Maharashtra Board Class 11 Maths Part 2 Chapter 6 Permutations and Combinations 6.7 Solutions

Question 1. Find n if \(^nC_8 = ^nC_{12}\)
Answer:
Solution:
\(^nC_8 = ^nC_{12}\)
If \(^nC_x = ^nC_y\), then either \(x = y\) or \(x = n-y\)
\( \therefore \) \(8 = 12\) or \(8 = n - 12\)
But \(8 = 12\) is not possible
\( \therefore \) \(8 = n - 12\)
\( \therefore \) \(n = 20\)
In simple words: When two combinations with the same 'n' are equal, either the 'r' values are the same, or one 'r' value is 'n' minus the other 'r' value. By applying this rule and eliminating the impossible case, we find the value of 'n'.

🎯 Exam Tip: Remember the property of combinations: if \(^nC_x = ^nC_y\), then either \(x=y\) or \(x=n-y\). This is a fundamental concept for solving such problems efficiently.

Question 2. Find n if \(^{23}C_{3n} = ^{23}C_{2n+3}\)
Answer:
Solution:
\(^{23}C_{3n} = ^{23}C_{2n+3}\)
If \(^nC_x = ^nC_y\), then either \(x = y\) or \(x = n-y\)
\( \therefore \) \(3n = 2n + 3\) or \(3n = 23 - (2n + 3)\)
\( \therefore \) \(n = 3\) or \(3n = 23 - 2n - 3\)
\( \therefore \) \(n = 3\) or \(5n = 20\)
\( \therefore \) \(n = 3\) or \(n = 4\)
In simple words: Using the property that if \(^nC_x = ^nC_y\), then \(x=y\) or \(x=n-y\), we set up two equations for 'n'. Solving these linear equations gives the possible values for 'n'.

🎯 Exam Tip: Always consider both possibilities when applying the combination equality property. Sometimes one solution is rejected based on context, but here both are valid.

Question 3. Find n if \(^{21}C_{6n} = ^{21}C_{n^2+5}\)
Answer:
Solution:
\(^{21}C_{6n} = ^{21}C_{n^2+5}\)
If \(^nC_x = ^nC_y\), then either \(x = y\) or \(x = n-y\)
\( \therefore \) \(6n = n^2 + 5\) or \(6n = 21 - (n^2 + 5)\)
\( \therefore \) \(n^2 - 6n + 5 = 0\) or \(6n = 21 - n^2 - 5\)
\( \therefore \) \(n^2 - 6n + 5 = 0\) or \(n^2 + 6n - 16 = 0\)
If \(n^2 - 6n + 5 = 0\) then \((n - 1)(n - 5) = 0\)
\( \therefore \) \(n = 1\) or \(n = 5\)
If \(n = 5\) then \(n^2 + 5 = 30 > 21\)
\( \therefore n \neq 5\)
\( \therefore n = 1\)
If \(n^2 + 6n - 16 = 0\) then \((n + 8)(n - 2) = 0\)
\(n = -8\) or \(n = 2\)
\(n \neq -8\)
\( \therefore n = 2\)
In simple words: We apply the combination equality rule to form two quadratic equations. Solving these equations gives several potential values for 'n', but some are rejected because the 'r' value (subscript) in \(^nC_r\) cannot exceed 'n', or 'n' cannot be negative. The valid solutions for 'n' are 1 and 2.

🎯 Exam Tip: When solving for 'n' in combinations, always check the validity of the solutions against the definition of combinations (\(n \ge r\) and \(n \ge 0\)). Negative values or values where \(r > n\) must be rejected.

Question 4. Find n if \(^{2n}C_{r-1} = ^{2n}C_{r+1}\)
Answer:
Solution:
\(^{2n}C_{r-1} = ^{2n}C_{r+1}\)
If \(^nC_x = ^nC_y\), then either \(x = y\) or \(x = n-y\)
\( \therefore \) \(r-1 = r+1\) or \(r - 1 = 2n - (r + 1)\)
But \(r - 1 = r + 1\) is not possible
\( \therefore \) \(r - 1 = 2n - r - 1\)
\( \therefore \) \(r + r = 2n\)
\( \therefore \) \(2r = 2n\)
\( \therefore \) \(r = n\)
In simple words: By equating the combination terms, we use the property \(^nC_x = ^nC_y \implies x=y\) or \(x=n-y\). The first case \(x=y\) leads to an impossible result. The second case, \(x=n-y\), simplifies to \(r=n\), which is the solution.

