Permutations and Combinations Class 11 NCERT Solutions & Concepts

The Core Logic: Arrangement vs. Selection

To excel in class 11 maths permutations and combinations, you must first understand the distinction between ordering things and choosing things.

1. The Fundamental Principle of Counting (FPC): This is the foundation of the entire chapter. The Multiplication Principle states that if an event A can occur in 'm' ways, and a second event B can occur in 'n' ways, then the total number of ways both events can occur in sequence is m n. The Addition Principle* applies when events are mutually exclusive; if event A occurs in 'm' ways and event B in 'n' ways, the choice of event A OR event B occurs in m + n ways.

1. Permutations (Arrangement): When the order of items matters, we talk about permutations. For instance, arranging the letters A, B, and C to form words. Here, ABC and BAC are treated as two entirely different arrangements. Mathematically, the number of permutations of 'n' distinct objects taken 'r' at a time is represented as nPr = n! / (n-r)!.

1. Combinations (Selection): When the order of items does not matter, we talk about combinations. For example, selecting a team of 3 students from a group of 5. Here, selecting {A, B, C} is identical to selecting {B, A, C}. The number of combinations of 'n' objects taken 'r' at a time is denoted as nCr = n! / (r! * (n-r)!).

Key Definitions and Notations

Factorial Notation (n!)

The product of the first n consecutive natural numbers. Represented as n! = n (n-1) (n-2) ... 3 2 1. By definition, 0! = 1.

Permutation (nPr)

An arrangement of a set of objects in a definite order. Formula: nPr = n! / (n - r)!, where 0 <= r <= n.

Combination (nCr)

A selection of items from a larger pool where the order of selection is irrelevant. Formula: nCr = n! / [r! * (n - r)!], where 0 <= r <= n.

A Systematic Strategy for NCERT Problems

1. Identify if Order Matters — Read the problem carefully. Ask yourself: 'Does changing the position of the selected elements create a new outcome?' If yes, use Permutations (nPr). If no, use Combinations (nCr).
2. Identify Constraints and Conditions — Look for phrases like 'together', 'never together', 'at least', or 'at most'. Always handle restricted items first before dealing with the remaining elements.
3. Apply Formulas and Simplify Factorials — Write down the correct mathematical expression using nPr or nCr. Expand the factorials carefully, cancelling out common terms in the numerator and denominator to simplify calculations.

Detailed Step-by-Step Solved Examples

• Example 1: Find the number of words that can be formed using all the letters of the word 'DAUGHTER' such that all vowels always occur together. Step 1: Identify vowels and consonants. The word 'DAUGHTER' has 8 letters: Vowels = {A, U, E} (3 letters), Consonants = {D, G, H, T, R} (5 letters). Step 2: Treat the vowels {A, U, E} as a single bundle/block. Now, we have this 1 vowel block + 5 individual consonants = 6 units to arrange. Step 3: Arrange these 6 units. The number of ways to arrange 6 units is 6! = 720 ways. Step 4: Arrange the elements inside the vowel bundle. The 3 vowels {A, U, E} can be arranged among themselves in 3! = 6 ways. Step 5: Apply the Multiplication Principle. Total arrangements = 6! 3! = 720 6 = 4320 ways. Final answer: 4320.
• Example 2: A committee of 5 members is to be formed from a group of 6 men and 4 women. In how many ways can this be done if the committee must contain at least 3 men? Step 1: Understand 'at least 3 men' in a 5-member committee. This means we can have 3 men, 4 men, or 5 men. Step 2: Calculate Case 1 (Exactly 3 men and 2 women). Ways to choose 3 men from 6 is 6C3. Ways to choose 2 women from 4 is 4C2. Total for Case 1 = 6C3 4C2 = [6! / (3!3!)] [4! / (2!2!)] = 20 6 = 120. Step 3: Calculate Case 2 (Exactly 4 men and 1 woman). Ways to choose 4 men from 6 is 6C4. Ways to choose 1 woman from 4 is 4C1. Total for Case 2 = 6C4 4C1 = 15 4 = 60. Step 4: Calculate Case 3 (Exactly 5 men and 0 women). Ways to choose 5 men from 6 is 6C5. Ways to choose 0 women from 4 is 4C0. Total for Case 3 = 6C5 4C0 = 6 * 1 = 6. Step 5: Apply the Addition Principle for mutually exclusive cases. Total ways = Case 1 + Case 2 + Case 3 = 120 + 60 + 6 = 186. Final answer: 186.

