MHT-CET Maths · Permutations and Combinations

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

Count numbers by fixing the constrained digit first (no leading zero, last digits for divisibility), and count figures from points by nCr minus the degenerate collinear choices.

All Permutations and Combinations notes MHT-CET Maths strategy

Why this matters

19 PYQs at 26% HARD, and the most mechanical page in the chapter. Digit stems test divisibility by 3 (digit sum), by 25 (last two digits) and the leading-zero exclusion; figure stems are nC2 handshakes and diagonals, nC3 triangles with collinear points removed, and the greatest number of intersections of lines and circles. Two of the HARD outliers — a gcd-with-36 count and triangles using no polygon side — are inclusion–exclusion in disguise.

Concept 1 of 4: Digit Counting: No Leading Zero, and Divisibility by the Last Digits or the Digit Sum

Fix the most constrained digit first. A leading position cannot be

0

; divisibility by

25

fixes the last two digits; divisibility by

3

constrains WHICH digits are used, via their sum, not where they go.

Definition

No leading zero: five-digit numbers from $\{0, 1, 2, 4, 5\}$ without repetition: $5! - 4! = 96$ (subtract those starting with $0$ ), or $4 \times 4!$ directly.
By 3: choose $5$ of $\{0,\dots,5\}$ (sum $15$ ) with digit-sum a multiple of $3$ : drop $0$ or drop $3$ . $\{1,2,3,4,5\}$ : $120$ ; $\{0,1,2,4,5\}$ : $96$ . Total $216$ .
By 25: last two digits $25$ or $75$ from $\{1,\dots,7\}$ ; first two from the remaining $5$ : $2 \times {}^5P_2 = 40$ .
Greater than a million from $2,3,0,3,4,2,3$ : seven digits, so every arrangement not starting with $0$ : $\dfrac{7!}{3!\,2!} - \dfrac{6!}{3!\,2!} = 420 - 60 = 360$ .

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

Worked example

How many four-digit numbers with distinct digits can be formed from

\{0, 1, 2, 3, 4\}

that are divisible by

5

Practise this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2025 · 22 April Shift II · Q141Moderate

Example 1 · Permutations and Combinations · Counting Numbers and Geometric Figures — Digits, Divisibility, Points and Polygons

If four digit numbers are formed by using the digits

1,2,3,4,5,6,7

without repetition, then out of these numbers, the numbers exactly divisible by 25 are

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.

Drill 2 more on digit counting: no leading zero, and divisibility by the last digits or the digit sum

Concept 2 of 4: Inclusion–Exclusion on Multiples: gcd Conditions

'gcd with

36

2

' means divisible by

2

, not by

4

, not by

3

. Count the multiples of

2

in range, then remove those divisible by

4

or by

3

, adding back the ones divisible by both.

Definition

Multiples of $d$ in $[a, b]$ : $\left\lfloor \dfrac{b}{d} \right\rfloor - \left\lfloor \dfrac{a - 1}{d} \right\rfloor$ . Three-digit multiples of $2$ : $450$ ; of $6$ : $150$ ; of $4$ : $225$ ; of $12$ : $75$ .
$\gcd(n, 36) = 2$ : $450 - (150 + 225 - 75) = 150$ .
The template: $|A \cup B| = |A| + |B| - |A \cap B|$ , and the wanted set is the base set minus $|A \cup B|$ .
Translate a gcd condition into 'divisible by these, not by those' before counting; $36 = 2^2\cdot3^2$ is the whole content of the stem.

Inclusion–exclusion

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

Worked example

How many two-digit numbers are divisible by neither

2

nor

5

Practise this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2025 · 19 April Shift II · Q139Hard

Example 2 · Permutations and Combinations · Counting Numbers and Geometric Figures — Digits, Divisibility, Points and Polygons

Total number of 3-digit numbers, whose g.c.d with 36 is 2 , is

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.

