Maharashtra Board Class 11 Maths Part 2 Chapter 6 Permutations and Combinations 6.1 Solutions

Question 1. A teacher wants to select the class monitor in a class of 30 boys and 20 girls. In how many ways can he select a student if the monitor can be a boy or a girl?
Answer: There are 30 boys and 20 girls in a class. The teacher wants to select a class monitor from these boys and girls. A boy can be selected in 30 ways and a girl can be selected in 20 ways.
. By using the fundamental principle of addition, in a number of ways either a boy or a girl is selected as a class monitor = 30 + 20 = 50.
In simple words: To select a monitor who can be either a boy or a girl, you add the number of ways to select a boy to the number of ways to select a girl.

🎯 Exam Tip: When choices are mutually exclusive (selecting a boy OR a girl), the total number of ways is found by adding the individual possibilities (addition principle).

Question 2. In question 1, in how many ways can the monitor be selected if the monitor must be a boy? What is the answer if the monitor must be a girl?
Answer: (i) Since there are 30 boys in the class
. A boy monitor can be selected in 30 ways.
(ii) Since there are 20 girls in the class
. A girl monitor can be selected in 20 ways.
In simple words: If the monitor must be a boy, there are 30 ways. If the monitor must be a girl, there are 20 ways.

🎯 Exam Tip: Clearly identify the constraints (e.g., "must be a boy" or "must be a girl") as these directly determine the number of available options for selection.

Question 3. A Signal is generated from 2 flags by putting one flag above the other. If 4 flags of different colours are available, how many different signals can be generated?
Answer: A signal is generated from 2 flags and there are 4 flags of different colours available.
. 1st flag can be any one of the available 4 flags.
. It can be selected in 4 ways. Now, 2nd flag is to be selected for which 3 flags are available for a different signal.
. 2nd flag can be anyone from these 3 flags.
. It can be selected in 3 ways.
. By using the fundamental principle of multiplication, Total number of ways in which a signal can be generated = 4 × 3 = 12
. 12 different signals can be generated.
In simple words: For a signal made of two different flags, you multiply the number of choices for the first flag by the number of choices for the second flag.

🎯 Exam Tip: When constructing an object with multiple distinct positions (like a signal with a top and bottom flag) and repetitions are not allowed, multiply the number of choices for each position sequentially (multiplication principle).

Question 4. How many two-letter words can be formed using letters from the word SPACE when repetition of letters
(i) is allowed
(ii) is not allowed
Answer: A two-letter word is to be formed out of the letters of the word SPACE.
(i) When repetition of the letters is allowed 1st letter can be selected in 5 ways 2nd letter can be selected in 5 ways
. By using the fundamental principle of multiplication, total number of 2-letter words = 5 × 5 = 25
(ii) When repetition of the letters is not allowed 1st letter can be selected in 5 ways 2nd letter can be selected in 4 ways
. By using the fundamental principle of multiplication, total number of 2-letter words = 5 × 4 = 20
In simple words: If letters can repeat, each position has all 5 options. If not, the second position has one fewer option than the first.

🎯 Exam Tip: Distinguish carefully between "repetition allowed" (choices remain constant) and "repetition not allowed" (choices decrease with each selection) as this significantly impacts the calculation.

Question 5. How many three-digit numbers can be formed from the digits 0, 1, 3, 5, 6 if repetitions of digits
(i) are allowed
(ii) are not allowed
Answer: The three-digit number is to be formed from the digits 0, 1, 3, 5, 6
(i) When repetition of digits is allowed: 100's place digit should be a non-zero number. Hence, it can be anyone from digits 1, 3, 5, 6
. 100's place digit can be selected in 4 ways. 0 can appear in 10's and unit's place and digits can be repeated.
. 10's place digit can be selected in 5 ways and the unit's place digit can be selected in 5 ways.
. By using the fundamental principle of multiplication, the total number of three-digit numbers = 4 × 5 × 5 = 100
(ii) When repetition of digits is not allowed: 100's place digit should be a non-zero number. Hence, it can be anyone from digits 1, 3, 5, 6
. 100's place digit can be selected in 4 ways 0 can appear in 10's and unit's place and digits can't be repeated.
. 10's place digit can be selected in 4 ways and the unit's place digit can be selected in 3 ways
. By using the fundamental principle of multiplication, total number of three-digit numbers = 4 × 4 × 3 = 48
In simple words: For three-digit numbers, the hundreds digit cannot be zero. Account for this restriction first, then consider if repetition is allowed or not for the remaining digits.

