RBSE Solutions Class 11 Maths Chapter 6 Permutations and Combinations Exercise 6.2

Question 1. Find the value of n, whereas:
(i) \( ^{2n}P_3 : ^nP_3 = 11 : 1 \)
(ii) \( ^{20}C_{n-2} = ^{20}C_{n+2} \)
(iii) \( ^nC_{10} = ^nC_{15} \)
Answer:
(i) We are given the ratio: \( ^{2n}P_3 : ^nP_3 = 11 : 1 \).
This can be written as:
\( \frac{^{2n}P_3}{^nP_3} = \frac{11}{1} \)
We know the formula for permutations: \( ^kP_r = \frac{k!}{(k-r)!} \).
So, \( ^{2n}P_3 = \frac{(2n)!}{(2n-3)!} = 2n(2n-1)(2n-2) \)
And, \( ^nP_3 = \frac{n!}{(n-3)!} = n(n-1)(n-2) \)
Substituting these into the ratio:
\( \frac{2n(2n-1)(2n-2)}{n(n-1)(n-2)} = 11 \)
We can simplify this equation by cancelling common terms. Since n cannot be 0, 1, or 2 for the permutations to be defined, we can divide by \( n \), \( n-1 \), and \( n-2 \). Also, \( (2n-2) = 2(n-1) \).
So, the equation becomes:
\( \frac{2n(2n-1)2(n-1)}{n(n-1)(n-2)} = 11 \)
Now, cancel \( n \) and \( (n-1) \) from both numerator and denominator:
\( \frac{2(2n-1)2}{n-2} = 11 \)
\( \frac{4(2n-1)}{n-2} = 11 \)
Now, cross-multiply:
\( 4(2n-1) = 11(n-2) \)
\( 8n - 4 = 11n - 22 \)
Bring n terms to one side and constants to the other:
\( 22 - 4 = 11n - 8n \)
\( 18 = 3n \)
Divide by 3 to find n:
\( n = \frac{18}{3} \)
\( n = 6 \)
Hence, the value of n is 6.

(ii) We are given the equation: \( ^{20}C_{n-2} = ^{20}C_{n+2} \)
We know that if \( ^kC_a = ^kC_b \), then either \( a = b \) or \( a + b = k \).
Here, \( k = 20 \), \( a = n-2 \), and \( b = n+2 \).
First possibility: \( a = b \)
\( n-2 = n+2 \)
\( -2 = 2 \)
This statement is false, so \( a \) cannot be equal to \( b \).
Second possibility: \( a + b = k \)
\( (n-2) + (n+2) = 20 \)
\( n-2+n+2 = 20 \)
\( 2n = 20 \)
Divide by 2 to find n:
\( n = \frac{20}{2} \)
\( n = 10 \)
Hence, the value of n is 10.

(iii) We are given the equation: \( ^nC_{10} = ^nC_{15} \)
Using the same property as above: if \( ^kC_a = ^kC_b \), then either \( a = b \) or \( a + b = k \).
Here, \( k = n \), \( a = 10 \), and \( b = 15 \).
First possibility: \( a = b \)
\( 10 = 15 \)
This statement is false.
Second possibility: \( a + b = k \)
\( 10 + 15 = n \)
\( 25 = n \)
Hence, the value of n is 25.
In simple words: For permutations, expand the terms, cancel common factors, and solve the resulting algebraic equation for n. For combinations, remember that if two combinations with the same top number (n or k) are equal, then either the bottom numbers are the same, or they add up to the top number.

🎯 Exam Tip: Always remember the property for combinations \( ^nC_r = ^nC_{n-r} \). This is crucial for solving equations like \( ^nC_a = ^nC_b \), leading to \( a = b \) or \( a+b = n \).

