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Combinatorics - Permutations: When Order Matters-advanced

Grade 9CBSE

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

🔑Concepts

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Fundamental Counting Principle: If one operation can be performed in mm ways and a second operation can be performed in nn ways, then the two operations in succession can be performed in m×nm \times n ways.

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Factorial Notation: The product of the first nn natural numbers is denoted by n!n!. By definition, 0!=10! = 1 and n!=n×(n−1)×(n−2)×⋯×1n! = n \times (n-1) \times (n-2) \times \dots \times 1.

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Definition of Permutation: A permutation is an arrangement of a number of objects in a specific order. In permutations, the order of elements is critical.

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Linear Permutation: The number of permutations of nn different objects taken rr at a time, where 0≤r≤n0 \leq r \leq n and objects do not repeat, is denoted by P(n,r)P(n, r) or nPr^{n}P_{r}.

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Permutations with Repetition: The number of permutations of nn objects, where pp objects are of one kind, qq are of another kind, and the rest are distinct, is given by n!p!q!\frac{n!}{p!q!}.

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Circular Permutations: The number of ways to arrange nn distinct objects in a circle is (n−1)!(n-1)! because one position must be fixed to break the symmetry.

📐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)!}

Number of arrangements of n objects where p,q,r are identical=n!p!q!r!\text{Number of arrangements of } n \text{ objects where } p, q, r \text{ are identical} = \frac{n!}{p!q!r!}

Circular Permutations of n objects=(n−1)!\text{Circular Permutations of } n \text{ objects} = (n-1)!

💡Examples

Problem 1:

How many 4-digit numbers can be formed using the digits 1,2,3,4,5,61, 2, 3, 4, 5, 6 if no digit is repeated?

Solution:

Here, the total number of digits n=6n = 6. We need to choose and arrange r=4r = 4 digits. Using the permutation formula: P(6,4)=6!(6−4)!P(6, 4) = \frac{6!}{(6-4)!} P(6,4)=6!2!P(6, 4) = \frac{6!}{2!} P(6,4)=6×5×4×3×2×12×1P(6, 4) = \frac{6 \times 5 \times 4 \times 3 \times 2 \times 1}{2 \times 1} P(6,4)=6×5×4×3=360P(6, 4) = 6 \times 5 \times 4 \times 3 = 360

Explanation:

Since the order of digits matters in a number and repetition is not allowed, we use the linear permutation formula P(n,r)P(n, r).

Problem 2:

Find the number of ways to arrange the letters of the word APPLEAPPLE.

Solution:

The word APPLEAPPLE contains n=5n = 5 letters. The letter PP is repeated 22 times. Using the formula for permutations with identical objects: Total arrangements=5!2!\text{Total arrangements} = \frac{5!}{2!} Total arrangements=1202=60\text{Total arrangements} = \frac{120}{2} = 60

Explanation:

When objects are identical (like the two 'P's), we divide the total permutations (5!5!) by the factorial of the count of identical items (2!2!) to avoid double-counting.

Problem 3:

In how many ways can 5 people be seated around a circular table?

Solution:

The number of people n=5n = 5. For a circular arrangement, the number of ways is given by (n−1)!(n-1)!. Ways=(5−1)!=4!\text{Ways} = (5-1)! = 4! 4!=4×3×2×1=244! = 4 \times 3 \times 2 \times 1 = 24

Explanation:

In a circle, shifting everyone one seat to the left results in the same relative arrangement. Therefore, we fix one person's position and arrange the remaining n−1n-1 people.