🎯 Exam Tip: Always address the most restrictive condition first, such as the hundreds digit not being zero for a number, before filling the remaining positions.

Question 6. How many three-digit numbers can be formed using the digits 2, 3, 4, 5, 6 if digits can be repeated?
Answer: A 3-digit number is to be formed from the digits 2, 3, 4, 5, 6 where digits can be repeated.
. The unit's place digit can be selected in 5 ways. 10's place digit can be selected in 5 ways. 100's place digit can be selected in 5 ways.
. By using fundamental principle of multiplication, the total number of 3-digit numbers = 5 × 5 × 5 = 125
In simple words: When repetition is allowed and no digit restrictions exist (like zero in the hundreds place), each position has the full set of choices.

🎯 Exam Tip: For problems involving repetition, the number of choices for each position remains constant and is independent of previous selections.

Question 7. A letter lock has 3 rings and each ring has 5 letters. Determine the maximum number of trials that may be required to open the lock.
Answer: A letter lock has 3 rings, each ring containing 5 different letters.
. A letter from each ring can be selected in 5 ways.
. By using fundamental principle of multiplication, the total number of trials that can be made = 5 × 5 × 5 = 125 Out of these 124 wrong attempts are made and in the 125th attempt, the lock gets opened, for a maximum number of trials.
. A maximum number of trials required to open the lock is 125.
In simple words: If a lock has multiple rings with choices, the total number of possible combinations is found by multiplying the choices for each ring. The maximum trials means trying all combinations until the correct one is found.

🎯 Exam Tip: The "maximum number of trials" in such a problem refers to the total number of unique combinations possible, assuming the last combination tried is the correct one.

Question 8. In a test that has 5 true/false questions, no student has got all correct answers and no sequence of answers is repeated. What is the maximum number of students for this to be possible?
Answer: For a set of 5 true/false questions, each question can be answered in 2 ways.
. By using fundamental principle of multiplication, the total number of possible sequences of answers = 2 × 2 × 2 × 2 × 2 = 32 Since no student has written all the correct answers.
. Total number of sequences of answers given by the students in the class = 32 - 1 = 31 Also, no student has given the same sequence of answers.
. Maximum number of students in the class = Number of sequences of answers given by the students = 31
In simple words: Calculate all possible answer combinations. Then, subtract one because no student got all answers correct, leaving the remaining unique sequences for different students.

🎯 Exam Tip: When a specific outcome is excluded (e.g., "no student got all correct answers"), subtract that one excluded possibility from the total number of possible outcomes.

Question 9. How many numbers between 100 and 1000 have 4 in the unit's place?
Answer: Numbers between 100 and 1000 are 3-digit numbers. A 3-digit number is to be formed from the digits 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 where the unit place digit is 4. Since Unit's place digit is 4.
. it can be selected in 1 way only. 10's place digit can be selected in 10 ways. For 3-digit number 100's place digit should be a non-zero number.
. 100's place digit can be selected in 9 ways.
. By using fundamental principle of multiplication, total number of numbers between 100 and 1000 which have 4 in the units place = 1 × 10 × 9 = 90
In simple words: To find three-digit numbers with 4 in the units place, fix the units digit as 4, allow all 10 digits for the tens place, and ensure the hundreds place is not zero (9 options).

🎯 Exam Tip: When a specific digit is required in a certain position, fix that position first, then apply rules for the remaining positions, remembering to account for restrictions like the hundreds digit not being zero.