Question 2. Find the value of \( ^{50}C_{11} + ^{50}C_{12} + ^{51}C_{13} - ^{52}C_{13} \).
Answer: We need to find the value of the expression: \( ^{50}C_{11} + ^{50}C_{12} + ^{51}C_{13} - ^{52}C_{13} \).
We will use the identity: \( ^nC_r + ^nC_{r-1} = ^{n+1}C_r \).
First, let's group the first two terms: \( (^{50}C_{11} + ^{50}C_{12}) \).
Using the identity, where \( n=50 \) and \( r=12 \):
\( ^{50}C_{11} + ^{50}C_{12} = ^{50+1}C_{12} = ^{51}C_{12} \)
Now, substitute this back into the expression:
The expression becomes: \( ^{51}C_{12} + ^{51}C_{13} - ^{52}C_{13} \).
Next, group the first two terms again: \( (^{51}C_{12} + ^{51}C_{13}) \).
Using the identity, where \( n=51 \) and \( r=13 \):
\( ^{51}C_{12} + ^{51}C_{13} = ^{51+1}C_{13} = ^{52}C_{13} \)
Now, substitute this back into the expression:
The expression becomes: \( ^{52}C_{13} - ^{52}C_{13} \).
When we subtract a term from itself, the result is zero.
\( ^{52}C_{13} - ^{52}C_{13} = 0 \)
Thus, the value of the given expression is 0.
In simple words: We add the first two terms using a special rule for combinations. This gives a new combination. Then we add that new combination to the next term, using the same rule. We keep doing this until we get two identical terms that cancel each other out, resulting in zero.

🎯 Exam Tip: Memorize and correctly apply the combination identity \( ^nC_r + ^nC_{r-1} = ^{n+1}C_r \). This identity helps simplify complex sums of combinations efficiently.

Question 3. In a triangle ABC, there are 3, 4 and 5 points on side AB, BC, CA respectively. How many triangles will be formed by these points?
Answer: To form a triangle, we need to choose 3 non-collinear points. A triangle has three vertices that do not lie on the same straight line.
First, let's find the total number of points available. On side AB there are 3 points, on BC there are 4 points, and on CA there are 5 points.
Total number of points \( = 3 + 4 + 5 = 12 \).
The total number of ways to choose 3 points from these 12 points is given by \( ^{12}C_3 \).
\( ^{12}C_3 = \frac{12!}{3!(12-3)!} = \frac{12!}{3!9!} \)
\( ^{12}C_3 = \frac{12 \times 11 \times 10 \times 9!}{3 \times 2 \times 1 \times 9!} \)
\( ^{12}C_3 = \frac{12 \times 11 \times 10}{3 \times 2 \times 1} \)
\( ^{12}C_3 = 2 \times 11 \times 10 \)
\( ^{12}C_3 = 220 \)
However, we cannot form a triangle if the three chosen points are collinear (lie on the same straight line). In this problem, points on each side of the triangle (AB, BC, CA) are collinear.
Number of ways to choose 3 points from side AB (3 points): \( ^3C_3 \)
\( ^3C_3 = \frac{3!}{3!(3-3)!} = \frac{3!}{3!0!} = 1 \) (Since \( 0! = 1 \))
Number of ways to choose 3 points from side BC (4 points): \( ^4C_3 \)
\( ^4C_3 = \frac{4!}{3!(4-3)!} = \frac{4!}{3!1!} = \frac{4 \times 3!}{3! \times 1} = 4 \)
Number of ways to choose 3 points from side CA (5 points): \( ^5C_3 \)
\( ^5C_3 = \frac{5!}{3!(5-3)!} = \frac{5!}{3!2!} = \frac{5 \times 4 \times 3!}{3! \times 2 \times 1} = \frac{5 \times 4}{2} = 10 \)
The total number of sets of 3 collinear points is the sum of these:
Total collinear sets \( = 1 + 4 + 10 = 15 \).
The number of triangles formed will be the total number of ways to choose 3 points minus the number of ways to choose 3 collinear points.
Number of triangles \( = \text{Total ways to choose 3 points} - \text{Ways to choose 3 collinear points} \)
Number of triangles \( = 220 - 15 \)
Number of triangles \( = 205 \)
Therefore, 205 triangles can be formed.
In simple words: First, count all possible ways to pick any three points from the total. Then, subtract the ways where the three chosen points fall on the same straight line (because these don't form a triangle). The remaining number is how many triangles can be made.

