Maharashtra Board Class 11 Maths Part 2 Chapter 6 Permutations and Combinations 6.3 Solutions

Question 1. Find n if \(^nP_6 : ^nP_3 = 120 : 1\)
Answer: Solution: \(^nP_6 : ^nP_3 = 120 : 1\)
\( \therefore \frac{n!}{(n-6)!} \div \frac{n!}{(n-3)!} = \frac{120}{1} \)
\( \therefore \frac{n!}{(n-6)!} \times \frac{(n-3)!}{n!} = 120 \)
\( \therefore \frac{(n-3)(n-4)(n-5)(n-6)!}{(n-6)!} = 120 \)
\( \therefore (n - 3) (n - 4) (n - 5) = 120 \)
\( \therefore (n - 3) (n - 4) (n - 5) = 6 \times 5 \times 4 \) Comparing on both sides, we get \(n - 3 = 6\)
\( \therefore n = 9\)
In simple words: This problem involves simplifying a ratio of permutations to find the value of 'n'. By expanding the permutation formula and canceling common terms, we set up an equation that allows us to compare factors and directly solve for 'n'.

🎯 Exam Tip: Remember to express permutations \(^nP_r\) as \(n! / (n-r)!\) and simplify common factors efficiently to solve such problems. Comparing terms after simplification is a key step.

Question 2. Find m and n if \((m+n)P_2 = 56\) and \((m-n)P_2 = 12\)
Answer: Solution: \((m+n)P_2 = 56\)
\( \therefore \frac{(m+n)!}{(m+n-2)!} = 56 \)
\( \therefore \frac{(m+n)(m+n-1)(m+n-2)!}{(m+n-2)!} = 56 \)
\( \therefore (m + n) (m + n - 1) = 8 \times 7 \) Comparing on both sides, we get \(m + n = 8\) .....(i) Also \((m-n)P_2 = 12\)
\( \therefore \frac{(m-n)!}{(m-n-2)!} = 12 \)
\( \therefore \frac{(m-n)(m-n-1)(m-n-2)!}{(m-n-2)!} = 12 \)
\( \therefore (m - n) (m - n - 1) = 4 \times 3 \) Comparing on both sides, we get \(m - n = 4\) .....(ii) Adding (i) and (ii), we get \(2m = 12\)
\( \therefore m = 6\) Substituting \(m = 6\) in (ii), we get \(6 - n = 4\)
\( \therefore n = 2\)
In simple words: This problem uses permutation definitions to form two simultaneous equations for 'm' and 'n'. We simplify the permutation expressions to get products of consecutive integers, then solve the system of linear equations to find the values of 'm' and 'n'.

🎯 Exam Tip: When given permutation equations, convert them into products of consecutive integers. Solving the resulting system of equations (often by addition or substitution) is crucial for finding the variables 'm' and 'n'.

Question 3. Find r if \(^{12}P_{r-2} : ^{11}P_{r-1} = 3 : 14\)
Answer: Solution: \(^{12}P_{r-2} : ^{11}P_{r-1} = 3 : 14\)
\( \therefore \frac{12!}{(12-(r-2))!} \div \frac{11!}{(11-(r-1))!} = \frac{3}{14} \)
\( \therefore \frac{12!}{(14-r)!} \times \frac{(12-r)!}{11!} = \frac{3}{14} \)
\( \therefore \frac{12 \times 11!}{(14-r)(13-r)(12-r)!} \times \frac{(12-r)!}{11!} = \frac{3}{14} \)
\( \therefore \frac{12}{(14-r)(13-r)} = \frac{3}{14} \)
\( \therefore (14 - r)(13 - r) = \frac{12 \times 14}{3} \)
\( \therefore (14 - r)(13 - r) = 4 \times 14 \)
\( \therefore (14 - r)(13 - r) = 56 \)
\( \therefore (14 - r)(13 - r) = 8 \times 7 \) Comparing on both sides, we get \(14 - r = 8\)
\( \therefore r = 6\)
In simple words: This problem asks us to find 'r' by solving a ratio of permutations. We expand the permutation expressions, simplify by canceling common factorials, and then cross-multiply to form a quadratic-like equation that can be solved by comparing factors.

