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

Question 1. Find the value of r if 56Cr+6 : 54Pr-1 = 30800 : 1
Answer: Solution:
56Pr+6:54Pr+3=30800:1
.. \(\frac{56P_{r+6}}{54P_{r+3}} = \frac{30800}{1}\)
.. \(\frac{56!}{(56-r-6)!} \cdot \frac{(54-r-3)!}{54!} = 30800\)
.. \(\frac{56!}{(50-r)!} \cdot \frac{(51-r)!}{54!} = 30800\)
.. \(\frac{56 \times 55 \times 54!}{(50-r)!} \cdot \frac{(51-r)(50-r)!}{54!} = 30800\)
.. \(51-r = \frac{30800}{56 \times 55}\)
.. \(51-r = 10\)
.. \(r = 51-10\)
.. \(r = 41\)
In simple words: This problem involves solving an equation using permutations. We set up the permutation ratio, expand the factorials, and simplify the equation to find the value of 'r'.

🎯 Exam Tip: Pay close attention to permutation formulas and algebraic manipulation of factorials to avoid calculation errors. Remember that nPr = n! / (n-r)!.

Question 2. How many words can be formed by writing letters in the word CROWN in a different order?
Answer: Solution:
Five Letters of the word CROWN are to be permuted.
Number of different words = \(5! = 120\)
In simple words: To find the number of ways to arrange distinct letters in a word, calculate the factorial of the number of letters.

🎯 Exam Tip: For permutations of distinct items, the number of arrangements is n!, where n is the number of items.

Question 3. Find the number of words that can be formed by using all the letters in the word REMAIN. If these words are written in dictionary order, what will be the 40th word?
Answer: Solution:
There are 6 letters A, E, I, M, N, R
The number of words starting with A = \(5!\)
The number of words starting with E = \(5!\)
The number of words starting with I = \(5!\)
The number of words starting with M = \(5!\)
The number of words starting with N = \(5!\)
The number of words starting with R = \(5!\)
Total number of words = \(6 \times 5! = 720\)
Number of words starting with AE = \(4! = 24\)
Number of words starting with AIE = \(3! = 6\)
Number of words starting with AIM = \(3! = 6\)
The number of words starting with AINE = \(2!\)
Total words = \(24 + 6 + 6 + 2 = 38\)
39th word is AINMER
40th word is AINMRE
In simple words: To find a specific word in dictionary order, list possible prefixes alphabetically and sum the permutations of remaining letters until you locate the target word's position.

🎯 Exam Tip: When finding words in dictionary order, systematically list prefixes, calculate permutations for remaining letters, and subtract from the target position to narrow down the exact word.

Question 4. Find the number of ways of distributing n balls in n cells. What will be the number of ways if each cell must be occupied?
Answer: Solution:
There are n balls and n cells
(i) Every ball can be put in any of the n cells.
Number of distributions = \(n \times n \times ... \times n = (n)^n\)
(ii) For filling the first cell, n balls are available.
The first cell is filled in n ways.
The second cell is filled in (n - 1) ways
The third cell is filled in (n - 2) ways and so on.
the nth cell is tilled in one way.
Required number = \(n(n - 1)(n - 2) ... 1 = n!\)
In simple words: If n balls are distributed into n cells with no restrictions, each ball has n choices, leading to n^n ways. If each cell must be occupied, it means each ball goes into a unique cell, which is an arrangement problem, calculated as n factorial.

🎯 Exam Tip: Distinguish between 'balls into cells' problems where repetition is allowed (\(n^k\)) and where items are distinct and arranged (\(n!\)). 'Each cell must be occupied' implies distinct placement for distinct items.

Question 5. Thane is the 20th station from C.S.T. If a passenger can purchase a ticket from any station to any other station, how many different tickets must be available at the booking window?
Answer: Solution:
Taking CST as the first station and Thane as 20th,
Let us name CST as A0 next station as A₁ and so on, Thane is A20
From station A0, 20 different journeys are possible
From station A₁, 20 different journeys are possible.
From station A20, 20 different journeys are possible.
Total number of different tickets of different journeys = \(21 \times 20 = 420\)
In simple words: When tickets can be purchased between any two distinct stations from a set of 'n' stations, the total number of unique tickets is found by calculating permutations of two stations, which is n multiplied by (n-1).