🎯 Exam Tip: When forming geometric figures like triangles or quadrilaterals using points, always remember to subtract the cases where chosen points are collinear, as these do not form the desired figure.

Question 4. A box contains two white, three black and four red balls. Determine the number of ways in which three balls can be selected from this box, in which at least one black ball is compulsory.
Answer: We have a box with:
White balls = 2
Black balls = 3
Red balls = 4
Total balls = \( 2 + 3 + 4 = 9 \).
We need to select 3 balls such that at least one black ball is compulsory. This means we can have 1 black ball, 2 black balls, or 3 black balls.
The remaining balls (white or red) are 2 white + 4 red = 6 balls.
Let's consider the possible cases:

Case 1: Exactly 1 Black ball
Choose 1 black ball from 3: \( ^3C_1 = 3 \)
Choose 2 other balls (from white or red) from 6: \( ^6C_2 = \frac{6 \times 5}{2 \times 1} = 15 \)
Total ways for Case 1 = \( ^3C_1 \times ^6C_2 = 3 \times 15 = 45 \)

Case 2: Exactly 2 Black balls
Choose 2 black balls from 3: \( ^3C_2 = 3 \)
Choose 1 other ball (from white or red) from 6: \( ^6C_1 = 6 \)
Total ways for Case 2 = \( ^3C_2 \times ^6C_1 = 3 \times 6 = 18 \)

Case 3: Exactly 3 Black balls
Choose 3 black balls from 3: \( ^3C_3 = 1 \)
Choose 0 other balls (from white or red) from 6: \( ^6C_0 = 1 \)
Total ways for Case 3 = \( ^3C_3 \times ^6C_0 = 1 \times 1 = 1 \)

Total number of ways to select 3 balls with at least one black ball is the sum of ways from these cases:
Total ways = \( 45 + 18 + 1 = 64 \)
Alternatively, we can find the total ways to select 3 balls from 9 without any restriction, and then subtract the ways where *no* black balls are selected.
Total ways to select 3 balls from 9 = \( ^9C_3 = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} = 3 \times 4 \times 7 = 84 \)
Ways to select 3 balls with *no* black balls: This means choosing 3 balls only from the white and red balls (2 white + 4 red = 6 balls).
Ways to select 3 balls from 6 (non-black) = \( ^6C_3 = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 20 \)
Number of ways with at least one black ball = Total ways - Ways with no black balls
Number of ways \( = 84 - 20 = 64 \)
Both methods yield the same result. The number of ways to select three balls with at least one black ball is 64.
In simple words: We need to pick three balls but always have at least one black one. We can do this by adding up the ways to pick exactly one black ball, exactly two black balls, and exactly three black balls. Or, we can find all possible ways to pick three balls and then subtract the ways where we pick no black balls at all. Both methods give the same answer.

🎯 Exam Tip: For "at least one" problems, it's often simpler to calculate the total possible outcomes and subtract the "none" case, rather than summing up multiple individual "at least" cases.

