Maharashtra Board Class 11 Maths Part 2 Chapter 6 Permutations and Combinations 6.2 Solutions

Exercise 6.2

Question 1. Evaluate:
(i) \(8!\)
Answer: \(8!\) \( = 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 \) \( = 40320 \)
In simple words: To evaluate a factorial, multiply the number by every positive integer less than it down to 1. Here, \(8!\) means multiplying all integers from 8 down to 1.

🎯 Exam Tip: Understanding factorial calculations is fundamental. Ensure accurate multiplication for full marks.

(ii) \(6!\)
Answer: \(6!\) \( = 6 \times 5 \times 4 \times 3 \times 2 \times 1 \) \( = 720 \)
In simple words: \(6!\) is the product of all positive integers from 6 down to 1, representing the number of ways to arrange 6 distinct items.

🎯 Exam Tip: Memorizing common factorials like \(5!\) or \(6!\) can save time in exams.

(iii) \(8! - 6!\)
Answer: \(8! - 6!\) \( = 8 \times 7 \times 6! - 6! \) \( = 6! (8 \times 7 - 1) \) \( = 6! (56 - 1) \) \( = 6! (55) \) \( = (6 \times 5 \times 4 \times 3 \times 2 \times 1) \times 55 \) \( = 720 \times 55 \) \( = 39,600 \)
In simple words: To calculate the difference between two factorials, express the larger factorial in terms of the smaller one to factor it out, then perform the arithmetic.

🎯 Exam Tip: Factoring out common factorials, especially the smaller one, simplifies calculations significantly.

(iv) \((8 - 6)!\)
Answer: \((8-6)!\) \( = 2! \) \( = 2 \times 1 \) \( = 2 \)
In simple words: First perform the subtraction inside the parentheses, then calculate the factorial of the resulting number.

🎯 Exam Tip: Always follow the order of operations (parentheses first) before calculating factorials.

Question 2. Compute:
(i) \( \frac{12!}{6!} \)
Answer: \( \frac{12!}{6!} \) \( = \frac{12 \times 11 \times 10 \times 9 \times 8 \times 7 \times 6!}{6!} \) \( = 12 \times 11 \times 10 \times 9 \times 8 \times 7 \) \( = 665280 \)
In simple words: When dividing factorials, expand the larger factorial until it includes the smaller factorial, then cancel out the common terms.

🎯 Exam Tip: This method avoids calculating very large numbers directly, making the computation much easier and less error-prone.

(ii) \( \left(\frac{12}{6}\right)! \)
Answer: \( \left(\frac{12}{6}\right)! \) \( = 2! \) \( = 2 \times 1 \) \( = 2 \)
In simple words: Simplify the fraction inside the parentheses first, then calculate the factorial of the resulting integer.

🎯 Exam Tip: Always prioritize operations within parentheses before applying the factorial operator.

(iii) \((3 \times 2)!\)
Answer: \((3 \times 2)!\) \( = 6! \) \( = 6 \times 5 \times 4 \times 3 \times 2 \times 1 \) \( = 720 \)
In simple words: First multiply the numbers inside the parentheses, then compute the factorial of the product.

🎯 Exam Tip: Correctly performing the inner operation first is crucial for accurate factorial evaluation.

(iv) \(3! \times 2!\)
Answer: \(3! \times 2!\) \( = (3 \times 2 \times 1) \times (2 \times 1) \) \( = 6 \times 2 \) \( = 12 \)
In simple words: Calculate each factorial separately, then multiply their results.

🎯 Exam Tip: Be careful not to confuse \((3 \times 2)!\) with \(3! \times 2!\); they yield different results.

Question 3. Compute:
(i) \( \frac{9!}{3!6!} \)
Answer: \( \frac{9!}{3!6!} \) \( = \frac{9 \times 8 \times 7 \times 6!}{(3 \times 2 \times 1) \times 6!} \) \( = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} \) \( = 3 \times 4 \times 7 \) \( = 84 \)
In simple words: Expand the numerator's factorial to cancel out the larger factorial in the denominator, then simplify the remaining terms.

🎯 Exam Tip: This technique is widely used in permutations and combinations to simplify calculations involving factorials.

