Chapter 6 — PERMUTATIONS AND COMBINATIONS

Theory

The most important identity in combinations is:

\[ ^nC_r = ^nC_{n-r} \]

This is called the symmetry property of combinations.

If \(^nC_a = ^nC_b\), then either:
1. \(a = b\), or
2. \(a + b = n\)

Solution Roadmap

Step 1: Apply symmetry property
Step 2: Form equation using \(a + b = n\)
Step 3: Find \(n\)
Step 4: Compute required combination

Solution

Given:

\[ ^nC_8 = ^nC_2 \]

Using symmetry property:

\[ 8 + 2 = n \Rightarrow n = 10 \]

Now compute:

\[ \begin{aligned} ^nC_2 &= ^{10}C_2 \\ &= \frac{10 \times 9}{2} \\ &= 45 \end{aligned} \]

Final Answer

\(^nC_2 = 45\)

Exam Significance

This is a very important shortcut-based problem.

• Direct CBSE board question type
• Frequently asked in JEE for quick solving
• Tests knowledge of symmetry property

Key takeaway: If \(nC_a = nC_b\), immediately check \(a + b = n\)

Theory

Combination formula:

\[ ^nC_r = \frac{n(n-1)(n-2)}{r!} \quad \text{(for small } r) \]

For ratio problems: expand only required terms and cancel early.

Key idea: \[ ^nC_3 = \frac{n(n-1)(n-2)}{6} \]

Solution Roadmap

Step 1: Convert ratio into fraction
Step 2: Use \( ^nC_3 = \frac{n(n-1)(n-2)}{6} \)
Step 3: Cancel common factors
Step 4: Solve resulting equation

Solution

(i)

\[ \frac{{}^{2n}C_3}{{}^{n}C_3} = 12 \] \[ \begin{aligned} \frac{(2n)(2n-1)(2n-2)}{n(n-1)(n-2)} &= 12 \end{aligned} \]

Simplify:

\[ \begin{aligned} (2n)(2n-1)2(n-1) &= 12n(n-1)(n-2) \\ 4n(n-1)(2n-1) &= 12n(n-1)(n-2) \end{aligned} \]

Cancel \(n(n-1)\):

\[ \begin{aligned} 4(2n-1) &= 12(n-2) \\ 2n-1 &= 3(n-2) \\ n &= 5 \end{aligned} \]

(ii)

\[ \frac{{}^{2n}C_3}{{}^{n}C_3} = 11 \] \[ \begin{aligned} \frac{(2n)(2n-1)(2n-2)}{n(n-1)(n-2)} &= 11 \end{aligned} \] \[ \begin{aligned} \frac{4(2n-1)}{(n-2)} &= 11 \\ 8n-4 &= 11(n-2) \\ 8n-4 &= 11n-22 \\ n &= 6 \end{aligned} \]

Final Answer

(i) \(n = 5\)
(ii) \(n = 6\)

Exam Significance

This is a high-speed simplification problem in combinations.

• Very common in JEE and boards
• Tests ability to avoid factorial expansion
• Focuses on algebraic reduction

Key takeaway: For \(nC_3\), directly use product form instead of factorials

Theory

A chord is formed by joining any two distinct points on a circle.

Since selecting 2 points does not depend on order, this is a combination problem.

\[ \text{Number of chords} = ^nC_2 \]

Solution Roadmap

Step 1: Identify total points → 21
Step 2: Select any 2 points
Step 3: Use combination formula
Step 4: Simplify

Solution

Total points = 21

Number of chords = number of ways to choose 2 points:

\[ \begin{aligned} {}^{21}C_2 &= \frac{21 \times 20}{2} \\ &= 210 \end{aligned} \]

Final Answer

\(210\)

Exam Significance

This is a standard geometric combination problem.

• Frequently asked in CBSE exams
• Common in JEE geometry-combinatorics crossover
• Foundation for polygon and diagonal problems

Key takeaway: Any pair of points ⇒ use combination \(nC_2\)

Theory

This is a selection problem, not arrangement, because order does not matter.

When selections are made from independent groups, we use: product of combinations.

\[ \text{Total ways} = ^nC_r \times ^mC_k \]

Here: Select boys and girls separately, then multiply.

Solution Roadmap

Step 1: Choose 3 boys from 5
Step 2: Choose 3 girls from 4
Step 3: Multiply (independent selections)

Solution

Choose 3 boys from 5:

\[ ^5C_3 = \frac{5 \times 4 \times 3}{3 \times 2 \times 1} = 10 \]

Choose 3 girls from 4:

\[ ^4C_3 = 4 \]

Total ways:

\[ \begin{aligned} \text{Total} &= ^5C_3 \times ^4C_3 \\ &= 10 \times 4 \ &= 40 \end{aligned} \]

Final Answer

\(40\)

Exam Significance

This is a multi-group selection problem.

• Common CBSE board question
• Important for JEE selection logic
• Foundation for probability and grouping problems

Key takeaway: Select from different groups separately, then multiply

Theory

This is a selection from multiple groups with fixed constraints.

When selections are made independently from different groups:

\[ \text{Total ways} = ^aC_x \times ^bC_y \times ^cC_z \]

Here, we must select exactly: 3 red + 3 white + 3 blue.

