RBSE Solutions Class 11 Maths Chapter 6 Permutations and Combinations Exercise 6.1

Question 1. Find the value of n, whereas
(i) \( ^{n-1}P_3 : ^{n+1}P_3 = 5 : 12 \)
(ii) \( ^nP_6 = 10 \cdot ^nP_5 \)
(iii) \( ^{56}P_{n+6} : ^{54}P_{n+3} = 30800 \)
(iv) \( ^{6+n}P_2 : ^{6-n}P_2 = 56:12 \)
Answer:
(i) We are given the ratio: \( ^{n-1}P_3 : ^{n+1}P_3 = 5 : 12 \)
This can be written as: \( \frac { (n-1)! }{ (n-1-3)! } \div \frac { (n+1)! }{ (n+1-3)! } = \frac { 5 }{ 12 } \)
\( \implies \frac { (n-1)! }{ (n-4)! } \times \frac { (n-2)! }{ (n+1)! } = \frac { 5 }{ 12 } \)
Now, we expand the factorials: \( (n+1)! = (n+1) \cdot n \cdot (n-1)! \) and \( (n-2)! = (n-2) \cdot (n-3) \cdot (n-4)! \)
\( \implies \frac { (n-1)! }{ (n-4)! } \times \frac { (n-2)(n-3)(n-4)! }{ (n+1)n(n-1)! } = \frac { 5 }{ 12 } \)
After cancelling common terms, we get:
\( \implies \frac { (n-2)(n-3) }{ n(n+1) } = \frac { 5 }{ 12 } \)
\( \implies 12(n^2 - 5n + 6) = 5(n^2 + n) \)
\( \implies 12n^2 - 60n + 72 = 5n^2 + 5n \)
\( \implies 12n^2 - 5n^2 - 60n - 5n + 72 = 0 \)
\( \implies 7n^2 - 65n + 72 = 0 \)
We can solve this quadratic equation for n:
\( n = \frac { -(-65) \pm \sqrt { (-65)^2 - 4(7)(72) } }{ 2(7) } \)
\( n = \frac { 65 \pm \sqrt { 4225 - 2016 } }{ 14 } \)
\( n = \frac { 65 \pm \sqrt { 2209 } }{ 14 } \)
\( n = \frac { 65 \pm 47 }{ 14 } \)
Two possible values for n are:
\( n = \frac { 65 + 47 }{ 14 } = \frac { 112 }{ 14 } = 8 \)
\( n = \frac { 65 - 47 }{ 14 } = \frac { 18 }{ 14 } = \frac { 9 }{ 7 } \)
Since n must be an integer for permutations, we choose \( n = 8 \).

(ii) We have: \( ^nP_6 = 10 \cdot ^nP_5 \)
Using the permutation formula \( ^nP_r = \frac{n!}{(n-r)!} \):
\( \frac{n!}{(n-6)!} = 10 \cdot \frac{n!}{(n-5)!} \)
\( \implies \frac{1}{(n-6)!} = \frac{10}{(n-5)!} \)
Since \( (n-5)! = (n-5) \cdot (n-6)! \), we can write:
\( \implies \frac{1}{(n-6)!} = \frac{10}{(n-5)(n-6)!} \)
\( \implies 1 = \frac{10}{n-5} \)
\( \implies n-5 = 10 \)
\( \implies n = 10 + 5 \)
\( \implies n = 15 \)

(iii) We are given: \( ^{56}P_{n+6} : ^{54}P_{n+3} = 30800 \)
This means: \( \frac{^{56}P_{n+6}}{^{54}P_{n+3}} = 30800 \)
Using the permutation formula \( ^nP_r = \frac{n!}{(n-r)!} \):
\( \frac{56!}{(56-(n+6))!} \div \frac{54!}{(54-(n+3))!} = 30800 \)
\( \implies \frac{56!}{(50-n)!} \times \frac{(51-n)!}{54!} = 30800 \)
We expand \( 56! = 56 \times 55 \times 54! \) and \( (51-n)! = (51-n) \times (50-n)! \)
\( \implies \frac{56 \times 55 \times 54!}{(50-n)!} \times \frac{(51-n)(50-n)!}{54!} = 30800 \)
After cancelling common terms:
\( \implies 56 \times 55 \times (51-n) = 30800 \)
\( \implies 3080 \times (51-n) = 30800 \)
\( \implies 51-n = \frac{30800}{3080} \)
\( \implies 51-n = 10 \)
\( \implies n = 51-10 \)
\( \implies n = 41 \)