(ii) \( \frac{6!-4!}{4!} \)
Answer: \( \frac{6!-4!}{4!} \) \( = \frac{6 \times 5 \times 4! - 4!}{4!} \) \( = \frac{4!(6 \times 5 - 1)}{4!} \) \( = 6 \times 5 - 1 \) \( = 30 - 1 \) \( = 29 \)
In simple words: Factor out the common factorial from the terms in the numerator, then cancel it with the denominator.

🎯 Exam Tip: Recognizing and factoring out common factorial terms is a key simplification strategy.

(iii) \( \frac{8!}{6!-4!} \)
Answer: \( \frac{8!}{6!-4!} \) \( = \frac{8 \times 7 \times 6 \times 5 \times 4!}{6 \times 5 \times 4! - 4!} \) \( = \frac{8 \times 7 \times 6 \times 5 \times 4!}{4!(6 \times 5 - 1)} \) \( = \frac{8 \times 7 \times 6 \times 5}{30 - 1} \) \( = \frac{1680}{29} \)
In simple words: Expand the factorials to find common terms in both numerator and denominator, then simplify the expression by canceling.

🎯 Exam Tip: Always simplify the denominator first by factoring out the smaller factorial before attempting to divide.

(iv) \( \frac{8!}{(6-4)!} \)
Answer: \( \frac{8!}{(6-4)!} \) \( = \frac{8!}{2!} \) \( = \frac{8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2!}{2!} \) \( = 8 \times 7 \times 6 \times 5 \times 4 \times 3 \) \( = 20160 \)
In simple words: Perform the subtraction in the denominator first, then expand the larger factorial in the numerator to cancel with the denominator's factorial.

🎯 Exam Tip: Correctly simplifying the denominator to a single factorial is the first crucial step here.

Question 4. Write in terms of factorials
(i) \(5 \times 6 \times 7 \times 8 \times 9 \times 10\)
Answer: \(5 \times 6 \times 7 \times 8 \times 9 \times 10\) \( = 10 \times 9 \times 8 \times 7 \times 6 \times 5 \) Multiplying and dividing by \(4!\), we get \( = \frac{10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4!}{4!} \) \( = \frac{10!}{4!} \)
In simple words: To express a product of consecutive integers as factorials, identify the largest number, then multiply and divide by the factorial of the number immediately preceding the smallest number in the product.

🎯 Exam Tip: This technique is useful for converting products into a compact factorial notation, often for further simplification or comparison.

(ii) \(3 \times 6 \times 9 \times 12 \times 15\)
Answer: \(3 \times 6 \times 9 \times 12 \times 15\) \( = (3 \times 1) \times (3 \times 2) \times (3 \times 3) \times (3 \times 4) \times (3 \times 5) \) \( = 3^5 \times (5 \times 4 \times 3 \times 2 \times 1) \) \( = 3^5 \times 5! \)
In simple words: Factor out the common multiplier from each term, then regroup the remaining numbers to form a factorial.

🎯 Exam Tip: Recognizing common factors is key to transforming a product into a factorial expression involving a power.

(iii) \(6 \times 7 \times 8 \times 9\)
Answer: \(6 \times 7 \times 8 \times 9\) \( = 9 \times 8 \times 7 \times 6 \) Multiplying and dividing by \(5!\), we get \( = \frac{9 \times 8 \times 7 \times 6 \times 5!}{5!} \) \( = \frac{9!}{5!} \)
In simple words: Rearrange the terms in descending order, then multiply and divide by the factorial of the missing consecutive numbers to complete the full factorial.

🎯 Exam Tip: This method is applicable when the given product is a part of a larger factorial sequence.

(iv) \(5 \times 10 \times 15 \times 20 \times 25\)
Answer: \(5 \times 10 \times 15 \times 20 \times 25\) \( = (5 \times 1) \times (5 \times 2) \times (5 \times 3) \times (5 \times 4) \times (5 \times 5) \) \( = 5^5 \times (5 \times 4 \times 3 \times 2 \times 1) \) \( = 5^5 \times 5! \)
In simple words: Identify and factor out the common multiplier from each term, then write the product of the remaining integers as a factorial.

🎯 Exam Tip: This problem is similar to part (ii), emphasizing the recognition of common factors to form factorial expressions.