Question 5. By 6 different coloured flags, taking one or more than one how many ways signal can be given?
Answer: We have 6 different coloured flags. We need to form signals by taking one or more flags.
When we form a signal, the order of flags matters, so this is a permutation problem. For example, a red-blue signal is different from a blue-red signal.
The number of flags we can use for a signal can be 1, 2, 3, 4, 5, or 6.
Number of signals using 1 flag = Permutations of 6 flags taken 1 at a time = \( ^6P_1 = 6 \)
Number of signals using 2 flags = Permutations of 6 flags taken 2 at a time = \( ^6P_2 = 6 \times 5 = 30 \)
Number of signals using 3 flags = Permutations of 6 flags taken 3 at a time = \( ^6P_3 = 6 \times 5 \times 4 = 120 \)
Number of signals using 4 flags = Permutations of 6 flags taken 4 at a time = \( ^6P_4 = 6 \times 5 \times 4 \times 3 = 360 \)
Number of signals using 5 flags = Permutations of 6 flags taken 5 at a time = \( ^6P_5 = 6 \times 5 \times 4 \times 3 \times 2 = 720 \)
Number of signals using 6 flags = Permutations of 6 flags taken 6 at a time = \( ^6P_6 = 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 720 \)
The total number of ways to give a signal is the sum of signals formed using 1, 2, 3, 4, 5, or 6 flags.
Total signals = \( ^6P_1 + ^6P_2 + ^6P_3 + ^6P_4 + ^6P_5 + ^6P_6 \)
Total signals = \( 6 + 30 + 120 + 360 + 720 + 720 \)
Total signals = \( 1956 \)
Therefore, 1956 different signals can be given.
In simple words: Since the flags are different and their order matters for a signal, we use permutations. We calculate how many unique signals can be made using one flag, then two flags, and so on, up to all six flags. Adding all these possibilities together gives the total number of signals.

🎯 Exam Tip: When the order of selection matters (like arranging flags to form a signal), use permutations. When the order doesn't matter (like choosing balls from a bag), use combinations.

Question 6. Number of diagonals in a polygon is 44 then find a number of its sides.
Answer: We are given that the number of diagonals in a polygon is 44.
The formula for the number of diagonals in a polygon with \( n \) sides is given by: \( \text{Number of diagonals} = \frac{n(n-3)}{2} \).
We set the formula equal to the given number of diagonals:
\( \frac{n(n-3)}{2} = 44 \)
Multiply both sides by 2:
\( n(n-3) = 44 \times 2 \)
\( n^2 - 3n = 88 \)
Rearrange the equation to form a standard quadratic equation:
\( n^2 - 3n - 88 = 0 \)
Now, we need to solve this quadratic equation for \( n \). We look for two numbers that multiply to -88 and add to -3. These numbers are -11 and 8.
So, we can factor the quadratic equation:
\( (n-11)(n+8) = 0 \)
This gives us two possible values for \( n \):
\( n-11 = 0 \implies n = 11 \)
\( n+8 = 0 \implies n = -8 \)
Since the number of sides of a polygon cannot be negative, we discard \( n = -8 \).
Therefore, the number of sides of the polygon is 11.
In simple words: We use the formula that connects the number of sides of a polygon to its number of diagonals. We put the given number of diagonals into this formula and solve the resulting equation to find how many sides the polygon has. We ignore any negative answers because a polygon cannot have negative sides.

🎯 Exam Tip: Remember the formula for the number of diagonals in a polygon, \( \frac{n(n-3)}{2} \). Always check for logical constraints, such as the number of sides being positive, when solving for \( n \).

Question 7. How many numbers can be formed taking 4 digits from digits 1, 2, 3, 4, 5, 6 whereas digit 4 and 5 are compulsory?
Answer: We have a set of 6 digits: {1, 2, 3, 4, 5, 6}.
We need to form 4-digit numbers. Digits 4 and 5 are compulsory, meaning they must be included in every number we form.
This means two of the four positions in our number are already taken by 4 and 5 (their positions will be determined later through permutation).
We need to choose 2 more digits from the remaining digits. The remaining digits are {1, 2, 3, 6}, which are 4 digits.
Number of ways to choose 2 digits from these 4 remaining digits = \( ^4C_2 \).
\( ^4C_2 = \frac{4!}{2!(4-2)!} = \frac{4 \times 3}{2 \times 1} = 6 \)
So, we have chosen a set of 4 digits which always include 4 and 5 (e.g., {4, 5, 1, 2}, {4, 5, 1, 3}, etc.).
Now, for each set of these 4 digits, we need to arrange them to form a number. Since the digits are distinct and order matters, we use permutations.
The number of ways to arrange 4 distinct digits = \( 4! \).
\( 4! = 4 \times 3 \times 2 \times 1 = 24 \)
To find the total number of numbers formed, we multiply the number of ways to choose the remaining 2 digits by the number of ways to arrange all 4 chosen digits.
Required number = (Ways to choose 2 digits from remaining 4) \( \times \) (Ways to arrange the 4 selected digits)
Required number = \( ^4C_2 \times 4! \)
Required number = \( 6 \times 24 \)
Required number = \( 144 \)
Therefore, 144 numbers can be formed.
In simple words: We must use digits 4 and 5. So, we first choose two more digits from the remaining available numbers. After selecting these two digits, we will have a total of four digits (including 4 and 5). Then, we arrange these four chosen digits in all possible ways to form different numbers.