Solution Roadmap

Step 1: Choose 3 red balls from 6
Step 2: Choose 3 white balls from 5
Step 3: Choose 3 blue balls from 5
Step 4: Multiply all selections

Solution

Choose 3 red balls from 6:

\[ ^6C_3 = 20 \]

Choose 3 white balls from 5:

\[ ^5C_3 = 10 \]

Choose 3 blue balls from 5:

\[ ^5C_3 = 10 \]

Total ways:

\[ \begin{aligned} \text{Total} &= ^6C_3 \times ^5C_3 \times ^5C_3 \\ &= 20 \times 10 \times 10 \\ &= 2000 \end{aligned} \]

Final Answer

\(2000\)

Exam Significance

This is a multi-category constrained selection problem.

• Very common in CBSE board exams
• Important for JEE grouping problems
• Foundation for probability distributions

Key takeaway: When fixed numbers are chosen from different groups → multiply combinations

Theory

A standard deck has:
• 4 aces
• 48 non-ace cards

For “exactly one ace”, we split the selection:

\[ \text{Total ways} = ^4C_1 \times ^{48}C_4 \]

This is a conditional selection problem: select required items first, then remaining.

Solution Roadmap

Step 1: Choose 1 ace from 4
Step 2: Choose remaining 4 cards from 48 non-aces
Step 3: Multiply both selections

Solution

Choose 1 ace from 4:

\[ ^4C_1 = 4 \]

Choose remaining 4 cards from 48 non-aces:

\[ ^{48}C_4 = \frac{48 \times 47 \times 46 \times 45}{24} = 194580 \]

Total ways:

\[ \begin{aligned} \text{Total} &= ^4C_1 \times ^{48}C_4 \\ &= 4 \times 194580 \\ &= 778320 \end{aligned} \]

Final Answer

\(778320\)

Exam Significance

This is a classic conditional combination problem.

• Very common in probability and JEE exams
• Tests “exactly k items” logic
• Foundation for card probability problems

Key takeaway: “Exactly k” ⇒ choose required first, then remaining from others

Theory

This is a selection with restriction problem.

Total players = 17
Bowlers = 5
Non-bowlers = 12

Since we need exactly 4 bowlers, we split the selection:

\[ \text{Total ways} = ^5C_4 \times ^{12}C_7 \]

Solution Roadmap

Step 1: Select required bowlers
Step 2: Select remaining players from non-bowlers
Step 3: Multiply independent selections

Bowlers → \(^5C_4\)
Non-bowlers → \(^{12}C_7\)
Combine selections

Solution

Choose 4 bowlers from 5:

\[ ^5C_4 = 5 \]

Choose remaining 7 players from 12 non-bowlers:

\[ ^{12}C_7 = 792 \]

Total ways:

\[ \begin{aligned} \text{Total} &= ^5C_4 \times ^{12}C_7 \\ &= 5 \times 792 \\ &= 3960 \end{aligned} \]

Final Answer

\(3960\)

Exam Significance

This is a fixed-category selection problem.

• Very common in CBSE exams
• Important for JEE grouping problems
• Used in probability (team selection cases)

Key takeaway: “Exactly k from a category” ⇒ split selection and multiply

Theory

This is a selection from multiple categories with fixed requirements.

Since order does not matter, we use combinations.

When selecting from different groups independently:

\[ \text{Total ways} = ^aC_x \times ^bC_y \]

Solution Roadmap

Step 1: Select 2 black balls from 5
Step 2: Select 3 red balls from 6
Step 3: Multiply independent selections

Black → \(^5C_2\)
Red → \(^6C_3\)
Multiply results

Solution

Choose 2 black balls from 5:

\[ ^5C_2 = \frac{5 \times 4}{2} = 10 \]

Choose 3 red balls from 6:

\[ ^6C_3 = \frac{6 \times 5 \times 4}{6} = 20 \]

Total ways:

\[ \begin{aligned} \text{Total} &= ^5C_2 \times ^6C_3 \\ &= 10 \times 20 \\ &= 200 \end{aligned} \]

Final Answer

\(200\)

Exam Significance

This is a basic multi-group selection problem.

• Very common in CBSE board exams
• Important for probability questions
• Foundation for conditional selection problems

Key takeaway: Select from each group separately, then multiply

Theory

This is a selection with compulsory elements.

When some items are compulsory:
• Include them first
• Choose remaining from the rest

\[ \text{Total ways} = ^{n-k}C_{r-k} \]

where \(k\) items are already fixed (compulsory).

Solution Roadmap

Step 1: Fix compulsory courses
Step 2: Count remaining courses to choose
Step 3: Select from remaining pool
Step 4: Apply combination

Solution

Total courses = 9
Compulsory courses = 2

Remaining courses = \(9 - 2 = 7\)

Total courses required = 5
Already selected = 2
Remaining to select = \(5 - 2 = 3\)

Number of ways:

\[ \begin{aligned} ^7C_3 &= \frac{7 \times 6 \times 5}{6} \\ &= 35 \end{aligned} \]

Final Answer

\(35\)

Exam Significance

This is a compulsory-element selection pattern.

• Very common in CBSE board exams
• Important for JEE selection constraints
• Used in scheduling and subset problems

Key takeaway: Fix compulsory items first, then choose remaining