Question 5. Evaluate: \( \frac{n!}{r!(n-r)!} \) for
(i) \(n = 8, r = 6\)
Answer: Given \(n = 8, r = 6\) \( \therefore \frac{n!}{r!(n-r)!} = \frac{8!}{6!(8-6)!} \) \( = \frac{8!}{6!2!} \) \( = \frac{8 \times 7 \times 6!}{6! \times 2 \times 1} \) \( = \frac{8 \times 7}{2 \times 1} \) \( = 4 \times 7 \) \( = 28 \)
In simple words: Substitute the values of \(n\) and \(r\) into the formula, then simplify the factorials by expanding and canceling common terms. This formula represents "n choose r" or combinations.

🎯 Exam Tip: This formula is the core of combination calculations. Practice simplifying it efficiently by canceling factorials.

(ii) \(n = 12, r = 12\)
Answer: Given \(n = 12, r = 12\) \( \therefore \frac{n!}{r!(n-r)!} = \frac{12!}{12!(12-12)!} \) \( = \frac{12!}{12!0!} \) Since \(0! = 1\), \( = \frac{12!}{12! \times 1} \) \( = 1 \)
In simple words: When \(n=r\), the formula simplifies to \(1/0!\), which equals 1, meaning there's only one way to choose all items from a set.

🎯 Exam Tip: Remember that \(0!\) is defined as 1, which is a crucial detail for problems like this.

Question 6. Find \(n\), if
(i) \( \frac{n}{8!} = \frac{3}{6!} + \frac{1}{4!} \)
Answer: \( \frac{n}{8!} = \frac{3}{6!} + \frac{1}{4!} \) To make the denominators common on the right side: \( \frac{n}{8!} = \frac{3}{6!} + \frac{1 \times (6 \times 5)}{4! \times (6 \times 5)} \) \( \frac{n}{8!} = \frac{3}{6!} + \frac{30}{6!} \) \( \frac{n}{8!} = \frac{33}{6!} \) Now, express \(8!\) in terms of \(6!\): \( \frac{n}{8 \times 7 \times 6!} = \frac{33}{6!} \) Multiply both sides by \(8 \times 7 \times 6!\): \( n = \frac{33}{6!} \times 8 \times 7 \times 6! \) \( n = 33 \times 8 \times 7 \) \( n = 33 \times 56 \) \( n = 1848 \)
In simple words: To find \(n\), first simplify the right side by finding a common denominator for the factorials, then isolate \(n\) by cross-multiplication and canceling common factorial terms.

🎯 Exam Tip: Expressing all factorials in terms of the smallest common factorial on one side (like \(6!\) or \(4!\)) helps in simplification.

(ii) \( \frac{n}{8!} = \frac{4}{6!} + \frac{3}{8!} \)
Answer: \( \frac{n}{8!} = \frac{4}{6!} + \frac{3}{8!} \) Bring terms with \(8!\) to one side: \( \frac{n}{8!} - \frac{3}{8!} = \frac{4}{6!} \) \( \frac{n-3}{8!} = \frac{4}{6!} \) Express \(8!\) in terms of \(6!\): \( \frac{n-3}{8 \times 7 \times 6!} = \frac{4}{6!} \) Multiply both sides by \(8 \times 7 \times 6!\): \( n-3 = \frac{4}{6!} \times 8 \times 7 \times 6! \) \( n-3 = 4 \times 8 \times 7 \) \( n-3 = 4 \times 56 \) \( n-3 = 224 \) \( n = 224 + 3 \) \( n = 227 \)
In simple words: Rearrange the equation to group terms with the same factorial, then simplify by canceling factorials to solve for \(n\).

🎯 Exam Tip: Consolidating terms with similar denominators is a crucial first step for solving such equations.