🎯 Exam Tip: Be quick to identify and discard impossible mathematical statements, like \(r-1 = r+1\). This saves time and focuses your effort on the correct path to the solution.

Question 5. Find n if \(^nC_{n-2} = 15\)
Answer:
Solution:
\(^nC_{n-2} = 15\)
\( \therefore ^nC_2 = 15\) .......[\( \because ^nC_r = ^nC_{n-r}\)]
\( \therefore \frac{n!}{(n-2)!2!} = 15\)
\( \therefore \frac{n(n-1)(n-2)!}{(n-2)!2 \times 1} = 15\)
\( \therefore \frac{n(n-1)}{2} = 15\)
\( \therefore n(n - 1) = 30\)
\( \therefore n(n - 1) = 6 \times 5\)
Comparing both sides, we get
\( \therefore n = 6\)
In simple words: We first simplify \(^nC_{n-2}\) to \(^nC_2\) using the property \(^nC_r = ^nC_{n-r}\). Then, we expand \(^nC_2\) using its factorial definition, leading to a quadratic equation. By inspection or solving the quadratic, we find that \(n=6\).

🎯 Exam Tip: Recognizing the property \(^nC_r = ^nC_{n-r}\) can simplify complex combination expressions, making them easier to solve. Also, for small numbers, direct comparison can be faster than solving a full quadratic equation.

Question 6. Find x if \(^nP_r = x \: ^nC_r\)
Answer:
Solution:
\(^nP_r = x(^nC_r)\)
\(x = \frac{^nP_r}{^nC_r}\)
\( = \frac{n!/(n-r)!}{n!/((n-r)!r!)}\)
\( = r!\)
In simple words: The problem asks to find the factor 'x' that relates permutations (\(^nP_r\)) and combinations (\(^nC_r\)). By substituting their respective factorial formulas and simplifying, we find that 'x' is equal to \(r!\).

🎯 Exam Tip: This question tests a fundamental relationship between permutations and combinations: \(^nP_r = r! \times ^nC_r\). Knowing this formula directly can save significant calculation time.

Question 7. Find r if \(^{11}C_4 + ^{11}C_5 + ^{12}C_6 + ^{13}C_7 = ^{14}C_r\)
Answer:
Solution:
\(^{11}C_4 + ^{11}C_5 + ^{12}C_6 + ^{13}C_7 = ^{14}C_r\)
\( \therefore (^{11}C_4 + ^{11}C_5) + ^{12}C_6 + ^{13}C_7 = ^{14}C_r\)
...[\( \because ^nC_r + ^nC_{r-1} = ^{n+1}C_r\)]
\( \therefore (^{12}C_5 + ^{12}C_6) + ^{13}C_7 = ^{14}C_r\)
\( \therefore (^{13}C_6 + ^{13}C_7) = ^{14}C_r\)
\( \therefore ^{14}C_7 = ^{14}C_r\)
If \(^nC_x = ^nC_y\), then either \(x = y\) or \(x = n-y\)
\( \therefore r = 7\) or \(r = 14-7\)
\( \therefore r = 7\)
In simple words: We repeatedly apply Pascal's Identity (\(^nC_r + ^nC_{r-1} = ^{n+1}C_r\)) to combine the sum of combination terms on the left side. This simplifies the entire expression to \(^{14}C_7\), which is then equated to \(^{14}C_r\) to find that \(r=7\).

🎯 Exam Tip: Pascal's Identity is crucial for simplifying sums of combinations. Recognize its pattern to efficiently solve such problems. Remember to check both possibilities for 'r' when \(^nC_x = ^nC_y\).