🎯 Exam Tip: Identify if the problem requires ordered selections (permutations, like tickets from A to B vs B to A) or unordered selections (combinations). Here, 'from any station to any other station' implies distinct start and end points, making it a permutation problem.

Question 6. English alphabet has 11 symmetric letters that appear the same when looked at in a mirror. These letters are A, H, I, M, O, T, U, V, W, X, and Y. How many symmetric three letters passwords can be formed using these letters?
Answer: Solution:
Number of 3 Letter passwords = \(11P3\)
= \(11 \times 10 \times 9\)
= \(990\)
In simple words: To form a password of a specific length using a given set of distinct symmetric letters where the order matters and repetition is not implied, calculate the number of permutations (nPr).

🎯 Exam Tip: Passwords typically involve distinct ordered arrangements, hence permutations (nPr) are used. If repetition were allowed, it would be \(n^r\).

Question 7. How many numbers formed using the digits 3, 2, 0, 4, 3, 2, 3 exceed one million?
Answer: Solution:
A number that exceeds one million is to be formed from the digits 3, 2, 0, 4, 3, 2, 3.
Then the numbers should be any number of 7 digits which can be formed from these digits.
Also among the given numbers 2 repeats twice and 3 repeats thrice.
.. Required number of numbers = Total number of arrangements possible among these digits – number of arrangements of 7 digits which begin with 0.
= \(\frac{7!}{2!3!} - \frac{6!}{2!3!}\)
= \(\frac{7 \times 6 \times 5 \times 4 \times 3!}{2 \times 3!} - \frac{6 \times 5 \times 4 \times 3!}{2 \times 3!}\)
= \(7 \times 6 \times 5 \times 2 - 6 \times 5 \times 2\)
= \(6 \times 5 \times 2(7 - 1)\)
= \(60 \times 6\)
= \(360\)
.. 360 numbers that exceed one million can be formed with the digits 3, 2, 0, 4, 3, 2, 3.
In simple words: To count numbers exceeding one million formed from a given set of digits, first calculate all possible distinct 7-digit numbers using those digits, then subtract any invalid numbers (those starting with zero) to get the final count.

🎯 Exam Tip: When forming numbers with repeated digits, use the permutation formula for repeated items (n! / (p!q!r!)). Remember that a number cannot start with zero if it's meant to be a full 'n'-digit number.

Question 8. Ten students are to be selected for a project from a class of 30 students. There are 4 students who want to be together either in the project or not in the project. Find the number of possible selections.
Answer: Solution:

Required number = \(26C6 + 26C10\)
In simple words: When a group of individuals has a condition to either be all selected or all excluded together, calculate the selections for each scenario (group included, group excluded) separately, then sum the results for the total possible combinations.

🎯 Exam Tip: For 'together or not together' conditions in selection problems, break it into mutually exclusive cases: Case 1: the special group is included (select remaining from remaining). Case 2: the special group is excluded (select from remaining). Sum the combinations from both cases.

Question 9. A student finds 7 books of his interest but can borrow only three books. He wants to borrow the Chemistry part II book only if Chemistry Part I can also be borrowed. Find the number of ways he can choose three books that he wants to borrow.
Answer: Solution:

Required Number = \(5 + 10 = 15\)
In simple words: To handle conditional choices, break the problem into cases: one where the condition is met (e.g., specific book borrowed, then choose remaining from available), and another where the condition is not met (e.g., specific book not borrowed, then choose all from the remaining pool).

🎯 Exam Tip: Conditional selection problems are best solved by case analysis. Identify the conditions, create distinct cases (e.g., 'A is chosen' and 'A is not chosen'), calculate combinations for each case, and sum them up.

Question 10. 30 objects are to be divided into three groups containing 7, 10, 13 objects. Find the number of distinct ways of doing so.
Answer: Solution:
Required number = \(30C7 \times 23C10 \times 13C13\)
In simple words: To divide a set of distinct objects into groups of specific sizes, calculate the combinations for the first group, then for the second group from the remaining objects, and so on, multiplying the results together.

🎯 Exam Tip: For partitioning distinct items into distinct groups of specified sizes, multiply the combinations at each step. (\(n C k_1 \times (n-k_1) C k_2 \times (n-k_1-k_2) C k_3 ...\)).