🎯 Exam Tip: Pay close attention to the indices \(r-2\) and \(r-1\) in the permutation formulas. Careful simplification of factorials, especially recognizing \((n)! = n \times (n-1)!\), is critical to avoid errors.

Question 4. Show that \((n + 1) ^nP_r = (n - r + 1) ^{n+1}P_r\).
Answer: Solution: L.H.S. \( = (n + 1) ^nP_r = (n + 1) \frac{n!}{(n-r)!} = \frac{(n+1)!}{(n-r)!} \) R.H.S.\( = (n - r + 1) ^{n+1}P_r = (n - r + 1) \frac{(n+1)!}{((n+1)-r)!} \)
\( = (n - r + 1) \frac{(n+1)!}{(n-r+1)!} \)
\( = (n - r + 1) \frac{(n+1)!}{(n-r+1)(n-r)!} \)
\( = \frac{(n+1)!}{(n-r)!} \)
\( \therefore \) L.H.S. = R.H.S.
In simple words: To prove this identity, we expand both the Left Hand Side and Right Hand Side using the standard permutation formula \(^kP_m = k! / (k-m)!\). By simplifying both sides, we show that they reduce to the same expression, thus proving the identity.

🎯 Exam Tip: For proving identities involving permutations, always start by expanding both sides using the factorial definition. Look for opportunities to cancel terms and manipulate factorials (e.g., \( (k)! = k \times (k-1)! \) ) to reach a common expression.

Question 5. How many 4 letter words can be formed using letters in the word MADHURI if
(i) letters can be repeated?
(ii) letters cannot be repeated?
Answer: Solution: There are 7 letters in the word MADHURI. (i) A 4 letter word is to be formed from the letters of the word MADHURI and repetition of letters is allowed.
\( \therefore \) 1st letter can be filled in 7 ways. 2nd letter can be filled in 7 ways. 3rd letter can be filled in 7 ways. 4th letter can be filled in 7 ways.
\( \therefore \) Total no. of ways a 4-letter word can be formed = \(7 \times 7 \times 7 \times 7 = 2401\)
\( \therefore \) 2401 four-lettered words can be formed when repetition of letters is allowed. (ii) When repetition of letters is not allowed, the number of 4-letter words formed from the letters of the word MADHURI is \(^7P_4 = \frac{7!}{(7-4)!} = \frac{7 \times 6 \times 5 \times 4 \times 3!}{3!} = 840 \)
\( \therefore \) 840 four-letter words can be formed when repetition of letters is not allowed. Alternate method: There are 7 letters in the word MADHURI. (i) Since letters can be repeated
\( \therefore \) In all places of a four-letter word, any one of seven letters M, A, D, H, U, R, I can appear.
\( \therefore \) Using the Multiplication theorem, we get Number of four-letter words with repetition of letters M, A, D, H, U, R, I = \(7 \times 7 \times 7 \times 7 = 2401\) (ii) Since the letters cannot be repeated therefore 1st, 2nd, 3rd, 4th places can be filled in 7, 6, 5, 4 ways respectively
\( \therefore \) Using the multiplication theorem, we get Number of four-letter words, with no repetition of letters M, A, D, H, U, R, I = \(7 \times 6 \times 5 \times 4 = 840\)
In simple words: This problem involves forming 4-letter words from 7 distinct letters. When repetition is allowed, each position has 7 choices. When repetition is not allowed, we use permutations, where the choices for each subsequent position decrease by one.

🎯 Exam Tip: Clearly distinguish between problems where repetition is allowed (using the multiplication principle directly) and where it's not (using permutations or the multiplication principle with decreasing choices). This is a fundamental concept in combinatorics.

