Chapter 6 — PERMUTATIONS AND COMBINATIONS

Theory

Factorial is defined as:

\[ n! = n \times (n-1) \times (n-2) \cdots 1 \]

Important identity used in simplification:

\[ n! = n \times (n-1)! \]

This identity helps in factoring expressions like \(4! - 3!\).

Solution Roadmap

Step 1: Expand factorial using definition
Step 2: Use identity \(n! = n \cdot (n-1)!\)
Step 3: Factor common terms
Step 4: Simplify numerically

Solution

(i)

\[ \begin{aligned} 8! &= 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 \\ &= 40320 \end{aligned} \]

(ii)

Using identity \(4! = 4 \times 3!\):

\[ \begin{aligned} 4! - 3! &= 4 \times 3! - 3! \\ &= 3!(4 - 1) \\ &= 3 \times 3! \\ &= 3 \times 3 \times 2 \times 1 \\ &= 18 \end{aligned} \]

Final Answer

(i) \(40320\)
(ii) \(18\)

Exam Significance

This is a fundamental factorial manipulation problem.

• Frequently asked in CBSE boards (direct evaluation)
• Essential for simplifying permutation formulas
• Helps in cancelling terms in JEE-level expressions

Key takeaway: Always try factoring using \(n! = n(n-1)!\) before expanding fully

Theory

Factorials grow very rapidly. Hence, expressions involving factorials should be simplified using identities instead of direct expansion.

Key identity:

\[ n! = n \times (n-1)! \]

This helps in taking common factors and simplifying expressions efficiently.

Solution Roadmap

Step 1: Express larger factorial in terms of smaller one
Step 2: Take common factor
Step 3: Simplify and compare both sides
Step 4: Avoid unnecessary large expansions

Solution

Simplify the left-hand side:

\[ \begin{aligned} 3! + 4! &= 3! + 4 \times 3! \\ &= 3!(1 + 4) \\ &= 3! \times 5 \\ &= 5 \times 3 \times 2 \times 1 \\ &= 30 \end{aligned} \]

Now evaluate the right-hand side:

\[ \begin{aligned} 7! &= 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 \\ &= 5040 \end{aligned} \]

Comparing both:

\[ 3! + 4! = 30 \quad \text{and} \quad 7! = 5040 \]

Hence,

\[ 3! + 4! \ne 7! \]

Final Answer

\(3! + 4! \ne 7!\)

Exam Significance

This question tests factorial simplification strategy, not calculation speed.

• Common CBSE board type question
• Builds habit of factoring instead of expanding
• Important for simplifying complex JEE expressions

Key takeaway: Always reduce factorial expressions using identities before comparing

Theory

Expressions involving factorials should be simplified using cancellation before expansion.

Key identity:

\[ n! = n \times (n-1)! \]

Also, this expression is of the form:

\[ \frac{n!}{(n-r)! \, r!} = {^nC_r} \]

Hence, \(\dfrac{8!}{6!2!} = {^8C_2}\), a combination value.

Solution Roadmap

Step 1: Identify cancellation opportunity
Step 2: Expand only required part of factorial
Step 3: Cancel common terms
Step 4: Simplify numerically

Solution

\[ \begin{aligned} \frac{8!}{6! \cdot 2!} &= \frac{8 \times 7 \times 6!}{6! \times 2 \times 1} \\ &= \frac{8 \times 7}{2 \times 1} \\ &= \frac{56}{2} \\ &= 28 \end{aligned} \]

\[ \text{Also, } ^8C_2 = 28 \]

Final Answer

\(28\)

Exam Significance

This is a core simplification pattern in permutations and combinations.

• Direct CBSE board question type
• Very common in JEE for fast simplification
• Helps avoid large factorial calculations

Key takeaway: Never expand fully — always cancel first

Theory

When dealing with factorials in denominators, the key idea is to: convert terms to a common factorial base.

Important identities:

\[ 7! = 7 \times 6!, \quad 8! = 8 \times 7! \]

This allows us to rewrite all terms in compatible forms for simplification.

Solution Roadmap

Step 1: Express all factorials in terms of \(6!\)
Step 2: Take LCM of denominators
Step 3: Simplify numerator
Step 4: Match with RHS and solve for \(x\)

Solution

\[ \begin{aligned} \frac{1}{6!} + \frac{1}{7!} &= \frac{1}{6!} + \frac{1}{7 \times 6!} \\ &= \frac{7 + 1}{7 \times 6!} \\ &= \frac{8}{7!} \end{aligned} \]

Now equate with RHS:

\[ \frac{8}{7!} = \frac{x}{8!} \]

Using \(8! = 8 \times 7!\):

\[ \begin{aligned} \frac{8}{7!} &= \frac{x}{8 \times 7!} \\ 8 \times 8 &= x \\ x &= 64 \end{aligned} \]

Final Answer

\(x = 64\)

Exam Significance

This is a high-frequency simplification pattern in factorial algebra.

• Common in CBSE board exams
• Very important for JEE algebraic simplifications
• Builds skill in handling factorial fractions

Key takeaway: Always convert factorials to a common base before adding or comparing

Theory

The expression

\[ \frac{n!}{(n-r)!} \]

represents permutation, i.e., number of ways to arrange \(r\) objects from \(n\) distinct objects:

\[ ^nP_r = \frac{n!}{(n-r)!} \]

Key idea: Only expand up to \(r\) terms and cancel the rest.

Solution Roadmap

Step 1: Substitute values of \(n\) and \(r\)
Step 2: Expand numerator only till cancellation
Step 3: Cancel common factorial terms
Step 4: Multiply remaining terms

Solution

(i) \(n=6, r=2\)

\[ \begin{aligned} \frac{6!}{(6-2)!} &= \frac{6!}{4!} \\ &= \frac{6 \times 5 \times 4!}{4!} \\ &= 6 \times 5 \\ &= 30 \end{aligned} \]

\[ ^6P_2 = 30 \]

(ii) \(n=9, r=5\)

\[ \begin{aligned} \frac{9!}{(9-5)!} &= \frac{9!}{4!} \\ &= \frac{9 \times 8 \times 7 \times 6 \times 5 \times 4!}{4!} \\ &= 9 \times 8 \times 7 \times 6 \times 5 \\ &= 15120 \end{aligned} \]

\[ ^9P_5 = 15120 \]

Final Answer

(i) \(30\)
(ii) \(15120\)

Exam Significance

This is a direct permutation evaluation pattern.

• Very common in CBSE board exams
• Foundation for all permutation problems
• Used heavily in JEE for fast computation

Key takeaway: \(\frac{n!}{(n-r)!}\) = multiply only first \(r\) descending terms