🎯 Exam Tip: When certain items are compulsory, first select those compulsory items (if not already selected), then choose the remaining items from the reduced set. Finally, arrange all selected items using permutations if order matters.

Question 8. In how many ways 6 '+' and 4 '-' sign can be kept in a line whereas any two '-' occur together?
Answer: We have 6 '+' signs and 4 '-' signs.
The condition is that any two '-' signs must occur together. This means that no two '-' signs can be separated by a '+' sign.
This implies all 4 '-' signs must be together. We can treat the group of 4 '-' signs as a single block or unit.
Let's represent this block as `(----)`.
Now we have 6 '+' signs and 1 block of `(----)`.
So, effectively we are arranging 7 items (6 '+' signs and 1 block).
The total number of items to arrange is \( 6 + 1 = 7 \).
We are arranging 7 items where 6 of them are identical ('+') and 1 is a distinct block ('----').
The number of ways to arrange these items is given by the formula for permutations with repetitions:
\( \frac{(\text{Total number of items})!}{(\text{Number of identical items})!} \)
Number of ways = \( \frac{7!}{6!1!} \)
\( \frac{7 \times 6!}{6! \times 1} = 7 \)
Alternatively, consider placing the 6 '+' signs first. This creates 7 possible slots where the block of 4 '-' signs can be placed:
\_ + \_ + \_ + \_ + \_ + \_ + \_
There are 7 positions (indicated by `_`) where the block of `(----)` can be placed. Since the `(----)` block is a single unit, we just choose one of these 7 positions.
Number of ways to choose 1 position out of 7 = \( ^7C_1 = 7 \).
Since all 6 '+' signs are identical and all 4 '-' signs within their block are identical, there's only 1 way to arrange them internally once their positions are chosen.
So, the total number of ways is 7.
In simple words: If no two '-' signs can be separated, it means all four '-' signs must stick together as one group. Now, we just need to arrange this one group of four '-' signs along with the six '+' signs. We can think of it as arranging seven items, where six are the same ('+') and one is a special block ('----'). This gives us 7 ways to arrange them.

🎯 Exam Tip: When "no two items" or "items must occur together" conditions are given, treat the group of restricted items as a single unit. Then, find the permutations of these units, considering internal arrangements if the items within the unit are distinct.