Question 10. How many numbers between 100 and 1000 have the digit 7 exactly once?
Answer: Numbers between 100 and 1000 are 3-digit numbers. A 3-digit number is to be formed from the digits 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, where exactly one of the digits is 7. When 7 is in the unit's place The unit's place digit is 7.
. it can be selected in 1 way only. 10's place digit can be selected in 9 ways. 100's place digit can be selected in 8 ways.
. total number of numbers which have 7 in the unit's place = 1 × 9 × 8 = 72 When 7 is in 10's place The unit's place digit can be selected in 9 ways. 10's place digit is 7
. it can be selected in 1 way only. 100's place digit can be selected in 8 ways.
. total number of numbers which have 7 in 10's place = 9 × 1 × 8 = 72 When 7 is in 100's place The unit's place digit can be selected in 9 ways. 10's place digit can be selected in 9 ways. 100's place digit is 7
. it can be selected in 1 way.
. total numbers which have 7 in 100's place = 9 × 9 × 1 = 81
. total number of numbers between 100 and 1000 having digit 7 exactly once = 72 + 72 + 81 = 225.
In simple words: To find numbers with '7' exactly once, consider cases where 7 is in the units, tens, or hundreds place separately, ensuring other digits are not 7 and the hundreds digit is not zero. Sum the results from these mutually exclusive cases.

🎯 Exam Tip: For "exactly once" problems, break it down into cases for each possible position of the unique digit, then sum the outcomes, ensuring no double counting.

Question 11. How many four-digit numbers will not exceed 7432 if they are formed using the digits 2, 3, 4, 7 without repetition?
Answer: Among many set's of digits, the greatest number is possible when digits are arranged in descending order.
. 7432 is the greatest number, formed from the digits 2, 3, 4, 7.
. Since a 4-digit number is to be formed from the digits 2, 3, 4, 7, where repetition of the digit is not allowed.
. 1000's place digit can be selected in 4 ways. 100's place digit can be selected in 3 ways. 10's place digit can be selected in 2 ways. The unit's place digit can be selected in 1 way.
. Total number of numbers not exceeding 7432 that can be formed from the digits 2, 3, 4, 7 = Total number of four-digit numbers formed from the digits 2, 3, 4, 7 = 4 × 3 × 2 × 1 = 24
In simple words: Since 7432 is the largest possible number using these digits without repetition, any permutation of these digits will either be 7432 itself or smaller than it. So, we just need to find all possible permutations.

🎯 Exam Tip: When the upper bound given is the largest possible permutation of the available digits without repetition, the problem simplifies to finding the total number of permutations of those digits.

Question 12. If numbers are formed using digits 2, 3, 4, 5, 6 without repetition, how many of them will exceed 400?
Answer: Case I: Three-digit numbers with 4 occurring in hundred's place: 100's place digit can be selected in 1 way. Ten's place can be filled by any one of the numbers 2, 3, 5, 6.
. 10's place digit can be selected in 4 ways. The unit's place digit can be selected in 3 ways.
. total number of numbers which have 4 in 100's place = 1 × 4 × 3 = 12 Case II: Three-digit numbers more than 500 100's place digit can be selected in 2 ways. 10's place digit can be selected in 4 ways. Unit's place digit can be selected in 3 ways.
. total number of three digit numbers more than 500 = 2 × 4 × 3 = 24 Case III: Number of four digit numbers formed from 2, 3, 4, 5, 6 Since, repetition of digits is not allowed
. total four digit numbers formed = 5 × 4 × 3 × 2 = 120 Case IV: Number of five digit numbers formed from 2, 3, 4, 5, 6 Since, repetition of digits is not allowed
. total five digit numbers formed = 5 × 4 × 3 × 2 × 1 = 120
. total number of numbers that exceed 400 = 12 + 24 + 120 + 120 = 276
In simple words: To find numbers exceeding 400, sum the counts for three-digit numbers starting with 4, three-digit numbers starting with 5 or 6, and all possible four-digit and five-digit numbers, as all of them will be greater than 400.

🎯 Exam Tip: Break down "exceeds N" problems into cases based on the number of digits and the leading digit, ensuring no repetition constraint is violated for each case.

