krit.club logo

Permutations and Combinations - Fundamental Principle of Counting

Grade 11CBSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

The Fundamental Principle of Multiplication states that if an event can occur in mm different ways, following which another event can occur in nn different ways, then the total number of ways of occurrence of both events in the specified order is m×nm \times n. This can be visualized as a rectangular grid where the mm rows represent choices for the first event and nn columns represent choices for the second event, with each intersection representing a unique outcome.

The Fundamental Principle of Addition applies when two tasks are mutually exclusive. If one task can be performed in mm ways and another task can be performed in nn ways, then either of the two tasks can be performed in m+nm + n ways. Visually, this is like looking at two separate, non-overlapping circles (sets) and counting the total number of items across both.

Factorial Notation (n!n!) represents the product of the first nn consecutive natural numbers. For any positive integer nn, n!=n×(n1)×(n2)××3×2×1n! = n \times (n-1) \times (n-2) \times \dots \times 3 \times 2 \times 1. By convention, 0!=10! = 1. This can be visualized as a descending ladder of numbers being multiplied together.

The 'Box Method' is a visual strategy used to solve counting problems. We draw a series of empty boxes, where each box represents a position or a sub-task. We then fill each box with the number of available choices for that position and multiply the numbers together to find the total combinations.

Tree Diagrams are visual tools used to list all possible outcomes of a sequence of events. Starting from a single point, branches are drawn for each choice of the first event; from the end of those branches, new branches are drawn for the choices of the second event, creating a branching structure that resembles a tree.

Repetition vs. Non-repetition: In many counting problems, it is crucial to identify if items can be reused. If repetition is allowed, the number of choices remains constant for each step; if repetition is not allowed, the number of choices decreases by 1 for each subsequent step, visualized as a shrinking pool of available objects.

Order of operations: When applying the Fundamental Principle of Counting, the order in which choices are made matters if the events are dependent. We always fill the positions with the most restrictive conditions first (e.g., if a number must be even, fill the units place first).

📐Formulae

Total ways (Multiplication Principle) = m×nm \times n

Total ways (Addition Principle) = m+nm + n

n!=n(n1)(n2)321n! = n(n-1)(n-2) \dots 3 \cdot 2 \cdot 1

n!=n×(n1)!n! = n \times (n-1)!

0!=10! = 1

Number of ways to fill rr places with nn things when repetition is allowed = nrn^r

💡Examples

Problem 1:

How many 3-digit even numbers can be formed from the digits 1,2,3,4,5,61, 2, 3, 4, 5, 6 if the digits can be repeated?

Solution:

  1. There are 3 positions to fill: Hundreds, Tens, and Units.
  2. Since the number must be even, the Units place can only be filled by digits 2,4,2, 4, or 66. Thus, there are 33 ways to fill the Units place.
  3. Since repetition is allowed, the Tens place can be filled by any of the 66 given digits. Thus, there are 66 ways to fill the Tens place.
  4. Similarly, the Hundreds place can be filled by any of the 66 given digits. Thus, there are 66 ways to fill the Hundreds place.
  5. Total ways = 6×6×3=1086 \times 6 \times 3 = 108.

Explanation:

We use the Multiplication Principle. We first address the constraint (even number means the last digit must be 2,4,2, 4, or 66) and then multiply the possibilities for each remaining position.

Problem 2:

Evaluate 8!6!×2!\frac{8!}{6! \times 2!}.

Solution:

  1. Expand the larger factorial in the numerator until it matches a factorial in the denominator: 8!=8×7×6!8! = 8 \times 7 \times 6!.
  2. Substitute this into the expression: 8×7×6!6!×2!\frac{8 \times 7 \times 6!}{6! \times 2!}.
  3. Cancel 6!6! from the numerator and denominator: 8×72!\frac{8 \times 7}{2!}.
  4. Expand 2!2!: 2×1=22 \times 1 = 2.
  5. Calculate the final value: 562=28\frac{56}{2} = 28.

Explanation:

This demonstrates the property n!=n×(n1)!n! = n \times (n-1)! to simplify expressions without calculating massive product values manually.