Question 9. By 8 students and 5 professors, we have to make a college council, taking 5 students and 2 professors. How many councils can be formed?
Answer: We need to form a college council.
Total students available = 8
Total professors available = 5
The council needs to have:
Number of students to be selected = 5
Number of professors to be selected = 2
Since the order in which we choose students or professors for the council does not matter, this is a combination problem.
Number of ways to select 5 students from 8 students = \( ^8C_5 \)
\( ^8C_5 = \frac{8!}{5!(8-5)!} = \frac{8!}{5!3!} = \frac{8 \times 7 \times 6 \times 5!}{5! \times 3 \times 2 \times 1} = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = 8 \times 7 = 56 \)
Number of ways to select 2 professors from 5 professors = \( ^5C_2 \)
\( ^5C_2 = \frac{5!}{2!(5-2)!} = \frac{5!}{2!3!} = \frac{5 \times 4 \times 3!}{2 \times 1 \times 3!} = \frac{5 \times 4}{2} = 10 \)
To find the total number of ways to form the council, we multiply the number of ways to select students by the number of ways to select professors (using the fundamental principle of counting).
Total number of councils = (Ways to select students) \( \times \) (Ways to select professors)
Total number of councils = \( ^8C_5 \times ^5C_2 \)
Total number of councils = \( 56 \times 10 \)
Total number of councils = \( 560 \)
Therefore, 560 different college councils can be formed.
In simple words: To form the council, we first figure out how many ways we can choose 5 students from the 8 available. Then, we find out how many ways we can choose 2 professors from the 5 available. Since these choices happen together, we multiply the two numbers to get the total number of different councils possible.

🎯 Exam Tip: When selecting items for a group (like a committee or council) where the order of selection doesn't matter, use combinations. If multiple independent selections are made, multiply the results for the total number of ways.

Question 10. In how many ways can one select a cricket team of 11 from 14 players in which at least 2 bowlers are compulsory whereas only 4 players can bowl. How many ways this team can be formed?
Answer: We need to select a cricket team of 11 players from a total of 14 players.
Total players = 14
Players who can bowl (bowlers) = 4
Players who cannot bowl (non-bowlers) = \( 14 - 4 = 10 \)
The condition is that at least 2 bowlers are compulsory in the team of 11. This means the team must have 2, 3, or 4 bowlers.

Case 1: Exactly 2 Bowlers in the team
Choose 2 bowlers from 4: \( ^4C_2 = \frac{4 \times 3}{2 \times 1} = 6 \)
Remaining players needed for the team = \( 11 - 2 = 9 \)
Choose 9 non-bowlers from 10: \( ^{10}C_9 = ^{10}C_{10-9} = ^{10}C_1 = 10 \)
Ways for Case 1 = \( ^4C_2 \times ^{10}C_9 = 6 \times 10 = 60 \)

Case 2: Exactly 3 Bowlers in the team
Choose 3 bowlers from 4: \( ^4C_3 = 4 \)
Remaining players needed for the team = \( 11 - 3 = 8 \)
Choose 8 non-bowlers from 10: \( ^{10}C_8 = ^{10}C_{10-8} = ^{10}C_2 = \frac{10 \times 9}{2 \times 1} = 45 \)
Ways for Case 2 = \( ^4C_3 \times ^{10}C_8 = 4 \times 45 = 180 \)

Case 3: Exactly 4 Bowlers in the team
Choose 4 bowlers from 4: \( ^4C_4 = 1 \)
Remaining players needed for the team = \( 11 - 4 = 7 \)
Choose 7 non-bowlers from 10: \( ^{10}C_7 = ^{10}C_{10-7} = ^{10}C_3 = \frac{10 \times 9 \times 8}{3 \times 2 \times 1} = 10 \times 3 \times 4 = 120 \)
Ways for Case 3 = \( ^4C_4 \times ^{10}C_7 = 1 \times 120 = 120 \)

Total number of ways to form the team = Sum of ways from all cases
Total ways = \( 60 + 180 + 120 = 360 \)
Therefore, the team can be formed in 360 ways.
In simple words: We need to pick 11 players for a cricket team from a group of 14, where only 4 are bowlers, and we must have at least 2 bowlers in the team. We break this down into different scenarios: choosing exactly 2 bowlers, exactly 3 bowlers, or exactly 4 bowlers. For each scenario, we pick the remaining players from the non-bowlers. Finally, we add up the ways from all these scenarios to get the total number of ways to form the team.

🎯 Exam Tip: When dealing with "at least" conditions involving different categories of items, systematically list all possible valid cases (e.g., 2, 3, or 4 bowlers) and calculate the combinations for each case. Then, sum these individual case results for the final answer.