Question 9. Find the differences between the largest values in the following:
(i) \(^{14}C_r - ^{12}C_r\)
Answer:
Solution:
Greatest value of \(^{14}C_r\)
Here \(n = 14\), which is even
Greatest value of \(^nC_r\) occurs at \(r = \frac{n}{2}\) if \(n\) is even
\( \therefore r = \frac{n}{2}\)
\( \therefore r = \frac{14}{2} = 7\)
\( \therefore \) Greatest value of \(^{14}C_r = ^{14}C_7 = \frac{14!}{7!7!}\)
\( = \frac{14 \times 13 \times 12 \times 11 \times 10 \times 9 \times 8 \times 7!}{7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 \times 7!}\)
\( = 3432\)
Also, for greatest value of \(^{12}C_r\)
\(n = 12\), which is even
\( \therefore r = \frac{12}{2} = 6\)
\( \therefore ^{12}C_6 = \frac{12!}{6!6!}\)
\( = \frac{12 \times 11 \times 10 \times 9 \times 8 \times 7 \times 6!}{6 \times 5 \times 4 \times 3 \times 2 \times 1 \times 6!}\)
\( = 924\)
\( \therefore \) Difference between the greatest values of \(^{14}C_r\) and \(^{12}C_r = 3432 - 924 = 2508\)
In simple words: To find the greatest value of \(^nC_r\) for an even 'n', 'r' is taken as \(n/2\). We calculate the greatest values for \(^{14}C_r\) (which is \(^{14}C_7\)) and \(^{12}C_r\) (which is \(^{12}C_6\)) separately using their factorial formulas, then find their difference.

🎯 Exam Tip: Remember the rule for finding the maximum value of \(^nC_r\): if 'n' is even, it's \(^nC_{n/2}\); if 'n' is odd, it's \(^nC_{(n-1)/2}\) or \(^nC_{(n+1)/2}\). Factorial calculations require careful simplification.

(ii) \(^{13}C_r - ^8C_r\)
Answer:
Solution:
Greatest value of \(^{13}C_r\)
Here \(n = 13\), which is odd
Greatest value of \(^nC_r\) occurs at \(r = \frac{n-1}{2}\) if \(n\) is odd
\( \therefore r = \frac{n-1}{2}\)
\( \therefore r = \frac{13-1}{2} = 6\)
\( \therefore \) Greatest value of \(^{13}C_r = ^{13}C_6 = \frac{13!}{6!7!}\)
\( = \frac{13 \times 12 \times 11 \times 10 \times 9 \times 8 \times 7!}{6 \times 5 \times 4 \times 3 \times 2 \times 1 \times 7!}\)
\( = 1716\)
Also, for greatest value of \(^8C_r\)
\(n = 8\), which is even
\( \therefore r = \frac{8}{2} = 4\)
\( \therefore ^8C_4 = \frac{8!}{4!4!}\)
\( = \frac{8 \times 7 \times 6 \times 5 \times 4!}{4 \times 3 \times 2 \times 1 \times 4!}\)
\( = 70\)
\( \therefore \) Difference between the greatest values of \(^{13}C_r\) and \(^8C_r = 1716 - 70 = 1646\)
In simple words: For \(^{13}C_r\) (odd 'n'), the greatest value occurs at \(r=(13-1)/2 = 6\). For \(^8C_r\) (even 'n'), it occurs at \(r=8/2 = 4\). We calculate \(^{13}C_6\) and \(^8C_4\) using factorial formulas and then find their difference.

🎯 Exam Tip: Correctly identifying 'r' for maximum combination value based on whether 'n' is odd or even is crucial. Simplify factorials by canceling common terms to avoid large number calculations.