(iv) We are given: \( ^{6+n}P_2 : ^{6-n}P_2 = 56 : 12 \)
This means: \( \frac{^{6+n}P_2}{^{6-n}P_2} = \frac{56}{12} = \frac{14}{3} \)
Using the permutation formula \( ^nP_r = n(n-1) \):
\( ^{6+n}P_2 = (6+n)(6+n-1) = (6+n)(5+n) \)
\( ^{6-n}P_2 = (6-n)(6-n-1) = (6-n)(5-n) \)
\( \implies \frac{(6+n)(5+n)}{(6-n)(5-n)} = \frac{14}{3} \)
\( \implies 3((6+n)(5+n)) = 14((6-n)(5-n)) \)
\( \implies 3(30 + 6n + 5n + n^2) = 14(30 - 6n - 5n + n^2) \)
\( \implies 3(30 + 11n + n^2) = 14(30 - 11n + n^2) \)
\( \implies 90 + 33n + 3n^2 = 420 - 154n + 14n^2 \)
Rearrange the terms to form a quadratic equation:
\( \implies 14n^2 - 3n^2 - 154n - 33n + 420 - 90 = 0 \)
\( \implies 11n^2 - 187n + 330 = 0 \)
Divide the entire equation by 11:
\( \implies n^2 - 17n + 30 = 0 \)
Factor the quadratic equation:
\( \implies (n-2)(n-15) = 0 \)
This gives two possible values for n: \( n = 2 \) or \( n = 15 \).
For permutations \( ^kP_r \) to be defined, \( k \) must be greater than or equal to \( r \).
For \( ^{6-n}P_2 \), we need \( 6-n \ge 2 \), which means \( n \le 4 \).
Comparing this condition with our solutions, \( n = 2 \) is the only valid solution because \( n=15 \) is greater than 4.
So, \( n=2 \).
In simple words: We used the permutation formula \( ^nP_r = \frac{n!}{(n-r)!} \) to set up equations for each part. Then we simplified these equations and solved for \( n \). For permutations, \( n \) must be a whole number, and the number of items must be greater than or equal to the number being chosen.

🎯 Exam Tip: Always remember that in permutations \( ^nP_r \), both \( n \) and \( r \) must be non-negative integers, and \( n \ge r \). These conditions are crucial for validating the solutions, especially when solving quadratic equations that yield multiple roots.

Question 2. Find the number of different words formed by letters of word ALLAHABAD.
Answer: The word ALLAHABAD has 9 letters in total.
Let's count the frequency of each letter:
A appears 4 times
L appears 2 times
H appears 1 time
B appears 1 time
D appears 1 time
The formula for permutations with repetitions is \( \frac{n!}{n_1! n_2! ... n_k!} \), where \( n \) is the total number of items, and \( n_1, n_2, ... n_k \) are the counts of repeated items.
So, the number of different words that can be formed is \( \frac{9!}{4! \cdot 2! \cdot 1! \cdot 1! \cdot 1!} \)
\( = \frac{9 \times 8 \times 7 \times 6 \times 5 \times 4!}{4! \times 2 \times 1} \)
\( = \frac{9 \times 8 \times 7 \times 6 \times 5}{2} \)
\( = 9 \times 4 \times 7 \times 6 \times 5 \)
\( = 7560 \)
Hence, 7560 different words can be formed.
In simple words: To find how many unique words you can make from ALLAHABAD, divide the factorial of the total letters by the factorial of each letter that repeats.

🎯 Exam Tip: When letters in a word are repeated, use the formula for permutations with repetitions. Make sure to count each letter's frequency carefully.

Question 3. How many words can be formed using letters of word TRIANGLE ? Out of these how many words begin with T and end with E ?
Answer: The word TRIANGLE has 8 letters. All the letters in TRIANGLE are distinct (no repetitions).
1. Total number of words that can be formed using all letters:
Since there are 8 distinct letters, the total number of words is \( 8! \).
\( 8! = 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 40320 \)

2. Number of words that begin with T and end with E:
If T is fixed at the first position and E is fixed at the last position, we are left with 6 remaining letters: R, I, A, N, G, L.
These 6 letters can be arranged in the middle 6 positions.
The number of ways to arrange these 6 distinct letters is \( 6! \).
\( 6! = 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 720 \)
So, 720 words can be formed that begin with T and end with E.
In simple words: First, count all possible arrangements of the word TRIANGLE. Then, if T must be at the start and E at the end, only arrange the letters in between.