Question 6. Determine the number of arrangements of letters of the word ALGORITHM if
(i) vowels are always together.
(ii) no two vowels are together.
(iii) Consonants are at even positions
(iv) O is first and T is last.
Answer: Solution: A word is to be formed using the letters of the word ALGORITHM. There are 9 letters in the word ALGORITHM. (i) When vowels are always together: There are 3 vowels in the word ALGORITHM. (i.e, A, I, O) Let us consider these 3 vowels as one unit. This unit with 6 other letters is to be arranged.
\( \therefore \) It becomes an arrangement of 7 things which can be done in \(^7P_7\) i.e., 7! ways and 3 vowels can be arranged among themselves in \(^3P_3\) i.e., 3! ways.
\( \therefore \) the total number of ways in which the word can be formed = \(7! \times 3!\) \( = 5040 \times 6 \) \( = 30240 \)
\( \therefore \) 30240 words can be formed if vowels are always together. (ii) When no two vowels are together: There are 6 consonants in the word ALGORITHM. They can be arranged among themselves in \(^6P_6\) i.e., 6! ways. Let consonants be denoted by C. _C_C_C_C_C_C_ 6 consonants create 7 gaps in which 3 vowels are to arranged.
\( \therefore \) 3 vowels can be filled in \(^7P_3\) \( = \frac{7!}{(7-3)!} \) \( = \frac{7 \times 6 \times 5 \times 4!}{4!} \) \( = 210 \) ways
\( \therefore \) total number of ways in which the word can be formed = \(6! \times 210\) \( = 720 \times 210 \) \( = 151200 \)
\( \therefore \) 151200 words can be formed if no two vowels are together. (iii) When consonants are at even positions: There are 4 even places and 6 consonants in the word ALGORITHM. 1st, 2nd, 3rd, 4th even places are filled in 6, 5, 4, 3 way respectively.
\( \therefore \) The number of ways to fill four even places by consonants = \(6 \times 5 \times 4 \times 3 = 360\) The remaining 5 letters (3 vowels and 2 consonants) can be arranged among themselves in \(^5P_5\) i.e., 5! ways.
\( \therefore \) Total number of ways the words can be formed In which even places are occupied by consonants = \(360 \times 5!\) \( = 360 \times 120 \) \( = 43200 \)
\( \therefore \) 43200 words can be formed if even positions are occupied by consonants. (iv) When beginning with O and ends with T: All the letters of the word ALGORITHM are to be arranged among themselves such that arrangement begins with O and ends with T. 7 letters other than O and T can be filled between O and T in \(^7P_7\) i.e., 7! ways = 5040 ways.
\( \therefore \) 5040 words beginning with O and ending with T can be formed.
In simple words: This question explores various arrangements of letters in "ALGORITHM" based on specific constraints: treating grouped vowels as one unit, placing vowels in gaps between consonants, filling even positions with consonants, and fixing letters at the start and end. Each scenario uses permutations and the multiplication principle.

🎯 Exam Tip: For arrangement problems with constraints, simplify by treating groups (like "vowels together") as single units. For "no two together" problems, arrange the "non-group" items first, then place the "group" items in the gaps created. Fixing positions reduces the number of available choices for those spots.

Question 7. In a group photograph, 6 teachers and principals are in the first row and 18 students are in the second row. There are 12 boys and 6 girls among the students. If the middle position is reserved for the principal and if no two girls are together, find the number of arrangements.
Answer: Solution: In 1st row middle seat is fixed for the principal. Also 1st row, 6 teachers can be arranged among themselves in \(^6P_6\) i.e., 6! ways. In the 2nd row, 12 boys can be arranged among themselves in \(^{12}P_{12}\) i.e., 12! ways. 13 gaps are created by 12 boys, in which 6 girls are to be arranged. together which can be done in \(^{13}P_6\) ways.
\( \therefore \) total number of arrangements \( = 6! \times 12! \times ^{13}P_6 \) .....[using Multiplications Principle] \( = 6! \times 12! \times \frac{13!}{(13-6)!} \) \( = 6! \times 12! \times \frac{13!}{7!} \) \( = \frac{6! \times 12! \times 13!}{7!} \)
\( = \frac{6! \times 12! \times 13 \times 12 \times 11 \times 10 \times 9 \times 8 \times 7!}{7!} \) \( = 6! \times 12! \times 13 \times 12 \times 11 \times 10 \times 9 \times 8 \)
In simple words: This problem requires arranging teachers, principal, boys, and girls with specific constraints. We first arrange teachers and boys independently. The principal occupies a fixed spot. For girls, we arrange the boys first, creating gaps where girls can be placed so no two are together. The total arrangements are found by multiplying the arrangements of each group.

