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Combinatorics - Combinatorics in the Modern World-advanced

Grade 9CBSE

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

🔑Concepts

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The Fundamental Principle of Counting (Product Rule) states that if one event can occur in mm ways and a second event can occur in nn ways, then the total number of ways both events can occur in sequence is m×nm \times n.

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Factorial Notation: The product of the first nn natural numbers is called nn factorial, denoted as n!n!. For example, 5!=5×4×3×2×1=1205! = 5 \times 4 \times 3 \times 2 \times 1 = 120.

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Permutations (PP) represent the number of ways to arrange a subset of items where the order of arrangement is important. This is used in modern applications like generating secure passwords or arranging data packets.

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Combinations (CC) represent the number of ways to select items from a group where the order does not matter. This is used in areas like network topology and lottery systems.

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The Pigeonhole Principle is an advanced combinatorial concept which states that if nn items are put into mm containers, with n>mn > m, then at least one container must contain more than one item.

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Modern applications of Combinatorics include Cryptography (designing codes), Coding Theory (detecting errors in digital communication), and Computational Biology (sequencing DNA).

📐Formulae

n!=n×(n−1)×(n−2)×⋯×1n! = n \times (n-1) \times (n-2) \times \dots \times 1

P(n,r)=n!(n−r)!P(n, r) = \frac{n!}{(n-r)!}

C(n,r)=(nr)=n!r!(n−r)!C(n, r) = \binom{n}{r} = \frac{n!}{r!(n-r)!}

0!=10! = 1

C(n,r)=C(n,n−r)C(n, r) = C(n, n-r)

💡Examples

Problem 1:

A cybersecurity expert is creating a 4-character password. The characters must be digits from 00 to 99. How many different passwords can be formed if repetition of digits is not allowed?

Solution:

Since the order of digits matters in a password and repetition is not allowed, we use permutations. Here, n=10n = 10 (digits 0−90-9) and r=4r = 4. Total passwords =P(10,4)= P(10, 4) P(10,4)=10!(10−4)!=10!6!P(10, 4) = \frac{10!}{(10-4)!} = \frac{10!}{6!} P(10,4)=10×9×8×7=5040P(10, 4) = 10 \times 9 \times 8 \times 7 = 5040

Explanation:

We choose 4 distinct digits out of 10 and arrange them. The first place has 10 options, the second has 9, the third has 8, and the fourth has 7.

Problem 2:

In a modern data network, a server needs to connect to 22 nodes out of a cluster of 1212 available nodes to perform a task. In how many ways can these 22 nodes be selected?

Solution:

Since the order in which nodes are selected does not change the connection pair, we use combinations. Here, n=12n = 12 and r=2r = 2. C(12,2)=12!2!(12−2)!=12!2!×10!C(12, 2) = \frac{12!}{2!(12-2)!} = \frac{12!}{2! \times 10!} C(12,2)=12×11×10!2×1×10!C(12, 2) = \frac{12 \times 11 \times 10!}{2 \times 1 \times 10!} C(12,2)=1322=66C(12, 2) = \frac{132}{2} = 66

Explanation:

We use the combination formula because selecting Node A then Node B is the same as selecting Node B then Node A.

Problem 3:

Calculate the value of 6!−4!6! - 4! using vertical subtraction.

Solution:

First, calculate the individual factorials: 6!=6×5×4×3×2×1=7206! = 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 720 4!=4×3×2×1=244! = 4 \times 3 \times 2 \times 1 = 24 Now, subtract: 720−24696\begin{array}{r} 720 \\ - 24 \\ \hline 696 \end{array}

Explanation:

Calculate the products for both factorials first, then perform standard subtraction.