Concept 3 of 4: Handshakes, Diagonals and Triangle Counts: Solve an nC2 or nC3 Equation

Every pair of people is one handshake, so

{}^nC_2 = 45

gives

n

. A polygon's diagonals are the pairs of vertices minus the sides:

{}^nC_2 - n = \dfrac{n(n-3)}{2}

. Triangles from

n

vertices are

{}^nC_3

, and the step

T_{n+1} - T_n = {}^nC_2

Definition

${}^nC_2 = 45 \Rightarrow n(n-1) = 90 \Rightarrow n = 10$ . Factor $90 = 10 \times 9$ rather than solving the quadratic.
Diagonals $\dfrac{n(n-3)}{2} = 54 \Rightarrow n^2 - 3n - 108 = 0 \Rightarrow (n - 12)(n + 9) = 0$ , $n = 12$ .
$T_n = {}^nC_3$ , so $T_{n+1} - T_n = {}^nC_2$ by Pascal; $= 21 \Rightarrow n = 7$ .
Intersections: $m$ lines meet in at most ${}^mC_2$ points; $c$ circles in at most $2\,{}^cC_2$ ; each line–circle pair in at most $2$ . $8$ lines and $4$ circles: $28 + 12 + 64 = 104$ .

Pairs and triples

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

Worked example

A polygon has

35

diagonals. How many sides does it have?

Practise this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2024 · 14th May Shift 2 · Q114Easy

Example 3 · Permutations and Combinations · Counting Numbers and Geometric Figures — Digits, Divisibility, Points and Polygons

In a regular polygon, the number of diagonals are 54, then the number of sides of the polygon are

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

Drill 3 more on handshakes, diagonals and triangle counts: solve an nc2 or nc3 equation

Concept 4 of 4: Triangles and Quadrilaterals From Points: Subtract the Collinear Choices

Any

3

n

points make a triangle unless they are collinear; any

4

make a quadrilateral unless

3

4

of them are collinear. Count

{}^nC_k

and subtract the degenerate picks. For 'triangles using no side of a polygon', subtract the triangles that use one or two sides.

Definition

$11$ points, $5$ collinear, quadrilaterals: ${}^{11}C_4 - {}^5C_3\,{}^6C_1 - {}^5C_4 = 330 - 60 - 5 = 265$ .
Triangles from $n$ points with $m$ collinear: ${}^nC_3 - {}^mC_3$ . Lines: ${}^nC_2 - {}^mC_2 + 1$ .
Regular $20$ -gon, triangles using no side: total ${}^{20}C_3 = 1140$ ; using two sides (three consecutive vertices) $20$ ; using exactly one side: $20$ sides $\times$ $16$ non-adjacent third vertices $= 320$ ; none: $1140 - 20 - 320 = 800$ .
General $n$ -gon, no side: ${}^nC_3 - n - n(n - 4) = \dfrac{n(n-4)(n-5)}{6}$ ; check $n = 20$ : $\dfrac{20 \cdot 16 \cdot 15}{6} = 800$ .

Degenerate subtraction

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

Worked example

There are

10

points in a plane, of which

4

are collinear. How many triangles have their vertices among these points?

Practise this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2025 · 25 April Shift II · Q137Moderate

Example 4 · Permutations and Combinations · Counting Numbers and Geometric Figures — Digits, Divisibility, Points and Polygons

There are 11 points in a plane of which 5 points are collinear. Then the total number of distinct quadrilaterals with vertices at these points is

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

Drill 1 more on triangles and quadrilaterals from points: subtract the collinear choices

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 (4)

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

Watch out for (4)

Keeping the leading zero
→ Digit Counting: No Leading Zero, and Divisibility by the Last Digits or the Digit Sum
Forgetting the add-back
→ Inclusion–Exclusion on Multiples: gcd Conditions
Counting the sides as diagonals
→ Handshakes, Diagonals and Triangle Counts: Solve an nC2 or nC3 Equation
Subtracting only the all-collinear picks
→ Triangles and Quadrilaterals From Points: Subtract the Collinear Choices

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