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Combinatorics - The Fundamental Principle of Counting (FPC)-advanced

Grade 9CBSE

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

🔑Concepts

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The Fundamental Principle of Counting (FPC) is the basic rule used to find the number of ways in which multiple events can occur together.

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The Multiplication Principle: If an event E1E_1 can occur in mm ways and another event E2E_2 can occur in nn ways, then the total number of ways the two events can occur in sequence is m×nm \times n.

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The Addition Principle: If an event E1E_1 can occur in mm ways and another event E2E_2 can occur in nn ways, and these events cannot happen at the same time (mutually exclusive), then there are m+nm + n ways for either E1E_1 or E2E_2 to occur.

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Factorial Notation: The product of the first nn natural numbers is called nn factorial, denoted as n!n!. By definition, 0!=10! = 1.

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Counting with Constraints: When solving advanced problems, always fill the positions with constraints (like 'must be even' or 'must start with 5') first.

📐Formulae

Total ways=n1×n2×n3⋯×nkTotal\ ways = n_1 \times n_2 \times n_3 \dots \times n_k

n!=n×(n−1)×(n−2)×⋯×3×2×1n! = n \times (n-1) \times (n-2) \times \dots \times 3 \times 2 \times 1

Total ways (with repetition)=nrTotal\ ways\ (with\ repetition) = n^r

Total ways (without repetition)=n!(n−r)!Total\ ways\ (without\ repetition) = \frac{n!}{(n-r)!}

💡Examples

Problem 1:

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

Solution:

We have three positions to fill: Hundreds, Tens, and Units.

  1. Units Place: Since the number must be even, the units place can only be filled by 2,4,2, 4, or 66. There are 33 ways.
  2. Tens Place: Since repetition is allowed, any of the 66 digits can be used. There are 66 ways.
  3. Hundreds Place: Similarly, any of the 66 digits can be used. There are 66 ways. Total ways =6×6×3=108= 6 \times 6 \times 3 = 108.

Explanation:

We apply the multiplication principle. The constraint (even number) is applied to the units digit first, then the remaining positions are filled based on the rule of repetition.

Problem 2:

Find the number of ways to arrange the letters of the word 'SMART' such that the word always starts with 'S' and ends with 'T'.

Solution:

The word 'SMART' has 55 distinct letters: S,M,A,R,TS, M, A, R, T.

  1. First position (fixed): Only 11 way ('S').
  2. Last position (fixed): Only 11 way ('T').
  3. Remaining 33 positions (the middle) must be filled by letters M,A,RM, A, R. Number of ways to fill the 2nd position =3= 3 (either M, A, or R). Number of ways to fill the 3rd position =2= 2. Number of ways to fill the 4th position =1= 1. Total ways =1×3×2×1×1=6= 1 \times 3 \times 2 \times 1 \times 1 = 6.

Explanation:

By fixing the first and last letters, we only need to calculate the permutations of the remaining 33 letters in the 33 available slots.

Problem 3:

In a class of 1010 students, in how many ways can a President, a Vice-President, and a Secretary be chosen if no student can hold more than one office?

Solution:

There are 33 positions to fill: President (PP), Vice-President (VPVP), and Secretary (SS). Number of ways to choose P=10P = 10. Number of ways to choose VP=9VP = 9 (since one student is already President). Number of ways to choose S=8S = 8 (since two students are already chosen). Total ways =10×9×8=720= 10 \times 9 \times 8 = 720.

Explanation:

This is a multiplication principle problem where repetition is not allowed because a student cannot hold multiple roles.