🎯 Exam Tip: Break down complex arrangement problems into smaller, independent sub-problems. Fixed positions reduce choices. For "no two together" constraints, always arrange the other elements first to create gaps for the constrained elements. The multiplication principle combines the results.

Question 8. Find the number of ways letters of the word HISTORY can be arranged if
(i) Y and T are together
(ii) Y is next to T.
Answer: Solution: There are 7 letters in the word HISTORY (i) When 'Y' and 'T' are together. Let us consider 'Y' and 'T' as one unit This unit with the other 5 letters is to be arranged.
\( \therefore \) The number of arrangements of one unit and 5 letters = \(^6P_6 = 6!\) Also, 'Y' and 'T' can be arranged among themselves in \(^2P_2\) i.e., 2! ways.
\( \therefore \) a total number of arrangements when Y and T are always together = \(6! \times 2!\) \( = 720 \times 2 \) \( = 1440 \)
\( \therefore \) 1440 words can be formed if Y and T are together. (ii) When 'Y' is next to 'T' Let us take this ('Y' next to 'T') as one unit. This unit with 5 other letters is to be arranged.
\( \therefore \) The number of arrangements of 6 letters and one unit = \(^6P_6 = 6!\) Also 'Y' has to be always next to 'T'. So they can be arranged in 1 way.
\( \therefore \) total number of arrangements possible when Y is next to T = \(6! \times 1 = 720\)
\( \therefore \) 720 words can be formed if Y is next to T.
In simple words: This problem involves arranging the letters of HISTORY with specific conditions on 'Y' and 'T'. If 'Y' and 'T' are together, we treat them as a single block and arrange the block and remaining letters, then account for permutations within the block. If 'Y' is next to 'T' in a specific order, they form a fixed unit.

🎯 Exam Tip: The distinction between "together" (allowing internal permutation) and "next to" (implying a fixed internal order, like YT but not TY) is crucial. Always count the number of elements (including combined units) and their internal arrangements. If specific order is implied, the internal arrangement might be 1 way.

Question 9. Find the number of arrangements of the letters in the word BERMUDA so that consonants and vowels are in the same relative positions.
Answer: Solution: There are 7 letters in the word "BERMUDA” out of which 3 are vowels (E, U, A) and 4 are consonants (B, R, M, D). If relative positions of consonants and vowels are not changed. 3 vowels can be arranged among themselves in \(^3P_3\) i.e., 3! ways. 4 consonants can be arranged among themselves in \(^4P_4\) i.e., 4! ways.
\( \therefore \) total no. of arrangements possible if relative positions of vowels and consonants are not changed = \(3! \times 4!\) \( = 6 \times 24 \) \( = 144 \)
In simple words: To arrange the letters of "BERMUDA" while keeping the relative positions of vowels and consonants unchanged, we count the number of vowels and consonants. Then, we independently arrange the vowels among themselves and the consonants among themselves, multiplying the results.

🎯 Exam Tip: When relative positions of categories (like vowels and consonants) are fixed, you can treat the arrangements within each category independently. Identify the letters belonging to each category and calculate their permutations separately, then multiply the results.

Question 10. Find the number of 4-digit numbers that can be formed using the digits 1, 2, 4, 5, 6, 8 if
(i) digits can be repeated
(ii) digits cannot be repeated
Answer: Solution: (i) A 4 digit number is to be made from the digits 1, 2, 4, 5, 6, 8 such that digits can be repeated.
\( \therefore \) The unit's place digit can be filled in 6 ways. 10's place digit can be filled in 6 ways. 100's place digit can be filled in 6 ways. 1000's place digit can be filled in 6 ways.
\( \therefore \) total number of numbers = \(6 \times 6 \times 6 \times 6 = 6^4 = 1296\)
\( \therefore \) 1296 four-digit numbers can be formed if repetition of digits is allowed. (ii) A 4 different digit number is to be made from the digits 1, 2, 4, 5, 6, 8 without repetition of digits.
\( \therefore \) 4 different digits are to be arranged from 6 given digits which can be done in \(^6P_4\) \( = \frac{6!}{(6-4)!} \) \( = \frac{6 \times 5 \times 4 \times 3 \times 2!}{2!} \) \( = 360 \) ways
\( \therefore \) 360 four-digit numbers can be formed, if repetition of digits is not allowed.
In simple words: This problem asks for the number of 4-digit numbers formed from 6 distinct digits, with and without repetition. If repetition is allowed, each of the four positions has 6 choices. If not, we use the permutation formula \(^nP_r\) or the multiplication principle with decreasing choices for each position.