(iii) \( \frac{1}{n!} = \frac{1}{4!} - \frac{4}{5!} \)
Answer: \( \frac{1}{n!} = \frac{1}{4!} - \frac{4}{5!} \) Express \(5!\) in terms of \(4!\) on the right side: \( \frac{1}{n!} = \frac{1}{4!} - \frac{4}{5 \times 4!} \) Find a common denominator: \( \frac{1}{n!} = \frac{5}{5 \times 4!} - \frac{4}{5 \times 4!} \) \( \frac{1}{n!} = \frac{5-4}{5 \times 4!} \) \( \frac{1}{n!} = \frac{1}{5 \times 4!} \) \( \frac{1}{n!} = \frac{1}{5!} \) \( \therefore n! = 5! \) \( \therefore n = 5 \)
In simple words: Simplify the right side by finding a common factorial denominator, then equate the denominators to find the value of \(n\).

🎯 Exam Tip: Working with factorials often involves expressing larger factorials as products involving smaller ones to simplify expressions.

Question 7. Find \(n\), if
(i) \((n + 1)! = 42 \times (n - 1)!\)
Answer: \((n + 1)! = 42(n - 1)!\) Expand \((n+1)!\) until \((n-1)!\): \( (n + 1) n (n - 1)! = 42(n - 1)! \) Divide both sides by \((n-1)!\) (assuming \(n-1 \ge 0 \implies n \ge 1\)): \( n(n + 1) = 42 \) \( n^2 + n - 42 = 0 \) Factorize the quadratic equation: \( n^2 + 7n - 6n - 42 = 0 \) \( n(n + 7) - 6(n + 7) = 0 \) \( (n + 7)(n - 6) = 0 \) So, \(n = -7\) or \(n = 6\). Since \(n\) must be a non-negative integer for factorials, we discard \(n = -7\). Therefore, \(n = 6\). Alternatively, by comparing: \( n(n + 1) = 6 \times 7 \) Comparing on both sides, we get \( n = 6 \)
In simple words: Expand the larger factorial to cancel out the smaller factorial, then solve the resulting algebraic equation for \(n\).

🎯 Exam Tip: Always check the validity of \(n\) in factorial problems; \(n\) must be a non-negative integer.

(ii) \((n + 3)! = 110 \times (n + 1)!\)
Answer: \((n + 3)! = 110 \times (n + 1)!\) Expand \((n+3)!\) until \((n+1)!\): \( (n + 3) (n + 2) (n + 1)! = 110 (n + 1)! \) Divide both sides by \((n+1)!\) (assuming \(n+1 \ge 0 \implies n \ge -1\)): \( (n + 3) (n + 2) = 110 \) Recognize that 110 is \(11 \times 10\). So, we have two consecutive integers. \( (n + 3) (n + 2) = 11 \times 10 \) Since \(n+3\) is greater than \(n+2\), we compare: \( n + 3 = 11 \) \( n = 11 - 3 \) \( n = 8 \) (Also, checking \(n+2=10 \implies n=8\), which is consistent.)
In simple words: Expand the larger factorial and cancel the common factorial term, then solve the quadratic equation or compare consecutive factors to find \(n\).

🎯 Exam Tip: Recognizing products of consecutive numbers (like \(110 = 11 \times 10\)) can simplify solving the equation significantly.

Question 8. Find \(n\), if:
(i) \( \frac{n!}{3!(n-3)!} : \frac{n!}{5!(n-5)!} = 5 : 3 \)
Answer: Given the ratio: \( \frac{\frac{n!}{3!(n-3)!}}{\frac{n!}{5!(n-5)!}} = \frac{5}{3} \) Rewrite the division as multiplication by the reciprocal: \( \frac{n!}{3!(n-3)!} \times \frac{5!(n-5)!}{n!} = \frac{5}{3} \) Cancel \(n!\): \( \frac{5!(n-5)!}{3!(n-3)!} = \frac{5}{3} \) Expand \(5!\) as \(5 \times 4 \times 3!\) and \((n-3)!\) as \((n-3)(n-4)(n-5)!\): \( \frac{5 \times 4 \times 3! \times (n-5)!}{3! \times (n-3)(n-4)(n-5)!} = \frac{5}{3} \) Cancel \(3!\) and \((n-5)!\): \( \frac{5 \times 4}{(n-3)(n-4)} = \frac{5}{3} \) \( \frac{20}{(n-3)(n-4)} = \frac{5}{3} \) Cross-multiply: \( 20 \times 3 = 5 \times (n-3)(n-4) \) \( 60 = 5 \times (n-3)(n-4) \) Divide by 5: \( 12 = (n-3)(n-4) \) We need two consecutive integers whose product is 12. These are 4 and 3. So, \( (n-3)(n-4) = 4 \times 3 \) Since \((n-3)\) is greater than \((n-4)\), we compare: \( n-3 = 4 \) \( n = 7 \) (Also, checking \(n-4=3 \implies n=7\), which is consistent.)
In simple words: Convert the ratio into a fraction, simplify by canceling common factorials, then solve the resulting algebraic equation to find \(n\).