🎯 Exam Tip: For problems with fixed positions, mentally "lock" those letters and calculate permutations only for the remaining free letters. If all letters are unique, use simple factorials.

Question 5. How many 6-digit numbers can be formed by using the digits 0, 1, 2, 3, 4, 5 ?
Answer: We need to form 6-digit numbers using the digits 0, 1, 2, 3, 4, 5, assuming no digit is repeated.
Total number of digits available is 6.
The total number of permutations of these 6 digits would be \( 6! = 720 \).
However, a 6-digit number cannot start with 0. Numbers that start with 0 are actually 5-digit numbers.
To find the number of arrangements where 0 is in the first position, we fix 0 at the first place. The remaining 5 digits (1, 2, 3, 4, 5) can be arranged in the other 5 positions in \( 5! \) ways.
\( 5! = 5 \times 4 \times 3 \times 2 \times 1 = 120 \).
The number of 6-digit numbers that can be formed is the total permutations minus the permutations where 0 is in the first place:
Number of 6-digit numbers = (Total permutations of 6 digits) - (Permutations with 0 at the first place)
\( = 6! - 5! \)
\( = 720 - 120 \)
\( = 600 \)
Therefore, 600 distinct 6-digit numbers can be formed.
In simple words: To make 6-digit numbers from 0, 1, 2, 3, 4, 5, first find all ways to arrange the digits. Then, subtract any arrangements that start with 0, because those are actually 5-digit numbers.

🎯 Exam Tip: When forming numbers, always account for the 'leading zero' constraint. The first digit cannot be zero unless specifically allowed for the context (e.g., a code).

Question 6. How many 3-digit numbers less than 1000 can be formed using the digits 1, 2, 3, 4, 5, 6, if no digits is repeated?
Answer: We need to form 3-digit numbers using the digits 1, 2, 3, 4, 5, 6 without repeating any digit.
Total number of available digits is 6 (1, 2, 3, 4, 5, 6).
We need to choose and arrange 3 digits out of these 6. This is a permutation problem \( ^nP_r \), where \( n=6 \) and \( r=3 \).
The number of 3-digit numbers is \( ^6P_3 \).
\( ^6P_3 = \frac{6!}{(6-3)!} = \frac{6!}{3!} \)
\( = \frac{6 \times 5 \times 4 \times 3!}{3!} \)
\( = 6 \times 5 \times 4 \)
\( = 120 \)
Hence, 120 different 3-digit numbers can be formed.
In simple words: We have 6 different numbers and want to pick 3 of them to make a 3-digit number, without using the same number twice. This is solved by multiplying the choices for each spot: 6 choices for the first digit, 5 for the second, and 4 for the third.

🎯 Exam Tip: For forming numbers with no repetition, think of it as filling slots. The number of choices decreases for each subsequent slot.

Question 7. How many ways 15 members of a committee can sit around the round table whereas secretary sit on one side and vice secretary on other side of Director?
Answer: Total members in the committee = 15.
The problem involves circular permutations with specific conditions for three members: Director (D), Secretary (S), and Vice Secretary (VS).
The condition "Secretary sits on one side and Vice Secretary on the other side of Director" means their relative positions are fixed. We can treat D, S, and VS as a fixed unit where S and VS are always next to D, on opposite sides. There are two ways to arrange S and VS around D (S-D-VS or VS-D-S). This gives us \( 2! = 2 \) arrangements for this block.
Now, consider this D-S-VS block as one unit. The remaining \( 15 - 3 = 12 \) members can be arranged around the table along with this block.
However, since the positions of S and VS are specified relative to D, we can consider the Director's position as fixed for a circular arrangement, which means we avoid the `(n-1)!` for the entire group. Instead, we arrange the remaining \( 15-1 = 14 \) people in relation to the Director.
If we fix the Director's position, the Secretary can sit in 2 specific spots next to the Director (left or right). Once the Secretary's spot is chosen, the Vice Secretary's spot is fixed on the other side of the Director.
The remaining 12 members can be arranged in the remaining 12 seats in \( 12! \) ways.
So, the total number of ways is \( 12! \times 2! \)
\( = 12! \times 2 \)
\( = 479,001,600 \times 2 \)
\( = 958,003,200 \)
In simple words: First, fix the Director. Then, the Secretary and Vice Secretary can sit on either side of the Director (2 options). After these three are placed, arrange the remaining 12 people in any order. Multiply these possibilities together.

🎯 Exam Tip: For circular permutations with specific relative positioning, fix one person's spot to convert it to a linear permutation problem, then account for the relative arrangements of the fixed group. Remember to multiply by the internal permutations of the constrained group.