🎯 Exam Tip: Always check if repetition is allowed. If yes, the number of choices remains constant for each position. If no, the number of choices decreases for subsequent positions, often solved using permutations \(^nP_r\).

Question 11. How many numbers can be formed using the digits 0, 1, 2, 3, 4, 5 without repetition so that the resulting numbers are between 100 and 1000?
Answer: Solution: A number between 100 and 1000 is a 3 digit number and is to be formed from the digits 0, 1, 2, 3, 4, 5, without repetition of digits.
\( \therefore \) 100's place digit must be a non-zero number which can be filled in 5 ways (1, 2, 3, 4, 5). 10's place digits can be filled in 5 ways (0 and the remaining 4 non-zero digits). Unit's place digit can be filled in 4 ways (remaining 4 digits).
\( \therefore \) total number of ways the number can be formed = \(5 \times 5 \times 4 = 100\)
\( \therefore \) 100 numbers between 100 and 1000 can be formed.
In simple words: To form 3-digit numbers (between 100 and 1000) using given digits without repetition, we first determine the choices for the hundreds place (cannot be 0). Then, for the tens place, we include 0 and the remaining digits. Finally, for the units place, we pick from the remaining distinct digits, multiplying the choices for each position.

🎯 Exam Tip: When forming numbers, be careful with the digit '0'. If '0' cannot be the first digit (e.g., in a 3-digit number), it restricts the choices for that position. Then, '0' can be used in subsequent positions. Always consider the "without repetition" constraint for decreasing choices.

Question 12. Find the number of 6-digit numbers using the digits 3, 4, 5, 6, 7, 8 without repetition. How many of these numbers are
(i) divisible by 5?
(ii) not divisible by 5?
Answer: Solution: A number of 6 different digits is to be formed from the digits 3, 4, 5, 6, 7, 8 which can be done in \(^6P_6\) i.e., \(6! = 720\) ways (i) If the number is divisible by 5, then The unit's place digit must be 5, and hence unit's place can be filled in 1 way. Other 5 digits can be arranged among themselves in \(^5P_5\) i.e., 5! ways
\( \therefore \) Total number of ways in which numbers divisible by 5 can be formed = \(1 \times 5! = 120\) (ii) If the number is not divisible by 5, then Unit's place can be any digit from 3, 4, 6, 7, 8 which can be selected in 5 ways. Other 5 digits can be arranged in \(^5P_5\) i.e., 5! ways
\( \therefore \) The total number of ways in which numbers not divisible by 5 can be formed = \(5 \times 5!\) \( = 5 \times 120 \) \( = 600 \)
In simple words: We form 6-digit numbers from given digits without repetition. For divisibility by 5, the unit's digit must be 5, fixing one position. For numbers not divisible by 5, the unit's digit can be any other digit, allowing 5 choices. The remaining 5 digits are arranged in 5! ways in both cases.

🎯 Exam Tip: For divisibility rules, fix the relevant digit(s) first. Then arrange the remaining digits. For "not divisible" cases, it's often easiest to calculate total arrangements and subtract the "divisible" arrangements. Alternatively, explicitly count the valid choices for the restricted position.