🎯 Exam Tip: Careful expansion and cancellation of factorial terms, especially those with variables like \((n-3)!\), are crucial for solving these types of problems.

(ii) \( \frac{n!}{3!(n-5)!} : \frac{n!}{5!(n-7)!} = 10 : 3 \)
Answer: Given the ratio: \( \frac{\frac{n!}{3!(n-5)!}}{\frac{n!}{5!(n-7)!}} = \frac{10}{3} \) Rewrite the division as multiplication by the reciprocal: \( \frac{n!}{3!(n-5)!} \times \frac{5!(n-7)!}{n!} = \frac{10}{3} \) Cancel \(n!\): \( \frac{5!(n-7)!}{3!(n-5)!} = \frac{10}{3} \) Expand \(5!\) as \(5 \times 4 \times 3!\) and \((n-5)!\) as \((n-5)(n-6)(n-7)!\): \( \frac{5 \times 4 \times 3! \times (n-7)!}{3! \times (n-5)(n-6)(n-7)!} = \frac{10}{3} \) Cancel \(3!\) and \((n-7)!\): \( \frac{5 \times 4}{(n-5)(n-6)} = \frac{10}{3} \) \( \frac{20}{(n-5)(n-6)} = \frac{10}{3} \) Cross-multiply: \( 20 \times 3 = 10 \times (n-5)(n-6) \) \( 60 = 10 \times (n-5)(n-6) \) Divide by 10: \( 6 = (n-5)(n-6) \) We need two consecutive integers whose product is 6. These are 3 and 2. So, \( (n-5)(n-6) = 3 \times 2 \) Since \((n-5)\) is greater than \((n-6)\), we compare: \( n-5 = 3 \) \( n = 8 \) (Also, checking \(n-6=2 \implies n=8\), which is consistent.)
In simple words: Simplify the ratio expression by canceling common factorial terms, then solve the resulting quadratic equation by finding two consecutive integers whose product matches the simplified value.

🎯 Exam Tip: Always ensure that the value of \(n\) obtained satisfies the condition that all factorial terms \((n-k)!\) are valid (i.e., \(n-k \ge 0\)).

Question 9. Find \(n\), if:
(i) \( \frac{(17-n)!}{(14-n)!} = 5! \)
Answer: \( \frac{(17-n)!}{(14-n)!} = 5! \) Expand \((17-n)!\) until \((14-n)!\): \( \frac{(17-n)(16-n)(15-n)(14-n)!}{(14-n)!} = 5! \) Cancel \((14-n)!\): \( (17-n)(16-n)(15-n) = 5! \) \( (17-n)(16-n)(15-n) = 5 \times 4 \times 3 \times 2 \times 1 \) \( (17-n)(16-n)(15-n) = 120 \) Notice that 120 can be written as \(6 \times 5 \times 4\). So, \( (17-n)(16-n)(15-n) = 6 \times 5 \times 4 \) Comparing the terms, since \(17-n > 16-n > 15-n\), we equate the largest terms: \( 17-n = 6 \) \( n = 17 - 6 \) \( n = 11 \) (Check: \(16-11=5\) and \(15-11=4\). This is consistent.)
In simple words: Expand the larger factorial until it can cancel with the smaller one, then solve the resulting equation by comparing the product of consecutive terms.

🎯 Exam Tip: Recognizing sequences of consecutive numbers in factorial equations allows for direct comparison and quicker solutions.

