krit.club logo

Combinatorics - Introduction: The Mathematics of Possibilities-advanced

Grade 9CBSE

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

🔑Concepts

•

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

•

The Addition Principle states that if one event can occur in mm ways and another event can occur in nn ways, and these events cannot occur simultaneously, then the number of ways either event can occur is m+nm + n.

•

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

•

Permutations (nPr^{n}P_{r}): These are the different arrangements that can be made by taking some or all of a number of things. Here, the order of arrangement is critical.

•

Combinations (nCr^{n}C_{r}): These are the different selections that can be made by taking some or all of a number of things, regardless of the order in which they are placed.

•

Permutations with Repetition: If there are nn objects where pp objects are of one kind, qq are of another kind, and rr are of a third kind, the number of arrangements is n!p!q!r!\frac{n!}{p!q!r!}.

•

Circular Permutations: The number of ways to arrange nn distinct objects around a circular table is (n−1)!(n - 1)!.

📐Formulae

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

nPr=n!(n−r)!^{n}P_{r} = \frac{n!}{(n-r)!}

nCr=n!r!(n−r)!^{n}C_{r} = \frac{n!}{r!(n-r)!}

nCr=nPrr!^{n}C_{r} = \frac{^{n}P_{r}}{r!}

nCr=nCn−r^{n}C_{r} = ^{n}C_{n-r}

nC0=nCn=1^{n}C_{0} = ^{n}C_{n} = 1

💡Examples

Problem 1:

In how many ways can the letters of the word TABLETABLE be arranged?

Solution:

The word TABLETABLE has 55 distinct letters. Number of arrangements = 5!5! 5!=5×4×3×2×1=1205! = 5 \times 4 \times 3 \times 2 \times 1 = 120 Therefore, there are 120120 ways.

Explanation:

Since all letters are distinct and the order of letters matters, we use the factorial of the total number of letters.

Problem 2:

A committee of 33 members is to be selected from a group of 88 people. In how many ways can this be done?

Solution:

We need to select r=3r = 3 people out of n=8n = 8. Using the combination formula: 8C3=8!3!(8−3)!^{8}C_{3} = \frac{8!}{3!(8-3)!} 8C3=8!3!×5!^{8}C_{3} = \frac{8!}{3! \times 5!} 8C3=8×7×63×2×1=56^{8}C_{3} = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = 56 There are 5656 ways to form the committee.

Explanation:

In a committee, the order of selection does not matter, so we use the combinations formula nCr^{n}C_{r}.

Problem 3:

How many 33-digit numbers can be formed using the digits 1,2,3,4,51, 2, 3, 4, 5 if repetition of digits is allowed?

Solution:

For a 33-digit number, we have 33 places to fill: Hundreds, Tens, and Units. Number of ways to fill Hundreds place = 55 Number of ways to fill Tens place = 55 Number of ways to fill Units place = 55 Total numbers = 5×5×5=1255 \times 5 \times 5 = 125

Explanation:

Since repetition is allowed, each of the 33 positions can be filled by any of the 55 available digits.

Problem 4:

Find the number of arrangements of the letters in the word APPLEAPPLE.

Solution:

Total number of letters n=5n = 5. The letter PP is repeated 22 times. Number of arrangements = 5!2!\frac{5!}{2!} =1202=60= \frac{120}{2} = 60

Explanation:

When objects are not distinct, we divide the total permutations by the factorial of the count of each repeated object.