(iii) \(^{15}C_r - ^{11}C_r\)
Answer:
Solution:
Greatest value of \(^{15}C_r\)
Here \(n = 15\), which is odd
Greatest value of \(^nC_r\) occurs at \(r = \frac{n-1}{2}\) if \(n\) is odd
\( \therefore r = \frac{n-1}{2}\)
\( \therefore r = \frac{15-1}{2} = 7\)
\( \therefore \) Greatest value of \(^{15}C_r = ^{15}C_7 = \frac{15!}{7!8!}\)
\( = \frac{15 \times 14 \times 13 \times 12 \times 11 \times 10 \times 9 \times 8!}{7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 \times 8!}\)
\( = 6435\)
Also, for greatest value of \(^{11}C_r\)
\(n = 11\), which is odd
\( \therefore r = \frac{11-1}{2} = 5\)
\( \therefore ^{11}C_5 = \frac{11!}{5!6!}\)
\( = \frac{11 \times 10 \times 9 \times 8 \times 7 \times 6!}{5 \times 4 \times 3 \times 2 \times 1 \times 6!}\)
\( = 462\)
Difference between the greatest values of \(^{15}C_r\) and \(^{11}C_r = 6435 - 462 = 5973\)
In simple words: We find the 'r' values that yield the maximum for \(^{15}C_r\) (odd 'n', so \(r=(15-1)/2 = 7\)) and \(^{11}C_r\) (odd 'n', so \(r=(11-1)/2 = 5\)). After calculating \(^{15}C_7\) and \(^{11}C_5\) using their factorial definitions, we compute their difference.

🎯 Exam Tip: Always state the rule for determining 'r' for the greatest combination value. Show the step-by-step factorial expansion and cancellation to arrive at the correct numerical result, minimizing calculation errors.

Question 10. In how many ways can a boy invite his 5 friends to a party so that at least three join the party?
Answer:
Solution:
Boy can invite = (3 or 4 or 5 friends)
Consider the following table:

\( \therefore \) Number of ways a boy can invite his friends to a party so that three or more join the party = \(10 + 5 + 1 = 16\)
In simple words: "At least three" means 3 friends, 4 friends, or 5 friends can join. We calculate the number of ways for each scenario using combinations (\(^5C_3\), \(^5C_4\), \(^5C_5\)) and then sum these possibilities to get the total number of ways.

🎯 Exam Tip: When "at least" conditions are given, break the problem into mutually exclusive cases (e.g., exactly 3, exactly 4, exactly 5) and sum the combinations for each case. Organize your work clearly, perhaps using a table.

Question 11. A group consists of 9 men and 6 women. A team of 6 is to be selected. How many possible selections will have at least 3 women?
Answer:
Solution:
There are 9 men and 6 women.
A team of 6 persons is to be formed such that it consists of at least 3 women.

\( \therefore \) Number of ways this can be done = \(1680 + 540 + 54 + 1 = 2275\)
\( \therefore \) 2275 teams can be formed if team consists of at least 3 women.
In simple words: To select a team of 6 with at least 3 women from 9 men and 6 women, we consider cases where there are exactly 3, 4, 5, or 6 women. For each case, we select the required number of women and the remaining members as men, multiply their combinations, and then sum the results from all cases.

🎯 Exam Tip: When forming teams with "at least" constraints, clearly define all possible combinations of men and women that satisfy the condition. Use combinations for selection and multiply the results for different groups (men and women).

Question 12. A committee of 10 persons is to be formed from a group of 10 women and 8 men How many possible committees will have at least 5 women? How many possible committees will have men in the majority?
Answer:
Solution:
(i) A committee of 10 persons is to be formed from 10 women and 8 men such that the committee contains at least 5 women
Consider the following table:

\( \therefore \) Number of committees = \(14112 + 14700 + 6720 + 1260 + 80 + 1 = 36873\)
\( \therefore \) At least 5 women are there in 36873 committees.

(ii) Number of committees with men in majority = Total number of committees - (Number of committees with women in majority + women and men equal in number)
Total committee members = 10 women + 8 men = 18 members. Team size is 10.
Total ways to form a committee of 10 from 18 persons = \(^{18}C_{10}\).
\(^{18}C_{10} = \frac{18 \times 17 \times 16 \times 15 \times 14 \times 13 \times 12 \times 11}{8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1} = 43758\)
Number of committees with women in majority + women and men equal in number: This sum is the result from part (i), which is 36873.
Number of committees with men in majority = \(43758 - 36873 = 6885\)
In simple words: For part (i), we list all combinations of men and women (totaling 10 people) where there are 5 or more women, calculate each combination, and sum them. For part (ii), we find the total number of ways to form a 10-person committee from 18 people. Then, to find committees with a male majority, we subtract the number of committees with a female majority or an equal number of men and women (which is the result from part (i)) from the total number of possible committees.