(ii) \( \frac{(15-n)!}{(13-n)!} = 12 \)
Answer: \( \frac{(15-n)!}{(13-n)!} = 12 \) Expand \((15-n)!\) until \((13-n)!\): \( \frac{(15-n)(14-n)(13-n)!}{(13-n)!} = 12 \) Cancel \((13-n)!\): \( (15-n)(14-n) = 12 \) We need two consecutive integers whose product is 12. These are 4 and 3. So, \( (15-n)(14-n) = 4 \times 3 \) Since \((15-n)\) is greater than \((14-n)\), we compare: \( 15-n = 4 \) \( n = 15 - 4 \) \( n = 11 \) (Check: \(14-11=3\). This is consistent.)
In simple words: Expand the factorial in the numerator to cancel the factorial in the denominator, then solve the resulting equation for \(n\) by matching products of consecutive integers.

🎯 Exam Tip: Simplify the factorial expressions first, then look for patterns of consecutive integers to solve for the variable \(n\).

Question 10. Find \(n\) if \( \frac{(2n)!}{7!(2n-7)!} : \frac{n!}{4!(n-4)!} = 24 : 1 \)
Answer: Given the ratio: \( \frac{\frac{(2n)!}{7!(2n-7)!}}{\frac{n!}{4!(n-4)!}} = \frac{24}{1} \) Rewrite the division as multiplication by the reciprocal: \( \frac{(2n)!}{7!(2n-7)!} \times \frac{4!(n-4)!}{n!} = 24 \) Expand terms to cancel: Numerator of first fraction: \( (2n)(2n-1)(2n-2)(2n-3)(2n-4)(2n-5)(2n-6)(2n-7)! \) Denominator of first fraction: \( 7 \times 6 \times 5 \times 4! \times (2n-7)! \) Numerator of second fraction: \( 4! (n-4)! \) Denominator of second fraction: \( n(n-1)(n-2)(n-3)(n-4)! \) So, substituting these expansions: \( \frac{(2n)(2n-1)(2n-2)(2n-3)(2n-4)(2n-5)(2n-6)(2n-7)!}{7 \times 6 \times 5 \times 4! \times (2n-7)!} \times \frac{4!(n-4)!}{n(n-1)(n-2)(n-3)(n-4)!} = 24 \) Cancel \((2n-7)!\) and \(4!\) and \((n-4)!\): \( \frac{(2n)(2n-1)(2n-2)(2n-3)(2n-4)(2n-5)(2n-6)}{7 \times 6 \times 5} \times \frac{1}{n(n-1)(n-2)(n-3)} = 24 \) Factor out 2 from even terms in the numerator: \( (2n) = 2 \times n \) \( (2n-2) = 2 \times (n-1) \) \( (2n-4) = 2 \times (n-2) \) \( (2n-6) = 2 \times (n-3) \) Substitute these: \( \frac{(2n)(2n-1)2(n-1)(2n-3)2(n-2)(2n-5)2(n-3)}{7 \times 6 \times 5 \times n(n-1)(n-2)(n-3)} = 24 \) Combine the factors of 2: \( \frac{2^4 n(n-1)(n-2)(n-3)(2n-1)(2n-3)(2n-5)}{7 \times 6 \times 5 \times n(n-1)(n-2)(n-3)} = 24 \) Cancel \(n(n-1)(n-2)(n-3)\): \( \frac{16 (2n-1)(2n-3)(2n-5)}{7 \times 6 \times 5} = 24 \) \( \frac{16 (2n-1)(2n-3)(2n-5)}{210} = 24 \) \( 16 (2n-1)(2n-3)(2n-5) = 24 \times 210 \) \( 16 (2n-1)(2n-3)(2n-5) = 5040 \) \( (2n-1)(2n-3)(2n-5) = \frac{5040}{16} \) \( (2n-1)(2n-3)(2n-5) = 315 \) We need three terms in arithmetic progression with a common difference of 2 whose product is 315. Notice that \(315 = 9 \times 35 = 9 \times 7 \times 5\). So, \( (2n-1)(2n-3)(2n-5) = 9 \times 7 \times 5 \) Comparing the terms: \( 2n-1 = 9 \) \( 2n = 10 \) \( n = 5 \) (Check: \(2(5)-3 = 7\) and \(2(5)-5 = 5\). This is consistent.)
In simple words: Expand the factorial terms in the given ratio expression, cancel common factors and variable terms, then simplify the equation to find a product of consecutive odd numbers to solve for \(n\).