Board Exam Traps and Best Practices

• The Factorial Trap: Remember that 0! is 1, not 0. Leaving 0! as 0 will break your formula when r = n.
• The Double-Counting Error: When calculating 'at least' problems, do not count a combination multiple times by selecting a core group and then picking randomly from the remainder. Always split 'at least' or 'at most' into distinct, mutually exclusive cases as shown in Example 2.
• Repeat Letters: If you are arranging letters of a word that has repeating characters (like 'ROOT' or 'MATHEMATICS'), remember to divide the total permutations by the factorials of the repetitions: n! / (p! q! r!).

Practice Questions with Solutions

• Q: How many 3-digit even numbers can be formed from the digits 1, 2, 3, 4, 5, 6 if the digits can be repeated? A: Step 1: Let the 3-digit number be represented by three blank spaces: _ _ _. Step 2: For the number to be even, the units place must be occupied by an even digit. The available even digits from the set are 2, 4, and 6 (3 options). So, the units place can be filled in 3 ways. Step 3: Since repetition of digits is allowed, the tens place can be filled by any of the 6 digits (6 ways). Step 4: Similarly, the hundreds place can be filled by any of the 6 digits (6 ways). Step 5: Apply the multiplication principle: Total numbers = 6 6 3 = 108. Final answer: 108.
• Q: If nC8 = nC2, find the value of nC2. A: Step 1: Recall the standard identity of combinations: If nCa = nCb, then either a = b or a + b = n. Step 2: Here, we have nC8 = nC2. Since 8 is not equal to 2, we must have 8 + 2 = n. Thus, n = 10. Step 3: Now, evaluate nC2 for n = 10. Step 4: 10C2 = 10! / (2! 8!) = (10 9) / (2 * 1) = 45. Final answer: 45.
• Q: In how many ways can the letters of the word 'PERMUTATIONS' be arranged if the words start with P and end with S? A: Step 1: Count the total letters in 'PERMUTATIONS'. There are 12 letters, where 'T' is repeated twice. Step 2: Fix the position of 'P' at the first place and 'S' at the last place. The structure is: P [10 remaining positions] S. Step 3: We now need to arrange the remaining 10 letters: E, R, M, U, T, A, I, O, N, T. Step 4: Among these 10 letters, 'T' is repeated 2 times. The rest are distinct. Step 5: Apply the permutation formula for identical items: Number of arrangements = 10! / 2!. Step 6: Calculate: 10! / 2 = 3,628,800 / 2 = 1,814,400. Final answer: 1,814,400.
• Q: Out of 12 points in a plane, 5 points are collinear. How many distinct triangles can be formed by joining these points? A: Step 1: To form a triangle, we need to choose any 3 non-collinear points. The total number of ways to choose 3 points out of 12 is 12C3. Step 2: Calculate 12C3 = 12! / (3! 9!) = (12 11 10) / (3 2 1) = 220. Step 3: Calculate the invalid triangles formed by choosing 3 points from the 5 collinear points, which only form a straight line instead of a triangle. The number of such selections is 5C3. Step 4: Calculate 5C3 = 5! / (3! 2!) = (5 * 4) / 2 = 10. Step 5: Subtract the collinear combinations from the total combinations: Total triangles = 12C3 - 5C3 = 220 - 10 = 210. Final answer: 210.

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