Question 8. There are 15 stations on a Railway line. For this, how many different tickets of a class should be printed such that any person from any station can buy the ticket of other station of this line?
Answer: Total number of stations on the railway line = 15.
A ticket is required from one station to another *different* station. The order matters because a ticket from Station A to Station B is different from a ticket from Station B to Station A.
This is a permutation problem where we choose 2 stations out of 15, and the order matters. We use the formula \( ^nP_r \), where \( n=15 \) (total stations) and \( r=2 \) (stations selected for a ticket - origin and destination).
Number of different tickets = \( ^{15}P_2 \)
\( ^{15}P_2 = \frac{15!}{(15-2)!} = \frac{15!}{13!} \)
\( = 15 \times 14 \)
\( = 210 \)
So, 210 different tickets should be printed.
In simple words: If you have 15 stations, a ticket means choosing a starting station and a different ending station. Since the journey direction matters, you multiply the number of choices for the first station by the number of choices for the second station.

🎯 Exam Tip: Identify whether order matters in a selection problem. If it does (like picking a start and end point for a journey), it's a permutation. If order doesn't matter (like picking items for a group), it's a combination.

Question 9. How many ways 10 different beads can be threads for making a garland whereas 4 special beads never remain separate?
Answer: Total number of different beads = 10.
The condition "4 special beads never remain separate" actually implies that these 4 special beads *always* remain together. If they "never remain separate", it means they always form a single block or group. (This is a common phrasing trick in permutation problems; "never separate" usually means "always together").
1. Treat the 4 special beads as a single unit. Now we have \( (10 - 4) \) individual beads plus this one unit of 4 beads, making a total of \( 6 + 1 = 7 \) units to arrange.
2. These 7 units are to be arranged to form a garland (circular arrangement). The number of ways to arrange \( n \) distinct items in a circle is \( (n-1)! \). So, the number of ways to arrange these 7 units is \( (7-1)! = 6! \).
3. The 4 special beads within their block can be arranged among themselves in \( 4! \) ways.
4. Since a garland can be viewed from both sides (clockwise and anti-clockwise arrangements are considered the same), we divide the total by 2.
Total number of ways = \( \frac{6! \times 4!}{2} \)
\( 6! = 720 \)
\( 4! = 24 \)
Total ways = \( \frac{720 \times 24}{2} \)
\( = \frac{17280}{2} \)
\( = 8640 \)
Therefore, there are 8640 ways to make the garland under the given condition.
In simple words: Imagine the 4 special beads as one big bead. Now you have 7 beads (6 regular + 1 big). Arrange these 7 beads in a circle. Then, remember that the 4 beads inside the "big bead" can also swap places. Finally, divide by 2 because a garland looks the same flipped around.

🎯 Exam Tip: Pay close attention to wording like "never remain separate" (implying "always together"). For garlands, always remember to divide by 2 if the items are distinct and there's no fixed orientation.

Question 11. How many words can be formed from letters of word SCHOOL, whereas both O's not come together?
Answer: The word SCHOOL has 6 letters: S, C, H, O, O, L.
Here, the letter O is repeated 2 times. The other letters (S, C, H, L) appear once.

1. Calculate the total number of words formed by arranging all 6 letters:
Total permutations = \( \frac{6!}{2!} \) (since O repeats 2 times)
\( = \frac{6 \times 5 \times 4 \times 3 \times 2 \times 1}{2 \times 1} \)
\( = \frac{720}{2} = 360 \)

2. Calculate the number of words where both O's *do* come together:
To do this, treat the two O's as a single block (OO). Now, we have 5 units to arrange: S, C, H, (OO), L.
These 5 distinct units can be arranged in \( 5! \) ways.
\( 5! = 5 \times 4 \times 3 \times 2 \times 1 = 120 \)

3. Calculate the number of words where both O's *do not* come together:
This is found by subtracting the words where O's are together from the total number of words.
Words (O's not together) = (Total words) - (Words where O's are together)
\( = 360 - 120 \)
\( = 240 \)
So, 240 words can be formed from the letters of SCHOOL where both O's do not come together.
In simple words: First, find all possible ways to arrange the letters in SCHOOL. Next, find all ways where the two 'O's stick together like one letter. Then, subtract the second number from the first to get the words where the 'O's are separated.

🎯 Exam Tip: For problems involving items that "do not come together," it's often easier to calculate the total arrangements and subtract the arrangements where they *do* come together.