Question 11. A student passes an examination if he secures a minimum in each of the 7 subjects. Find the number of ways a student can fail.
Answer: Solution:
Every subject a student may pass or fail.
.. Total number of outcomes = \(2^7 = 128\)
This number includes one case when the student passes in all subjects.
.. Required number = \(128 - 1 = 127\)
In simple words: To find the number of ways a student can fail an exam with multiple subjects, consider that for each subject, there are two outcomes (pass or fail). Calculate the total possible outcomes for all subjects, then subtract the single outcome where the student passes all subjects.

🎯 Exam Tip: Problems involving 'at least one' or 'fail' can often be solved by finding the total possible outcomes and subtracting the complement (the 'all pass' or 'none fail' case). For 'pass/fail' binary outcomes, powers of 2 are often involved.

Question 12. Nine friends decide to go for a picnic in two groups. One group decides to go by car and the other group decides to go by train. Find the number of different ways of doing so if there must be at least 3 friends in each group.
Answer: Solution:

Required number = \(9C3+9C4+9C5+9C6\)
= \((9C4+9C3) + (9C6+9C5)\)
= \(10C4+ 10C6\)
= \(\frac{10!}{6!4!} + \frac{10!}{4!6!}\)
= \(210+210 = 420\)
In simple words: When dividing items into two groups with a minimum size constraint for each, list all possible size distributions that meet the constraint, calculate the combinations for each distribution, and then sum these combinations to get the total number of ways.

🎯 Exam Tip: For 'at least' conditions in group formation, enumerate all valid combinations of group sizes. Remember to use combinations (nCr) when the order within the groups doesn't matter, and Pascal's Identity (\(nCr + nC(r-1) = (n+1)Cr\)) can simplify calculations.

Question 13. Five balls are to be placed in three boxes, where each box can contain upto five balls. Find the number of ways if no box is to remain empty.
Answer: Solution:
Let boxes be named as I, II, III
Let sets A, B, C represent cases in which boxes I, II, III remain empty
Then \(A \cup B \cup C\) represent the cases in which at least one box remains empty.
Then we use method of indirect counting
Required number = Total number of distributions – \(n(A \cup B \cup C)\) .....(i)
\(n(A \cup B \cup C)\) represent the number of undesirable cases
Total number of distributions = \(3 \times 3 \times 3 \times 3 \times 3 = 3^5 = 243\) .....(ii)
\(n(A \cup B \cup C) = n(A) + n(B) + n(C) - n(A \cap B) - n(B \cap C) - n(C \cap A) + n(A \cap B \cap C)\) .....(iii)
In box I is empty then every ball has two places (boxes) to go.
Similarly for box II and III.
.. \(n(A) + n(B) + n(C) = 3 \times 2^5\) ......(iv)
If boxes I and II remain empty then all balls go to box III
Similarly we would have two more cases.
.: \(n(A \cap B) + n(B \cap C) + n(C \cap A) = 3 \times 1^5\) ......(v)
∴ \(n(A \cap B \cap C) = 0\) .......(vi) [as all boxes cannot be empty]
Substitute from (iv), (v), (vi) to (iii) to get
\(n(A \cup B \cup C) = 3 \times 2^5 - 3 \times 1^5\)
= \(96 - 3\)
= \(93\)
Substitute \(n(A \cup B \cup C)\) and from (ii) to (i), we get
Required number = \(243 - 93 = 150\)
In simple words: To find the number of ways to distribute distinct items into distinct containers such that no container remains empty, use the Principle of Inclusion-Exclusion. Calculate the total ways without restrictions, then subtract cases where one or more containers are empty, adding back cases counted multiple times.

🎯 Exam Tip: Problems requiring 'no box remains empty' or 'onto functions' often use Inclusion-Exclusion. The formula is: Total - (at least 1 empty) + (at least 2 empty) - ... . Be precise with the power and factorial terms.

Question 14. A hall has 12 lamps and every lamp can be switched on independently. Find the number of ways of illuminating the hall.
Answer: Solution:
Every lamp is either ON or OFF.
There are 12 lamps
Number of instances = \(2^{12}\)
This number includes the case in which all 12 lamps are OFF.
.. Required Number = \(2^{12} - 1 = 4095\)
In simple words: For a set of independent switches, each having two states (on/off), the total number of combinations is 2 raised to the power of the number of switches. To find ways of 'illuminating' (meaning at least one lamp is on), subtract the single case where all lamps are off.