🎯 Exam Tip: This problem requires meticulous algebraic manipulation and careful cancellation of terms. Pay close attention to factoring out powers of 2 and identifying patterns of consecutive odd numbers.

Question 11. Show that \( \frac{n!}{r!(n-r)!} + \frac{n!}{(r-1)!(n-r+1)!} = \frac{(n+1)!}{r!(n-r+1)!} \)
Answer: L.H.S. \( = \frac{n!}{r!(n-r)!} + \frac{n!}{(r-1)!(n-r+1)!} \) To add these fractions, we need a common denominator. We know \(r! = r \times (r-1)!\) and \((n-r+1)! = (n-r+1) \times (n-r)!\). So, the common denominator will be \(r!(n-r+1)!\). L.H.S. \( = \frac{n!}{r \times (r-1)! \times (n-r)!} + \frac{n!}{(r-1)! \times (n-r+1) \times (n-r)!} \) Factor out common terms \( \frac{n!}{(r-1)!(n-r)!} \): L.H.S. \( = \frac{n!}{(r-1)!(n-r)!} \left[ \frac{1}{r} + \frac{1}{n-r+1} \right] \) Combine the terms inside the bracket: L.H.S. \( = \frac{n!}{(r-1)!(n-r)!} \left[ \frac{(n-r+1) + r}{r(n-r+1)} \right] \) L.H.S. \( = \frac{n!}{(r-1)!(n-r)!} \left[ \frac{n+1}{r(n-r+1)} \right] \) Rearrange the terms: L.H.S. \( = \frac{(n+1) \times n!}{r \times (r-1)! \times (n-r+1) \times (n-r)!} \) Recognize that \((n+1) \times n! = (n+1)!\) and \(r \times (r-1)! = r!\) and \((n-r+1) \times (n-r)! = (n-r+1)!\): L.H.S. \( = \frac{(n+1)!}{r!(n-r+1)!} \) This is equal to R.H.S. Hence, shown.
In simple words: To prove the identity, express the terms on the left-hand side with a common factorial denominator, combine them, and then use factorial properties to simplify the expression into the right-hand side.

🎯 Exam Tip: This identity is an important property of combinations, often represented as \( \binom{n}{r} + \binom{n}{r-1} = \binom{n+1}{r} \). Understanding factorial expansion is key to proving it.

Question 12. Show that \( \frac{9!}{3!6!} + \frac{9!}{4!5!} = \frac{10!}{4!6!} \)
Answer: L.H.S. \( = \frac{9!}{3!6!} + \frac{9!}{4!5!} \) Express the denominators in terms of common factorials: \(6! = 6 \times 5!\) and \(4! = 4 \times 3!\). L.H.S. \( = \frac{9!}{3! \times 6 \times 5!} + \frac{9!}{4 \times 3! \times 5!} \) Factor out \( \frac{9!}{3!5!} \): L.H.S. \( = \frac{9!}{3!5!} \left[ \frac{1}{6} + \frac{1}{4} \right] \) Combine the fractions inside the bracket: L.H.S. \( = \frac{9!}{3!5!} \left[ \frac{4+6}{24} \right] \) L.H.S. \( = \frac{9!}{3!5!} \left[ \frac{10}{24} \right] \) L.H.S. \( = \frac{9!}{3!5!} \times \frac{5}{12} \) Now, let's manipulate this to match R.H.S. \( = \frac{10!}{4!6!} \). R.H.S. \( = \frac{10 \times 9!}{(4 \times 3!) \times (6 \times 5!)} \) R.H.S. \( = \frac{10 \times 9!}{4! \times 6!} \) Let's go back to L.H.S. and arrange it to match R.H.S.: L.H.S. \( = \frac{9!}{3!5!} \times \frac{10}{24} \) Multiply and divide by \(4\) in \(3!\) and \(6\) in \(5!\) to get \(4!6!\): L.H.S. \( = \frac{9! \times 10}{ (3! \times 4) \times (5! \times 6) } \) L.H.S. \( = \frac{10 \times 9!}{4! \times 6!} \) L.H.S. \( = \frac{10!}{4!6!} \) This is equal to R.H.S. Hence, shown.
In simple words: To prove this identity, find a common denominator for the left-hand side terms by expressing factorials appropriately, simplify the expression, and then show it equals the right-hand side using factorial definitions.

