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Combinatorics - The Permutation Formula-advanced

Grade 9CBSE

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

🔑Concepts

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The factorial of a non-negative integer nn, denoted by n!n!, is the product of all positive integers less than or equal to nn. By definition, 0!=10! = 1.

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A permutation is an arrangement of a set of objects in a specific order. The order of selection matters in permutations.

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The number of permutations of nn different objects taken rr at a time, where 0≤r≤n0 \le r \le n and objects do not repeat, is denoted by nPr^n P_r or P(n,r)P(n, r).

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When all nn objects are distinct and arranged in a line, the total number of permutations is n!n!.

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

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In Circular Permutations, the number of ways to arrange nn distinct objects around a circle is (n−1)!(n-1)! because one position is fixed to break the symmetry.

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Restricted Permutations: When mm specific objects must always be together, we treat them as a single 'block'. We arrange the remaining (n−m)(n-m) objects plus the 1 block, then multiply by the internal arrangements of the mm objects (m!m!).

📐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)!}

nPn=n!^n P_n = n!

nP0=1^n P_0 = 1

Permutations with identical items=n!p1!p2!…pk!\text{Permutations with identical items} = \frac{n!}{p_1! p_2! \dots p_k!}

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

💡Examples

Problem 1:

How many 4-letter words (with or without meaning) can be formed using the letters of the word 'LOGARITHMS' if repetition of letters is not allowed?

Solution:

The word 'LOGARITHMS' has 10 distinct letters. We need to arrange 4 letters out of 10. Using the permutation formula nPr^n P_r: n=10,r=4n = 10, r = 4 10P4=10!(10−4)!^{10} P_4 = \frac{10!}{(10-4)!} 10P4=10!6!^{10} P_4 = \frac{10!}{6!} 10P4=10×9×8×7×6!6!^{10} P_4 = \frac{10 \times 9 \times 8 \times 7 \times 6!}{6!} 10P4=10×9×8×7=5040^{10} P_4 = 10 \times 9 \times 8 \times 7 = 5040

Explanation:

Since the letters are distinct and the order of letters in a word matters, we apply the permutation formula for n=10n=10 and r=4r=4.

Problem 2:

Find the number of arrangements of the letters of the word 'MATHEMATICS'.

Solution:

Total number of letters in 'MATHEMATICS' is n=11n = 11. Frequency of repeating letters: M appears 2 times (p=2p=2) A appears 2 times (q=2q=2) T appears 2 times (r=2r=2) Other letters (H, E, I, C, S) appear 1 time each. Total arrangements = 11!2!2!2!\frac{11!}{2! 2! 2!} Total arrangements = 399168002×2×2\frac{39916800}{2 \times 2 \times 2} Total arrangements = 399168008=4989600\frac{39916800}{8} = 4989600

Explanation:

When objects are not distinct (some are identical), we divide the total factorial by the factorials of the counts of each repeating object.

Problem 3:

In how many ways can 5 boys and 3 girls be seated in a row such that all the girls sit together?

Solution:

Treat the 3 girls as a single 'block' or unit. Number of units to arrange = 5 boys + 1 block of girls = 6 units. Ways to arrange 6 units = 6!=7206! = 720. Inside the block, the 3 girls can be arranged among themselves in 3!3! ways. Ways to arrange 3 girls = 3!=63! = 6. Total arrangements = 720×6=4320720 \times 6 = 4320.

Explanation:

We use the 'String Method' where objects that must stay together are tied into one unit, arranged, and then the internal arrangements of that unit are factored in using the multiplication principle.

The Permutation Formula-advanced Class 9 Notes & Examples