🎯 Exam Tip: When items have two independent states (e.g., on/off, yes/no), the total number of outcomes is \(2^n\). If the problem specifies 'at least one' of a certain state, subtract the case where all items are in the complementary state.

Question 15. How many quadratic equations can be formed using numbers from 0, 2, 4, 5 as coefficients if a coefficient can be repeated in an equation?
Answer: Solution:
Let the quadratic equation be \(ax^2 + bx + c = 0, a \neq 0\)

.. Required number = \(3 \times 4 \times 4 = 48\)
In simple words: To form quadratic equations with given coefficients and specific constraints (like 'a' cannot be zero), count the available choices for each coefficient independently, then multiply these choices to find the total number of distinct equations.

🎯 Exam Tip: When coefficients can be repeated, treat each coefficient's selection as an independent event. The constraint \(a \neq 0\) is critical and directly limits the choices for the leading coefficient.

Question 16. How many six-digit telephone numbers can be formed if the first two digits are 45 and no digit can appear more than once?
Answer: Solution:
Let the telephone number be 45abcd

.. Required number = \(8P4 = 1680\)
In simple words: When forming a multi-digit number with specific leading digits and no digit repetition, subtract the used digits from the total available, then calculate the permutations for the remaining positions using the remaining distinct digits.

🎯 Exam Tip: Always account for the digits already used and the 'no repetition' rule. The number of choices decreases for each subsequent position. Permutations (nPr) are suitable when order matters and repetition is not allowed.

Question 17. A question paper has 6 questions. How many ways does a student have if he wants to solve at least one question?
Answer: Solution:
Every question is 'SOLVED' or 'NOT SOLVED'.
There are 6 question.
Number of outcomes = \(2^6\)
This number includes the case when the student solves NONE of the question.
Required number = \(2^6 - 1\)
= \(64 - 1\)
= \(63\)
In simple words: For 'at least one' scenarios with binary choices (solve/not solve), find the total possible outcomes (2 to the power of number of questions) and subtract the single case where none are solved.

🎯 Exam Tip: The 'at least one' principle is frequently tested. Calculate total possibilities and subtract the case where the condition is *not* met at all. For independent binary choices, this often involves \(2^n - 1\).

Question 18. Find the number of ways of dividing 20 objects in three groups of sizes 8, 7 and 5.
Answer: Solution:
Select 8 object out of 20 in \(20C8\) ways
Select 7 object from remaining 12 in \(12C7\) ways and 5 objects from remaining 5 in \(5C5\) ways
Required number is = \(20C8 \times 12C7 \times 5C5\)
In simple words: To divide distinct objects into distinct groups of specified sizes, calculate the combinations for the first group, then for the second group from the remaining objects, and so on, multiplying the results.

🎯 Exam Tip: This is a partitioning problem for distinct objects. The number of ways is calculated by taking combinations for each group sequentially from the remaining pool: \(n C k_1 \times (n-k_1) C k_2 \times (n-k_1-k_2) C k_3\).

Question 19. There are 8 doctors and 4 lawyers in a panel. Find the number of ways for selecting a team of 6 if at least one doctor must be in the team.
Answer: Solution:
There are 8 doctors and 4 lawyers.
We need to select a team of 6 which contains at least one doctor.
Since, there are only 4 lawyers any team of 6 will contain at least two doctors.
Required number = \(12C6 = 924\)
In simple words: When a selection problem has a 'minimum' constraint that is automatically satisfied by the selection size and available groups, the problem simplifies to selecting the total number of items from the overall available pool.

🎯 Exam Tip: Always cross-check conditions against the total numbers available. If a 'minimum' requirement is inherently met (e.g., 'at least one doctor' is always true if minimum doctors are 2), then the condition becomes non-restrictive, and you simply calculate total combinations.

Question 20. Four parallel lines intersect another set of five parallel lines. Find the number of distinct parallelograms that can be formed.
Answer: Solution:
We need 2 lines from each set.
Required number = \(4C2 \times 5C2\)
= \(6 \times 10\)
= \(60\)
In simple words: To form a parallelogram from two sets of parallel lines, you need to choose two lines from the first set and two lines from the second set. The total number of parallelograms is the product of these two combinations.

🎯 Exam Tip: A parallelogram is formed by selecting two parallel lines from one set and two parallel lines from another intersecting set. This is a direct application of combinations for each set, multiplied together.