Question 13. A code word is formed by two distinct English letters followed by two non-zero distinct digits. Find the number of such code words. Also, find the number of such code words that end with an even digit.
Answer: Solution: (i) There is a total of 26 alphabets. A code word contains 2 English alphabets.
\( \therefore \) 2 alphabets can be filled in \(^{26}P_2\) \( = \frac{26!}{(26-2)!} \) \( = \frac{26 \times 25 \times 24!}{24!} \) \( = 650 \) ways Also, alphabets to be followed by two distinct non-zero digits from 1 to 9 which can be filled in \(^9P_2\) \( = \frac{9!}{(9-2)!} \) \( = \frac{9 \times 8 \times 7!}{7!} \) \( = 72 \) ways
\( \therefore \) Total number of a code words = \(650 \times 72 = 46800\) (ii) There are total 26 alphabets. A code word contains 2 English alphabets.
\( \therefore \) 2 alphabets can be filled in \(^{26}P_2\) \( = \frac{26!}{(26-2)!} \) \( = \frac{26 \times 25 \times 24!}{24!} \) \( = 650 \) ways For a code word to end with an even integer, the digit in the unit's place should be an even number between 1 to 9 which can be filled in 4 ways (2, 4, 6, 8). Also, 10's place can be filled in 8 ways (from the remaining 8 non-zero digits).
\( \therefore \) Total number of codewords = \(650 \times 4 \times 8 = 20800\) ways
\( \therefore \) 20800 codewords end with an even integer.
In simple words: This problem involves forming code words with distinct letters and distinct non-zero digits. First, calculate arrangements for two distinct letters. Then, calculate arrangements for two distinct non-zero digits. Multiply these to get the total. For code words ending with an even digit, fix the last digit as one of the 4 even digits (2,4,6,8) and then select the preceding distinct digit from the remaining 8 non-zero digits.

🎯 Exam Tip: Break down code word problems into independent selections for different character types (letters, digits). For "distinct" elements, use permutations. When a specific condition (like "ends with an even digit") is applied, address that restricted position first, then fill the remaining positions with distinct elements.

Question 14. Find the number of ways in which 5 letters can be posted in 3 post boxes if any number of letters can be posted in a post box.
Answer: Solution: There are 5 letters and 3 post boxes and any number of letters can be posted in all three post boxes.
\( \therefore \) Each letter can be posted in 3 ways.
\( \therefore \) Total number of ways in which 5 letters can be posted = \(3 \times 3 \times 3 \times 3 \times 3 = 3^5 = 243\)
In simple words: In this problem, each of the 5 distinct letters can be posted into any of the 3 post boxes independently. Since repetition of post boxes is allowed for each letter, we multiply the number of choices for each letter.

🎯 Exam Tip: For "n items into k bins" problems where any item can go into any bin (with replacement), the formula is \(k^n\). Clearly identify what the 'items' are and what the 'bins' are. Here, letters are items and post boxes are bins.

Question 15. Find the number of arranging 11 distinct objects taken 4 at a time so that a specified object
(i) always occurs
(ii) never occurs
Answer: Solution: There are 11 distinct objects and 4 are to be taken at a time. (i) The number of permutations of n distinct objects, taken r at a time, when one specified object will always occur is \(r \times ^{n-1}P_{r-1}\) Here, \(r = 4, n = 11\)
\( \therefore \) The number of permutations of 4 out of 11 objects when a specified object occurs. \( = 4 \times ^{11-1}P_{4-1} \) \( = 4 \times ^{10}P_3 \) \( = 4 \times \frac{10!}{(10-3)!} \) \( = 4 \times \frac{10!}{7!} \) \( = 4 \times \frac{10 \times 9 \times 8 \times 7!}{7!} \) \( = 4 \times 10 \times 9 \times 8 \) \( = 2880 \)
\( \therefore \) There are 2880 permutations of 11 distinct objects, taken 4 at a time, in which one specified object always occurs. (ii) When one specified object does not occur then 4 things are to be arranged from the remaining 10 things, which can be done in \(^{10}P_4\) ways \( = 10 \times 9 \times 8 \times 7 \) \( = 5040 \) ways
\( \therefore \) There are 5040 permutations of 11 distinct objects, taken 4 at a time, in which one specified object never occurs.
In simple words: This problem asks for arrangements of 4 out of 11 distinct objects under two conditions. If a specified object always occurs, we fix its presence, arrange the remaining (r-1) objects from the remaining (n-1) choices, and multiply by 'r' positions for the specified object. If a specified object never occurs, we simply select and arrange 'r' objects from the remaining (n-1) objects.

🎯 Exam Tip: For problems involving specific objects: if an object *must* be included, fix its position (or consider its multiple possible positions) and arrange the rest. If an object *must not* be included, remove it from the total pool of objects before calculating permutations.