Question 13. How many numbers formed with the digits 0, 1, 2, 5, 7, 8 will fall between 13 and 1000 if digits can be repeated?
Answer: Case I: 2-digit numbers more than 13, less than 20, formed from the digits 0, 1, 2, 5, 7, 8 Number of such numbers = 3 Case II: 2-digit numbers more than 20 formed from 0, 1, 2, 5, 7, 8 Ten's place digit is selected from 2, 5, 7, 8.
. Ten's place digit can be selected in 4 ways. Unit's place digit is anyone from 0, 1, 2, 5, 7, 8
. The unit's place digit can be selected in 6 ways. Using the multiplication principle, the number of such numbers (repetition allowed) = 4 × 6 = 24 Case III: 3-digit numbers formed from 0, 1, 2, 5, 7, 8 100's place digit is anyone from 1, 2, 5, 7, 8.
. 100's place digit can be selected in 5 ways. As digits can be repeated, the 10's place and unit's place digits are selected from 0, 1, 2, 5, 7, 8
. 10's place and unit's place digits can be selected in 6 ways each. Using multiplication principle, the number of such numbers (repetition allowed) = 5 × 6 × 6 = 180 All cases are mutually exclusive and exhaustive.
. Required number = 3 + 24 + 180 = 207
In simple words: Break the problem into 2-digit numbers (more than 13, less than 20, and more than 20) and 3-digit numbers. Calculate each case ensuring digits can repeat, then sum them up.

🎯 Exam Tip: For range-bound problems (e.g., "between 13 and 1000"), systematically consider different digit counts (2-digit, 3-digit) and leading digit constraints, then sum the valid results.

Question 14. A school has three gates and four staircases from the first floor to the second floor. How many ways does a student have to go from outside the school to his classroom on the second floor?
Answer: A student can go inside the school from outside in 3 ways and from the first floor to the second floor in 4 ways.
. Number of ways to choose gates = 3 Number of ways to choose staircase = 4
. By using fundamental principle of multiplication, number of ways in which a student has to go from outside the school to his classroom = 4 × 3 = 12
In simple words: To find the total ways to complete a sequence of independent actions, multiply the number of ways for each action.

🎯 Exam Tip: Use the multiplication principle when multiple independent choices are made in sequence to form a complete path or outcome.

Question 15. How many five-digit numbers formed using the digit 0, 1, 2, 3, 4, 5 are divisible by 3 if digits are not repeated?
Answer: For a number to be divisible by 3. The sum of digits must be divisible by 3. Given 6 digits are 0, 1,2, 3, 4, 5. Sum of 1, 2, 3, 4, 5 = 15, which is divisible by 3.
. There are two cases of 5 digit numbers formed from 0, 1, 2, 3, 4, 5 and divisible by 3. Either 3 is selected in 5 digits (and 0 not selected) or 3 is not selected in 5 digits (and 0 is selected) Case I: 3 is not selected (and 0 is selected) i.e., the digits are 0, 1, 2, 4, 5. 10000's place digit can be selected in 4 ways (as 0 cannot appear). As digits are not repeated, 1000's place digit can be selected in 4 ways. 100's place digit can be selected in 3 ways. 10's place digit can be selected in 2 ways. The unit's place digit can be selected in 1 way.
. Using multiplication theorem, Number of 5-digit number formed from 0, 1, 2, 4, 5 (with no repetition of digits) = 4 × 4 × 3 × 2 × 1 = 96 Case II: 3 is selected (and 0 is not selected) i.e., 1, 2, 3, 4, 5 10000's place digit can be selected in 5 ways. 1000's place digit can be selected in 4 ways. 100's place digit can be selected in 3 ways. 10's place digit can be selected in 2 ways. The unit's place digit can be selected in 1 way. Using multiplication theorem, Number of 5-digit numbers formed from 1, 2, 3, 4, 5 = 5 × 4 × 3 × 2 × 1 = 120 Both the cases are mutually exclusive and exhaustive.
. Required number = 96 + 120 = 216
In simple words: For a number to be divisible by 3, its sum of digits must be divisible by 3. Identify combinations of 5 digits from the given set whose sum is divisible by 3. Since the sum of all 6 digits (0,1,2,3,4,5) is 15 (divisible by 3), any set of 5 digits will result in a sum divisible by 3 if the excluded digit is either 0 or 3. Calculate the permutations for each valid set of 5 digits, ensuring the leading digit is not zero, and sum the results.

🎯 Exam Tip: For divisibility rules, especially by 3, first check the sum of all available digits. Then, form subsets of digits that maintain this divisibility property, and calculate permutations for each subset while respecting constraints like non-repetition and non-zero leading digits.