NCERT Solutions Class 11 Maths Permutations and Combinations Miscellaneous Exercise

The Conceptual Framework: Permutation vs. Combination

To conquer the miscellaneous exercise, you must develop a systematic approach to reading questions. The biggest obstacle students face is deciding whether to apply ${^nP_r}$ (Permutation) or ${^nC_r}$ (Combination).

Always ask yourself: Does the order of arrangement matter?

• If yes (e.g., forming words from letters, building numbers from digits, arranging people in a queue), you are dealing with Permutations.
• If no (e.g., choosing a committee from a group, selecting cards from a deck, grouping points to form triangles), you are dealing with Combinations.

Once the core operation is decided, use the Fundamental Principle of Counting (FPC). If operations $A$ and $B$ must occur together, multiply their possibilities ($A \times B$). If they are mutually exclusive alternatives, add them ($A + B$).

Key Definitions and Formulate Reference

Permutation ($^nP_r$)

An arrangement of $r$ objects chosen from $n$ distinct objects in a specific order. Formula: $^nP_r = \frac{n!}{(n-r)!}$ where $0 \le r \le n$.

Combination ($^nC_r$)

A selection of $r$ objects chosen from $n$ distinct objects where order of selection does not matter. Formula: $^nC_r = \frac{n!}{r!(n-r)!}$.

Complementary Selection

The property that selecting $r$ objects is equivalent to leaving behind $n-r$ objects. Mathematically, $^nC_r = ^nC_{n-r}$.

Step-by-Step Worked Miscellaneous Problems

1. Problem 1: Committee Selection with Constraints — Determine the number of ways of selecting a committee of 5 members from 6 gentlemen and 4 ladies if the committee must contain at least one lady. Step 1: Identify total available people: 6 gentlemen, 4 ladies. Committee size required = 5. Step 2: Define 'at least one lady' cases. This can be broken into mutually exclusive cases: - Case 1: 1 Lady and 4 Gentlemen - Case 2: 2 Ladies and 3 Gentlemen - Case 3: 3 Ladies and 2 Gentlemen - Case 4: 4 Ladies and 1 Gentleman Step 3: Calculate combinations for each case using $^nC_r$: - Case 1: $^4C_1 \times ^6C_4 = 4 \times 15 = 60$ - Case 2: $^4C_2 \times ^6C_3 = 6 \times 20 = 120$ - Case 3: $^4C_3 \times ^6C_2 = 4 \times 15 = 60$ - Case 4: $^4C_4 \times ^6C_1 = 1 \times 6 = 6$ Step 4: Sum the cases: Total ways = $60 + 120 + 60 + 6 = 246$.
2. Problem 2: Word Arrangement with Vowels Together — Find the number of words that can be formed using all letters of the word 'DAUGHTER' so that all vowels occur together. Step 1: Analyze the word 'DAUGHTER'. Total letters = 8. Vowels = {A, U, E} (3 vowels). Consonants = {D, G, H, T, R} (5 consonants). Step 2: Group the vowels together as a single block: [AUE]. Now treat this block as 1 entity along with the 5 consonants. Total entities to arrange = $1 + 5 = 6$ entities. Step 3: Arrange these 6 entities. This can be done in $6! = 720$ ways. Step 4: Arrange the vowels inside their block. Since the 3 vowels {A, U, E} are distinct, they can be arranged among themselves in $3! = 6$ ways. Step 5: Multiply the results by Multiplication Principle: Total arrangements = $6! \times 3! = 720 \times 6 = 4320$ ways.

Common Exam Traps and Board Tips

1. The 'At Least' Trap: When asked for 'at least' condition, it is often simpler to calculate the complement. $\text{Required Ways} = \text{Total Ways} - \text{Unwanted Ways}$. For example, if choosing 5 people with 'at least 1 lady', you can calculate: $\text{Total selections of 5} - \text{Selections with 0 ladies}$. Let's verify: $^{10}C_5 - ^6C_5 = 252 - 6 = 246$. It saves enormous time and prevents calculation errors!

1. Identical vs. Distinct Objects: Always check if letters or objects are repeating (like in 'ASSASSINATION'). If objects are identical, remember to divide the permutation total by the factorials of the repetitions of those identical items ($p!q!r!$).

Practice Questions with Solutions

• Q: How many 4-digit numbers can be formed using the digits 1, 2, 3, 4, 5 if repetition of digits is not allowed? A: Step 1: Identify variables. We have $n = 5$ digits and we need to choose and arrange $r = 4$ digits. Step 2: Since repetition is not allowed and order matters for forming numbers, we use Permutations. Step 3: Calculate using the formula $^nP_r$: $^5P_4 = \frac{5!}{(5-4)!} = \frac{120}{1} = 120$. Final answer: There are 120 such 4-digit numbers.
• Q: If $^nC_9 = ^nC_8$, find the value of $^nC_{17}$. A: Step 1: Use the property of combinations: If $^nC_x = ^nC_y$, then either $x = y$ or $x + y = n$. Step 2: Here, $9 \ne 8$, so we must have $n = 9 + 8 = 17$. Step 3: Substitute $n = 17$ into $^nC_{17}$: $^{17}C_{17} = 1$. Final answer: The value of $^{17}C_{17}$ is 1.
• Q: In how many ways can a team of 3 boys and 3 girls be selected from 5 boys and 4 girls? A: Step 1: Identify that this is a selection problem (order does not matter), hence we use Combinations. Step 2: Number of ways to select 3 boys from 5 boys = $^5C_3$. $^5C_3 = \frac{5 \times 4}{2 \times 1} = 10$. Step 3: Number of ways to select 3 girls from 4 girls = $^4C_3$. $^4C_3 = ^4C_1 = 4$. Step 4: Use the multiplication principle to find the combined selection: Total ways = $^5C_3 \times ^4C_3 = 10 \times 4 = 40$. Final answer: There are 40 ways to select the team.
• Q: Find the number of words with or without meaning which can be made using all the letters of the word 'AGAIN'. A: Step 1: Count the total letters in 'AGAIN'. Total letters = 5. Step 2: Identify repetitions. The letter 'A' is repeated 2 times. Other letters {G, I, N} appear once. Step 3: Apply the permutation formula for identical objects: Total permutations = $\frac{n!}{p!} = \frac{5!}{2!} = \frac{120}{2} = 60$. Final answer: The number of words that can be made is 60.

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