Permutations and Combinations — MHT-CET Maths Formulas and Traps

Fundamental Principle, nPr and nCr — Definitions and Identities

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}

Common traps

Adding stages

{}^4P_2 + {}^6P_3

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

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.

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.

Arrangements with Constraints — Together, Never Together, Fixed Positions and Repeated Letters

Learn this subtopic in the notes

Repeated Letters: Divide n! by k! for Each Repeat

Permutations of a multiset

\frac{n!}{p!\,q!\,r!\cdots}

Fixed Positions: Fill the Constrained Slots First, Then the Rest

Constrained slots first

\text{ways} = (\text{fill constrained slots}) \times (\text{arrange what remains})

Together: Glue the Group Into One Block, Then Arrange Inside It

Block method

(\text{blocks})! \times \prod (\text{size of each block})!

Never Together: Total Minus Together, or Place Them in the Gaps

Two routes to 'apart'

\text{apart} = \text{total} - \text{together} \qquad \text{or} \qquad (\text{others})! \times {}^{k+1}P_r \ (\text{or } {}^{k+1}C_r \text{ if identical})

Common traps

Counting the letters wrong

Every word question is lost at the census. CALCULATE is nine letters, not eight; BARRACK has two R's AND two A's. Write the tally before any factorial.

Assuming four adjacent pairs remain

After

B_1

takes seat

2

, seats

1

and

3

are no longer adjacent. The count of adjacent pairs is

2

, not

3

or

4

; option (B)

12

is what

3

pairs would give.

Forgetting the inside of the block

4! = 24

counts the blocks only; without

3! \times 4!

inside, the answer is

24

instead of

3456

. Conversely, a block of identical letters has NO inside factor.

Using nPr for identical items in gaps

Two identical M's into

5

gaps is

{}^5C_2 = 10

, not

{}^5P_2 = 20

. The doubled answer

80

is option (C) on the MANAMA stem.

Selections with Conditions — At Least, At Most, Included and Excluded

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At Least and At Most: List the Cases and Add

Casework

\sum_{\text{cases}} {}^{a}C_{i}\,{}^{b}C_{k-i}

The Complement: Unrestricted Total Minus the Forbidden Selections

Complement

\text{allowed} = \text{total} - \text{forbidden}

Select the Team, Then Choose the Captain: Multiply by the Team Size

Team then role

(\text{number of teams}) \times (\text{team size})

Common traps

Dropping a case

'At least

2

from each of two sections when choosing

6

' has THREE cases:

(2,4)

,

(3,3)

,

(4,2)

. Missing

(4,2)

gives

275

, which is option (B).

Subtracting the wrong thing

For '

A

and

B

not together', subtract committees with BOTH — not committees with either. Subtracting 'contains

A

' and 'contains

B

' removes too much.

Stopping at the team count

95

is offered as option (A) on the linguistic-club stem. The captain multiplies it by

4

.

Circular Arrangements

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(n − 1)! Around a Table, and Girls Apart via the Gaps Between Boys

Circular permutations

(n - 1)! \qquad \text{girls apart: } (b - 1)! \times {}^{b}P_{g}

Never Together Around a Table: Total Minus the Glued Block

Glued block on a circle

(n - 1)! - (n - k)!\,k!

Alternating Seats With a Couple Together, and Colouring a Circle

Cycle colourings

P(C_n, k) = (k-1)^n + (-1)^n (k-1) \qquad P(C_5, 3) = 32 - 2 = 30

Common traps

Using n + 1 gaps on a circle

Six boys in a ROW make

7

gaps; around a TABLE they make

6

.

{}^7P_5

instead of

{}^6P_5

is how the wrong options are built.

Gluing on a circle with row arithmetic

A block among

10

people on a circle leaves

8

units, seated in

7!

ways, not

8!

. Using

8! \times 3!

subtracts too much and lands on a non-option.

Treating the parents' block as a row block

In the alternating stem the parents' block has no free

2!

— the alternation decides which side the father sits. The official answer is

4! \times 4! = 576

;

1152

is the doubled count.

Counting Numbers and Geometric Figures — Digits, Divisibility, Points and Polygons

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Digit Counting: No Leading Zero, and Divisibility by the Last Digits or the Digit Sum

Digit rules

3 \mid n \iff 3 \mid \text{digit sum} \qquad 25 \mid n \iff \text{last two digits} \in \{00, 25, 50, 75\} \qquad \text{leading digit} \ne 0

Inclusion–Exclusion on Multiples: gcd Conditions

Inclusion–exclusion

|A \cup B| = |A| + |B| - |A \cap B|

Handshakes, Diagonals and Triangle Counts: Solve an nC2 or nC3 Equation

Pairs and triples

\text{handshakes} = {}^nC_2 \qquad \text{diagonals} = \frac{n(n-3)}{2} \qquad {}^{n+1}C_3 - {}^nC_3 = {}^nC_2

Triangles and Quadrilaterals From Points: Subtract the Collinear Choices

Degenerate subtraction

\triangle = {}^nC_3 - {}^mC_3 \qquad \text{no-side triangles of an } n\text{-gon} = \frac{n(n-4)(n-5)}{6}

Common traps

Keeping the leading zero

\{0,1,2,4,5\}

gives

96

numbers, not

120

; the total is

216

, and

240

— option (B) — is the count that forgot the zero.

Forgetting the add-back

Subtracting the multiples of

4

and of

6

without adding back the multiples of

12

gives

75

, not

150

. Every two-condition removal has an intersection to restore.

Counting the sides as diagonals

{}^nC_2

includes the

n

sides. Setting

{}^nC_2 = 54

has no integer solution; the diagonals are

{}^nC_2 - n

.

Subtracting only the all-collinear picks

For quadrilaterals,

3

collinear points plus any fourth is ALSO degenerate.

330 - 5 = 325

is option (D); the answer is

265

.

Permutations and Combinations formulas

Fundamental Principle, nPr and nCr — Definitions and Identities

The Fundamental Principle: Multiply Stages, Add Alternatives

nPr and nCr: Ordered Versus Unordered, and When They Are Defined

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

Common traps

Arrangements with Constraints — Together, Never Together, Fixed Positions and Repeated Letters

Repeated Letters: Divide n! by k! for Each Repeat

Fixed Positions: Fill the Constrained Slots First, Then the Rest

Together: Glue the Group Into One Block, Then Arrange Inside It

Never Together: Total Minus Together, or Place Them in the Gaps

Common traps

Selections with Conditions — At Least, At Most, Included and Excluded

At Least and At Most: List the Cases and Add

The Complement: Unrestricted Total Minus the Forbidden Selections

Select the Team, Then Choose the Captain: Multiply by the Team Size

Common traps

Circular Arrangements

(n − 1)! Around a Table, and Girls Apart via the Gaps Between Boys

Never Together Around a Table: Total Minus the Glued Block

Alternating Seats With a Couple Together, and Colouring a Circle

Common traps

Counting Numbers and Geometric Figures — Digits, Divisibility, Points and Polygons

Digit Counting: No Leading Zero, and Divisibility by the Last Digits or the Digit Sum

Inclusion–Exclusion on Multiples: gcd Conditions

Handshakes, Diagonals and Triangle Counts: Solve an nC2 or nC3 Equation

Triangles and Quadrilaterals From Points: Subtract the Collinear Choices

Common traps

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