Home Class 11 Applied Maths Chapter 8: Combinatorics Exercise 8.4: Combinations
Chapter 8 • Unit IV

Exercise 8.4: Combinations

Comprehensive guide and practice questions for CBSE Class 11 Applied Mathematics curriculum (2026-27).

8.4 Combinations: Selection Theory, Committee Formation, and Geometric Counting

In Applied Mathematics, corporate governance, portfolio asset selection, and statistical sampling, a combination represents a collection of items chosen from a broader pool where order does not matter. Choosing a project team of 4 engineers from a pool of 10 is fundamentally a combination problem, as the order of selection has no bearing on team composition.

Fundamental Formulae & Core Theorems

  • Definition: The number of combinations (selections) of \(n\) distinct objects taken \(r\) at a time is denoted by \(^n C_r\) or \(\binom{n}{r}\): \[ ^n C_r = \frac{n!}{r! (n - r)!} = \frac{^n P_r}{r!} \quad (0 \le r \le n) \]
  • Boundary Values: \[ ^n C_0 = 1, \quad ^n C_n = 1, \quad ^n C_1 = n \]
  • Complementary Selection Theorem: Selecting \(r\) objects to include is mathematically identical to selecting \((n - r)\) objects to reject: \[ ^n C_r = ^n C_{n - r} \]
  • Equivalence Property: \[ ^n C_x = ^n C_y \implies x = y \quad \text{or} \quad x + y = n \]
  • Pascal's Identity (Addition Formula): \[ ^n C_r + ^n C_{r - 1} = {}^{n + 1} C_r \]
  • Sum of All Combinations: The total number of ways to choose any number of items from \(n\) distinct objects (from 0 to \(n\)): \[ \sum_{r=0}^n \binom{n}{r} = \binom{n}{0} + \binom{n}{1} + \dots + \binom{n}{n} = 2^n \] Ways to select at least one object: \(2^n - 1\).

Combinatorial Geometry Formulary

Geometric Entity Conditions on \(n\) Points in a Plane Formula
Straight Lines No 3 points collinear \[ \binom{n}{2} = \frac{n(n - 1)}{2} \]
Straight Lines (Collinear Case) \(m\) points out of \(n\) are collinear \[ \binom{n}{2} - \binom{m}{2} + 1 \]
Triangles No 3 points collinear \[ \binom{n}{3} = \frac{n(n - 1)(n - 2)}{6} \]
Triangles (Collinear Case) \(m\) points out of \(n\) are collinear \[ \binom{n}{3} - \binom{m}{3} \]
Diagonals of an \(n\)-gon A polygon with \(n\) vertices \[ \binom{n}{2} - n = \frac{n(n - 3)}{2} \]
Parallelograms Intersecting \(m\) parallel lines with \(p\) parallel lines \[ \binom{m}{2} \times \binom{p}{2} \]

Comprehensive Corporate Worked Example: Audit Committee Selection

Scenario: A corporate board of directors consists of 7 Senior Partners and 5 Independent Directors. An internal audit committee of 5 members is to be formed.

In how many ways can the committee be constituted if:

  1. There are no restrictions on composition?
  2. The committee must consist of exactly 3 Senior Partners and 2 Independent Directors?
  3. The committee must contain at least 3 Independent Directors?

Solution:

  1. No restrictions: Choose 5 members from the total 12 directors: \[ \binom{12}{5} = \frac{12 \times 11 \times 10 \times 9 \times 8}{5 \times 4 \times 3 \times 2 \times 1} = 792 \text{ ways.} \]
  2. Exactly 3 Senior Partners and 2 Independent Directors: \[ \binom{7}{3} \times \binom{5}{2} = \frac{7 \times 6 \times 5}{6} \times \frac{5 \times 4}{2} = 35 \times 10 = 350 \text{ ways.} \]
  3. At least 3 Independent Directors (Mutually exclusive cases):
    • Case A (3 Independent, 2 Senior): \(\binom{5}{3} \times \binom{7}{2} = 10 \times 21 = 210\) ways.
    • Case B (4 Independent, 1 Senior): \(\binom{5}{4} \times \binom{7}{1} = 5 \times 7 = 35\) ways.
    • Case C (5 Independent, 0 Senior): \(\binom{5}{5} \times \binom{7}{0} = 1 \times 1 = 1\) way.
    By the Addition Principle: \[ \text{Total Ways} = 210 + 35 + 1 = 246 \text{ ways.} \]
Previous: Ex 8.3 Chapter Hub