🎯 Exam Tip: For "at least" questions, systematically enumerate all valid cases. For "majority" questions, consider using the complementary counting principle: Total ways - (ways for the opposite majority + ways for equal numbers). This often simplifies calculations.

Question 13. A question paper has two sections. Section I has 5 questions and section II has 6 questions. A student must answer at least two questions from each section among 6 questions he answers. How many different choices does the student have in choosing questions?
Answer:
Solution:
There are 11 questions, out of which 5 questions are from section I and 6 questions are from section II.
The student has to select 6 questions taking at least 2 questions from each section.
Consider the following table:

\( \therefore \) Number of choices = \(150 + 200 + 75 = 425\)
\( \therefore \) In 425 ways students can select 6 questions, taking at least 2 questions from each section.
In simple words: To select 6 questions with at least 2 from each section, we identify all valid combinations of questions from Section I (5 questions) and Section II (6 questions). These are (2 from Sec I, 4 from Sec II), (3 from Sec I, 3 from Sec II), and (4 from Sec I, 2 from Sec II). We calculate the combinations for each case and sum them up.

🎯 Exam Tip: For multi-section selection problems with "at least" conditions, determine the minimum and maximum questions possible from each section to reach the total required. List all valid distributions of questions across sections and calculate combinations for each, then sum them.

Question 14. There are 3 wicketkeepers and 5 bowlers among 22 cricket players. A team of 11 players is to be selected so that there is exactly one wicketkeeper and at least 4 bowlers in the team. How many different teams can be formed?
Answer:
Solution:
There are 22 cricket players, of which 3 are wicketkeepers and 5 are bowlers.
Remaining players = \(22 - (3 + 5) = 14\). These 14 are general players.
A team of 11 players is to be chosen such that exactly one wicketkeeper and at least 4 bowlers are to be included in the team.
Consider the following table:

\( \therefore \) Number of ways a team of 11 players can be selected = \(45045 + 6006 = 51051\)
In simple words: We must select exactly one wicketkeeper from 3. For bowlers, "at least 4" means either 4 or 5 bowlers (since there are only 5 available). The remaining players needed to make a team of 11 are selected from the general players. We calculate the combinations for each case (1 WK, 4 B, 6 P; and 1 WK, 5 B, 5 P) and sum the results.

🎯 Exam Tip: Break down the total pool of players into distinct categories (wicketkeepers, bowlers, general players). Satisfy exact conditions first, then "at least" conditions by considering all valid cases, ensuring the total team size is met in each case.

Question 15. Five students are selected from 11. How many ways can these students be selected if
(i) two specified students are selected?
(ii) two specified students are not selected?
Answer:
Solution:
5 students are to be selected from 11 students
(i) When 2 specified students are included
then remaining 3 students can be selected from \((11 - 2) = 9\) students.
\( \therefore \) Number of ways of selecting 3 students from 9 students = \(^9C_3\)
\( = \frac{9!}{3! \times 6!}\)
\( = \frac{9 \times 8 \times 7 \times 6!}{3 \times 2 \times 1 \times 6!}\)
\( = 84\)
\( \therefore \) Selection of students is done in 84 ways when 2 specified students are selected.

(ii) When 2 specified students are not included then 5 students can be selected from the remaining \((11 - 2) = 9\) students
\( \therefore \) Number of ways of selecting 5 students from 9 students = \(^9C_5\)
\( = \frac{9!}{5!4!}\)
\( = \frac{9 \times 8 \times 7 \times 6 \times 5!}{5! \times 4 \times 3 \times 2 \times 1}\)
\( = 126\)
\( \therefore \) Selection of students is done in 126 ways when 2 specified students are not selected.
In simple words: For part (i), if 2 specific students are already chosen, we just need to select 3 more from the remaining 9 students. For part (ii), if 2 specific students are excluded, we need to select all 5 students from the remaining 9 students. Both scenarios involve straightforward combination calculations.

🎯 Exam Tip: When certain individuals are 'specified' (must be included or excluded), adjust the total number of items 'n' and the number of items to choose 'r' accordingly before applying the combination formula. This is a common pattern in selection problems.