed by two different digits from

\{1, 2, 3, 4, 5\}

. How many passwords are there?

Practise this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2023 · 16th May Shift 1 · Q132Moderate

Example 1 · Permutations and Combinations · Fundamental Principle, nPr and nCr — Definitions and Identities

Eight chairs are numbered 1 to 8. Two women and three men wish to occupy one chair each. First the women choose chairs from amongst the chairs marked 1 to 4, and then the men select the chairs from amongst the remaining. The number of possible arrangements is

Adding stages

{}^4P_2 + {}^6P_3

is offered as an option. Stages that BOTH happen multiply; only mutually exclusive alternatives add.

Drill 1 more on the fundamental principle: multiply stages, add alternatives

Concept 2 of 3: nPr and nCr: Ordered Versus Unordered, and When They Are Defined

{}^nC_r

picks

r

n

with no regard to order;

{}^nP_r = {}^nC_r \cdot r!

then lines them up. Both need

0 \le r \le n

with

n, r

whole numbers — which is itself a question when

n

and

r

are expressions in

x

Definition

${}^nP_r = \dfrac{n!}{(n - r)!}$ , ${}^nC_r = \dfrac{n!}{r!\,(n - r)!}$ , ${}^nP_r = r!\,{}^nC_r$ .
Defined only for whole numbers $n \ge r \ge 0$ . The domain of $f(x) = {}^{7 - x}P_{x - 1}$ : $x - 1 \ge 0$ , $7 - x \ge x - 1$ , $x$ an integer — so $x \in \{1, 2, 3, 4\}$ , a four-element set, not an interval.
'Always include $6$ , always exclude $5$ ' from $25$ for a team of $11$ : the pool shrinks to $25 - 6 - 5 = 14$ and the choice to $11 - 6 = 5$ : ${}^{14}C_5 = 2002$ .
Small values worth knowing cold: ${}^nC_2 = \dfrac{n(n-1)}{2}$ , ${}^nC_3 = \dfrac{n(n-1)(n-2)}{6}$ , ${}^nC_1 = n$ , ${}^nC_0 = {}^nC_n = 1$ .

Definitions

{}^nP_r = \frac{n!}{(n-r)!} \qquad {}^nC_r = \frac{n!}{r!\,(n-r)!} \qquad {}^nP_r = r!\,{}^nC_r \qquad (0 \le r \le n)

Worked example

Find the domain of

f(x) = {}^{9 - x}C_{x - 2}

Practise this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2025 · 19 April Shift II · Q136Moderate

Example 2 · Permutations and Combinations · Fundamental Principle, nPr and nCr — Definitions and Identities

The domain of the function

f(x) =\ ^{7 -x}P_{x- 1}

Reporting an interval as the domain of nPr

1 \le x \le 4

is right as an inequality but the domain is the INTEGERS in it,

\{1, 2, 3, 4\}

. Option (A) '

\mathbb{R}

' and option (B) '

\mathbb{R} - \{1\}

' exist for readers who forget

n

and

r

must be whole numbers.

Drill 1 more on npr and ncr: ordered versus unordered, and when they are defined

Concept 3 of 3: Identities: Pascal's Rule, Symmetry and the Ratio of Consecutive Coefficients

Three identities cover every equation-style stem:

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

(choosing

r

to keep is choosing

n - r

to drop), Pascal's

{}^nC_r + {}^nC_{r-1} = {}^{n+1}C_r

, and the ratio

\dfrac{{}^nC_r}{{}^nC_{r-1}} = \dfrac{n - r + 1}{r}

Definition

Symmetry: ${}^{n+4}C_{n+1} = {}^{n+4}C_3$ . Rewrite a large lower index as $n - r$ before doing anything else.
Pascal: ${}^{n+4}C_3 - {}^{n+3}C_3 = {}^{n+3}C_2 = \dfrac{(n+3)(n+2)}{2}$ . Set equal to $15(n+2)$ : $n + 3 = 30$ , $n = 27$ . Also ${}^nC_4 + {}^nC_5 = {}^{n+1}C_5$ .
Ratio: $\dfrac{{}^{n+1}C_5}{{}^{n+1}C_4} = \dfrac{n + 1 - 4}{5} = \dfrac{n - 3}{5}$ ; ' $> 1$ ' gives $n > 8$ .
Greatest coefficient: ${}^nC_r$ is largest at $r = \dfrac{n}{2}$ ( $n$ even) or at both $r = \dfrac{n \pm 1}{2}$ ( $n$ odd). $\dfrac{{}^{10}C_5}{{}^{11}C_6} = \dfrac{252}{462} = \dfrac{6}{11}$ .
Two particular items: chosen together in ${}^{n-2}C_{r-2}$ ways, both left out in ${}^{n-2}C_r$ ways. Ratio $2 : 3$ with $r = 5$ : $\dfrac{{}^{n-2}C_3}{{}^{n-2}C_5} = \dfrac{5 \cdot 4}{(n-5)(n-6)} = \dfrac23 \Rightarrow (n-5)(n-6) = 30 \Rightarrow n = 11$ .

The three identities

{}^nC_r = {}^nC_{n-r} \qquad {}^nC_r + {}^nC_{r-1} = {}^{n+1}C_r \qquad \frac{{}^nC_r}{{}^nC_{r-1}} = \frac{n - r + 1}{r}

Worked example

{}^{n}C_{4} = {}^{n}C_{6}

, find

{}^{n}C_{2}

Practise this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2025 · 21 April Shift II · Q133Moderate

Example 3 · Permutations and Combinations · Fundamental Principle, nPr and nCr — Definitions and Identities

\ ^{n+ 4}C_{n+ 1}-\ ^{n+ 3}C_{n}= 15(n+ 2)

, then

n=

Expanding the factorials

{}^{n+4}C_{n+1} - {}^{n+3}C_n

written out in factorials is a swamp. Symmetry turns the lower indices into

3

, Pascal collapses the difference to one term, and the equation is quadratic-free.

Drill 3 more on identities: pascal's rule, symmetry and the ratio of consecutive coefficients

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.

Formulas (3)

The Fundamental Principle: Multiply Stages, Add Alternatives
Fundamental principle
$\text{stages: } m \times n \qquad \text{alternatives: } m + n$
nPr and nCr: Ordered Versus Unordered, and When They Are Defined
Definitions
${}^nP_r = \frac{n!}{(n-r)!} \qquad {}^nC_r = \frac{n!}{r!\,(n-r)!} \qquad {}^nP_r = r!\,{}^nC_r \qquad (0 \le r \le n)$
Identities: Pascal's Rule, Symmetry and the Ratio of Consecutive Coefficients
The three identities
${}^nC_r = {}^nC_{n-r} \qquad {}^nC_r + {}^nC_{r-1} = {}^{n+1}C_r \qquad \frac{{}^nC_r}{{}^nC_{r-1}} = \frac{n - r + 1}{r}$

Watch out for (3)

Adding stages
→ The Fundamental Principle: Multiply Stages, Add Alternatives
Reporting an interval as the domain of nPr
→ nPr and nCr: Ordered Versus Unordered, and When They Are Defined
Expanding the factorials
→ Identities: Pascal's Rule, Symmetry and the Ratio of Consecutive Coefficients

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Related notes

Selections with Conditions — the moves that use nCr with at-least and at-most →