🎯 Exam Tip: This identity is a specific case of Pascal's identity. Convert all factorials to common terms for easier simplification and comparison.

Question 13. Find the value of:
(i) \( \frac{8!+5(4!)}{4!-12} \)
Answer: \( \frac{8!+5(4!)}{4!-12} \) Express \(8!\) in terms of \(4!\): \(8! = 8 \times 7 \times 6 \times 5 \times 4!\). \( = \frac{(8 \times 7 \times 6 \times 5 \times 4!) + 5(4!)}{4! - 12} \) Factor out \(4!\) from the numerator: \( = \frac{4!(8 \times 7 \times 6 \times 5 + 5)}{4! - 12} \) Calculate \(8 \times 7 \times 6 \times 5 = 1680\). Calculate \(4! = 4 \times 3 \times 2 \times 1 = 24\). \( = \frac{4!(1680 + 5)}{24 - 12} \) \( = \frac{4!(1685)}{12} \) Substitute \(4! = 24\): \( = \frac{24 \times 1685}{12} \) \( = 2 \times 1685 \) \( = 3370 \)
In simple words: To evaluate the expression, expand the larger factorial in terms of the smaller one, factor out common terms, calculate numerical values, and then simplify the fraction.

🎯 Exam Tip: Be careful with the order of operations, especially when mixing factorials and subtraction/addition. Factoring helps prevent errors.

(ii) \( \frac{5(26!)+(27!)}{4(27!)-8(26!)} \)
Answer: \( \frac{5(26!)+(27!)}{4(27!)-8(26!)} \) Express \(27!\) in terms of \(26!\): \(27! = 27 \times 26!\). \( = \frac{5(26!) + 27(26!)}{4(27 \times 26!) - 8(26!)} \) Factor out \(26!\) from both numerator and denominator: \( = \frac{26!(5 + 27)}{26!(4 \times 27 - 8)} \) Cancel \(26!\): \( = \frac{5 + 27}{4 \times 27 - 8} \) \( = \frac{32}{108 - 8} \) \( = \frac{32}{100} \) Simplify the fraction: \( = \frac{8}{25} \)
In simple words: Convert all larger factorials to terms of the smallest factorial in the expression, factor out the common factorial, and then simplify the resulting numerical fraction.

🎯 Exam Tip: This type of problem highlights the efficiency of factoring out common factorial terms to simplify complex expressions quickly.

Question 14. Show that \( \frac{(2n)!}{n!} = 2^n (2n - 1) (2n - 3)...5.3.1 \)
Answer: L.H.S. \( = \frac{(2n)!}{n!} \) Expand \((2n)!\) into its terms: \( (2n)! = (2n)(2n-1)(2n-2)(2n-3)(2n-4)...(6)(5)(4)(3)(2)(1) \) Group the even and odd terms: \( (2n)! = [ (2n)(2n-2)(2n-4)...(4)(2) ] \times [ (2n-1)(2n-3)...(5)(3)(1) ] \) Factor out 2 from each even term: The number of even terms from \(2n\) down to \(2\) is \(n\). So, \( (2n)(2n-2)(2n-4)...(4)(2) = 2^n \times (n)(n-1)(n-2)...(2)(1) \) This is \( 2^n \times n! \) Substitute this back into the expansion of \((2n)!\): \( (2n)! = [ 2^n \times n! ] \times [ (2n-1)(2n-3)...(5)(3)(1) ] \) Now, substitute this into the L.H.S. expression: L.H.S. \( = \frac{2^n \times n! \times (2n-1)(2n-3)...5.3.1}{n!} \) Cancel \(n!\): L.H.S. \( = 2^n (2n - 1)(2n - 3)...5.3.1 \) This is equal to R.H.S. Hence, shown.
In simple words: To prove the identity, expand the factorial \( (2n)! \) into its even and odd components. Factor out a \(2\) from each even term to reveal \(2^n n!\), then cancel \(n!\) from the denominator.

🎯 Exam Tip: This identity is useful for simplifying expressions involving double factorials or products of odd numbers. The key is separating even and odd terms in the factorial expansion.