MHT-CET Maths · Permutations and Combinations

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

Choose a committee or a question set under a condition: list the cases for 'at least' and 'at most' and add the products of nCr terms, or subtract the forbidden selections from the unrestricted total.

All Permutations and Combinations notes MHT-CET Maths strategy

Why this matters

7 PYQs at 43% HARD. These are committee and question-paper stems — a team with at most one boy, at least two from each section, two members who refuse to serve together — and the same linguistic-club question was set in two 2024 shifts. There is one method, listing cases, and one shortcut, the complement; the HARD ones just have more cases, or a leader to choose after the team.

Concept 1 of 3

At Least and At Most: List the Cases and Add

Intuition

'At least

2

girls' in a committee of

5

means

2

3

4

5

girls — separate, mutually exclusive cases. Count each as a product of

{}^nC_r

factors and add.

Definition

$8$ boys, $5$ girls, committee of $5$ with at least $2$ girls AND at most $2$ boys: the cases are $(3G, 2B)$ , $(4G, 1B)$ , $(5G, 0B)$ : ${}^5C_3\,{}^8C_2 + {}^5C_4\,{}^8C_1 + {}^5C_5\,{}^8C_0 = 280 + 40 + 1 = 321$ .
$11$ questions in sections of $6$ and $5$ , choose $6$ with at least $2$ from each: $(2, 4)$ , $(3, 3)$ , $(4, 2)$ : $75 + 200 + 150 = 425$ .
Committee of $11$ from $8$ males + $5$ females with at least $6$ males: $(6, 5)$ , $(7, 4)$ , $(8, 3)$ : $28 + 40 + 10 = 78$ . 'At least $3$ females' lists the SAME three cases, so $m = n = 78$ .
Two conditions at once ('at least $2$ girls' and 'at most $2$ boys') intersect the case lists; write both ranges and keep only the pairs that satisfy both.

Casework

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

Worked example

From

6

men and

4

women, a committee of

4

with at least

3

women is chosen. How many ways?

Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 1Permutations and CombinationsMODERATE

If a question paper consists of 11 questions divided into two sections I and II. Section I consists of 6 questions and section II consists of 5 questions, then the number of different ways a student can select 6 questions, taking at least 2 questions from each section, is

[Q128 · Shift 1 · 2022]

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).

Drill 2 more on at least and at most: list the cases and add

Concept 2 of 3

The Complement: Unrestricted Total Minus the Forbidden Selections

Intuition

When the forbidden case is a single, easy count ('

A

and

B

both on the team'), count everything and subtract it. The complement of 'at least one' is 'none', which is one case.

Definition

$2$ boys of $5$ and $3$ girls of $7$ with girls $A$ , $B$ refusing to serve together: total ${}^5C_2\,{}^7C_3 = 350$ ; both $A$ and $B$ in: ${}^5C_2 \cdot {}^5C_1 = 50$ ; answer $300$ .
'At least one from each of $3$ sections, $5$ questions, $5$ per section': list the distributions $(1,1,3)$ in $3$ orders and $(1,2,2)$ in $3$ orders: $3 \cdot 5 \cdot 5 \cdot 10 + 3 \cdot 5 \cdot 10 \cdot 10 = 750 + 1500 = 2250$ . Here the complement (a section left empty) needs inclusion–exclusion, so direct casework is quicker.
Rule of thumb: complement when the forbidden event is ONE simple case; casework when the allowed event is a short list.

Complement

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

Worked example

From

10

people a committee of

4

is chosen. Two particular people refuse to serve together. How many committees?

Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 2Permutations and CombinationsMODERATE

Consider a group of 5 boys and 7 girls. The number of different teams, consisting of 2 boys and 3 girls that can be formed from this group if there are two specific girls A and B, who refuse to be the members of the same team, is

[Q141 · 9th May Shift 2 · 2023]

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.

Drill 1 more on the complement: unrestricted total minus the forbidden selections

Concept 3 of 3

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

Intuition

A role picked from the selected members is a second stage: teams

\times

(ways to pick the role-holder from the team). For a captain among

4

members that is

\times 4

, applied to the TOTAL of all cases.

Definition

$6$ girls, $4$ boys, team of $4$ with at most one boy, then a captain: teams ${}^6C_4 + {}^4C_1\,{}^6C_3 = 15 + 80 = 95$ ; captain $\times 4$ : $380$ .
If the role-holder is constrained (the captain must be a girl), the multiplier changes per case; do it case by case.
Two roles from the team: $\times {}^4P_2$ if the roles differ, $\times {}^4C_2$ if they are the same kind.

Team then role

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

Worked example

From

7

players a team of

5

is chosen and then a captain and a vice-captain are named from the team. How many ways?

Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 3Permutations and CombinationsMODERATE

A linguistic club consists of 6 girls and 4 boys. A team of 4 members is to be selected from this group including the selection of a leader (from among these 4 members) for the team. If the team has to include at most one boy, the number of ways of selecting the team is

[Q118 · 14th May Shift 1 · 2024]

Stopping at the team count

95

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

4

Drill 1 more on select the team, then choose the captain: multiply by the team size

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)

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

Watch out for (3)

Dropping a case
→ At Least and At Most: List the Cases and Add
Subtracting the wrong thing
→ The Complement: Unrestricted Total Minus the Forbidden Selections
Stopping at the team count
→ Select the Team, Then Choose the Captain: Multiply by the Team Size

Drill every past-year question on this subtopic

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

Drill the 7 questions

Drill every past-year question on this subtopic

Related notes