NCERT Exercise 7.4: Permutations and Combinations Class 11

The Core Concept of Combinations

In combinatorics, a combination is a selection of items from a collection, such that the order of selection does not matter. To understand this intuitively, suppose we have three letters: A, B, and C. If we want to arrange two letters, the order matters: AB, BA, BC, CB, AC, and CA are 6 distinct arrangements (permutations). However, if we only want to select a group of two letters, selecting A and B is identical to selecting B and A. Therefore, there are only 3 possible selections (combinations): AB, BC, and AC. Mathematically, the number of combinations of $n$ distinct objects taken $r$ at a time is denoted by $^nC_r$ and calculated using the formula: $^nC_r = \frac{n!}{r!(n-r)!}$ where $0 \le r \le n$. This links back to permutations via the relation $^nP_r = ^nC_r \times r!$, meaning we first select $r$ objects and then arrange them in $r!$ ways.

Key Algebraic Properties of Combinations

Complementary Selection

The identity $^nC_r = ^nC_{n-r}$. Selecting $r$ objects out of $n$ is mathematically equivalent to rejecting or leaving behind $(n-r)$ objects.

Condition for Equal Combinations

If $^nC_a = ^nC_b$, then either $a = b$ or $a + b = n$. This property is highly tested in CBSE board exams to solve for unknown variables.

Pascal's Rule

The formula $^nC_r + ^nC_{r-1} = ^{n+1}C_r$. This allows us to combine two combination expressions with identical $n$ values and consecutive $r$ values.

Step-by-Step Worked NCERT Examples

1. Solving for n when combinations are equal — Problem: If $^nC_8 = ^nC_2$, find the value of $n$ and determine $^nC_2$. Step 1: Apply the property of equal combinations. If $^nC_a = ^nC_b$, then $a + b = n$ (since $8 \neq 2$). Step 2: Calculate $n = 8 + 2 = 10$. Step 3: Substitute $n = 10$ to find $^{10}C_2$. Step 4: Evaluate using the formula: $^{10}C_2 = \frac{10!}{2!(10-2)!} = \frac{10 \times 9}{2 \times 1} = 45$. Final Answer: $n = 10$ and $^nC_2 = 45$.
2. Selecting with group restrictions — Problem: A committee of 3 persons is to be chosen from a group of 2 men and 3 women. In how many ways can this be done? How many of these committees would consist of 1 man and 2 women? Step 1: To find the total ways, combine the groups. Total people = $2 + 3 = 5$. We need to select any 3. Total ways = $^5C_3 = \frac{5!}{3!2!} = 10$ ways. Step 2: For the restricted case, we must choose 1 man AND 2 women. Number of ways to choose 1 man from 2 = $^2C_1 = 2$. Step 3: Number of ways to choose 2 women from 3 = $^3C_2 = ^3C_1 = 3$. Step 4: Use the Fundamental Product Principle: Total ways = $^2C_1 \times ^3C_2 = 2 \times 3 = 6$ ways.

Common Exam Traps and Calculation Shortcuts

• The Permutation vs Combination Dilemma: Before writing a formula, ask: 'Does changing the order of selected elements create a new outcome?' If yes, use $^nP_r$. If no, use $^nC_r$.
• Factorial Cancellation Shortcut: Do not expand factorials completely. Always expand the larger factorial in the numerator down to the value of the largest factorial in the denominator, then cancel them out directly.
• The AND/OR Rule: Remember that 'AND' represents independent tasks happening together, requiring multiplication (e.g., choosing a batsman AND a bowler). 'OR' represents mutually exclusive alternatives, requiring addition (e.g., choosing either 3 boys OR 3 girls).

Practice Questions with Solutions

• Q: Find $n$ if $^{2n}C_3 : ^nC_3 = 12 : 1$. A: Step 1: Write down the given ratio in equation form: $\frac{^{2n}C_3}{^nC_3} = 12$ Step 2: Expand both combinations using the formula: $\frac{\frac{(2n)!}{3!(2n-3)!}}{\frac{n!}{3!(n-3)!}} = 12$ Step 3: Cancel out the common factor $3!$ from the denominators: $\frac{(2n)(2n-1)(2n-2)(2n-3)!}{(2n-3)!} \times \frac{(n-3)!}{n(n-1)(n-2)(n-3)!} = 12$ Step 4: Simplify the factorials: $\frac{2n(2n-1)2(n-1)}{n(n-1)(n-2)} = 12$ Step 5: Cancel the non-zero variables $n$ and $(n-1)$: $\frac{4(2n-1)}{n-2} = 12$ Step 6: Cross-multiply and solve for $n$: $4(2n-1) = 12(n-2) \implies 8n - 4 = 12n - 24 \implies 4n = 20 \implies n = 5$. Final answer: $n = 5$
• Q: A bag contains 5 black and 6 red balls. Determine the number of ways in which 2 black and 3 red balls can be selected. A: Step 1: Identify the groups and required selections. We need 2 black balls out of 5, and 3 red balls out of 6. Step 2: Calculate the number of ways to select the black balls: $^5C_2 = \frac{5 \times 4}{2 \times 1} = 10$ Step 3: Calculate the number of ways to select the red balls: $^6C_3 = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 20$ Step 4: Since we need to select black AND red balls simultaneously, multiply the possibilities: Total Ways = $^5C_2 \times ^6C_3 = 10 \times 20 = 200$. Final answer: 200 ways
• Q: How many chords can be drawn through 21 points on a circle? A: Step 1: Understand the geometry of a chord. A chord is formed by joining any two distinct points on the circumference of a circle. Step 2: The order of selecting the points does not matter (chord AB is the same as chord BA). This is a selection problem. Step 3: We need to choose 2 points out of 21. Use the combination formula: $^{21}C_2 = \frac{21!}{2!(21-2)!} = \frac{21 \times 20}{2 \times 1} = 210$. Final answer: 210 chords
• Q: In how many ways can a student choose a program of 5 courses if 9 courses are available and 2 specific courses are compulsory for every student? A: Step 1: Understand the constraints. Out of 9 total courses, 2 specific courses are compulsory. They must be included in the selection. Step 2: Therefore, the student has no choice for those 2 courses. They are already selected. Step 3: The student now only needs to choose the remaining courses: Courses left to select = $5 - 2 = 3$ courses. Step 4: Find the remaining courses available to select from: Available courses remaining = $9 - 2 = 7$ courses. Step 5: Apply combinations to find the selection ways of choosing 3 courses out of 7: $^7C_3 = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = 35$ ways. Final answer: 35 ways

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