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Permutations and Combinations - Permutations

Grade 11CBSE

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

🔑Concepts

Fundamental Principle of Counting (Multiplication Principle): If one event can occur in mm different ways, and following it, another event can occur in nn different ways, then the total number of ways both events can occur in the specified order is m×nm \times n. Visual: Imagine a tree diagram where the first choice creates mm main branches, and each of those branches splits into nn smaller sub-branches, resulting in m×nm \times n total paths.

Factorial Notation (n!n!): The product of the first nn natural numbers is called nn factorial, denoted by n!n!. It is defined as n!=n×(n1)×(n2)××1n! = n \times (n-1) \times (n-2) \times \dots \times 1. By convention, 0!=10! = 1. Visual: Think of this as a sequence of numbers decreasing by one at each step, representing the diminishing number of choices available as you fill positions.

Definition of Permutation: A permutation is an arrangement of a specific number of objects in a definite order. Unlike combinations, the order of arrangement is critical. Visual: If you have two colored blocks, Red (RR) and Blue (BB), the arrangement (R,B)(R, B) is visually and mathematically distinct from (B,R)(B, R) in permutations.

Permutations of nn Distinct Objects taken rr at a time: When we arrange rr objects out of nn available distinct objects without replacement, the number of permutations is denoted by nPr^nP_r. Visual: Imagine rr empty boxes in a row; you have nn choices for the first box, n1n-1 for the second, and you continue until all rr boxes are filled.

Permutations when Repetition is Allowed: If the objects can be reused (replacement allowed), the number of permutations of nn objects taken rr at a time is nrn^r. Visual: Imagine a digital lock with 3 dials, where each dial can be any digit from 0-9. Every dial always has 10 options regardless of what the previous dial was set to.

Permutations of Objects Not All Distinct: The number of permutations of nn objects, where pp objects are of one kind, qq objects are of a second kind, and the rest are all different, is given by n!p!q!\frac{n!}{p!q!}. Visual: If you arrange the letters in 'BOO', the two 'O's are identical. Swapping them doesn't change the look of the word, so we divide the total arrangements (3!3!) by the ways to arrange the identical items (2!2!) to remove duplicates.

Constraints in Permutations: Sometimes certain items must stay together or certain items must never be together. Visual: If items A and B must be together, we 'tie' them into a single block. This block and the remaining items are then arranged like a single unit, and then A and B are arranged within their own internal block.

📐Formulae

n!=n×(n1)×(n2)××3×2×1n! = n \times (n-1) \times (n-2) \times \dots \times 3 \times 2 \times 1

0!=10! = 1

nPr=n!(nr)!^nP_r = \frac{n!}{(n-r)!}, where 0rn0 \le r \le n

nPn=n!^nP_n = n!

Number of permutations of nn objects with repetition allowed: nrn^r

Permutations of nn objects where p1p_1 are of one kind, p2p_2 of another, ..., pkp_k of the kthk^{th} kind: n!p1!p2!pk!\frac{n!}{p_1! p_2! \dots p_k!}

💡Examples

Problem 1:

How many 3-letter words (with or without meaning) can be formed using the letters of the word 'EQUATION', using each letter exactly once?

Solution:

Step 1: Identify the total number of distinct objects. The word 'EQUATION' has 8 distinct letters, so n=8n = 8. \ Step 2: Identify the number of objects to be arranged. We need to form 3-letter words, so r=3r = 3. \ Step 3: Apply the permutation formula nPr=n!(nr)!^nP_r = \frac{n!}{(n-r)!}. \ 8P3=8!(83)!=8!5!^8P_3 = \frac{8!}{(8-3)!} = \frac{8!}{5!}. \ Step 4: Calculate the value: 8×7×6×5!5!=8×7×6=336\frac{8 \times 7 \times 6 \times 5!}{5!} = 8 \times 7 \times 6 = 336.

Explanation:

Since all letters are distinct and the order of letters matters in a word, we use the standard linear permutation formula for nn distinct objects taken rr at a time.

Problem 2:

Find the number of permutations of the letters of the word 'STATISTICS'.

Solution:

Step 1: Count the total number of letters in 'STATISTICS'. n=10n = 10. \ Step 2: Identify the frequency of repeating letters: \ 'S' appears 3 times (p=3p=3). \ 'T' appears 3 times (q=3q=3). \ 'I' appears 2 times (r=2r=2). \ 'A' and 'C' appear 1 time each. \ Step 3: Apply the formula for permutations with identical objects: n!p!q!r!\frac{n!}{p!q!r!}. \ Total permutations = 10!3!×3!×2!\frac{10!}{3! \times 3! \times 2!}. \ Step 4: Calculate: 36288006×6×2=362880072=50400\frac{3628800}{6 \times 6 \times 2} = \frac{3628800}{72} = 50400.

Explanation:

Because some letters (S, T, I) are repeated, we must divide the total arrangements of 10 letters (10!10!) by the factorials of the counts of each repeating letter to avoid counting identical-looking arrangements multiple times.