Maharashtra Board Class 11 Maths Part 2 Chapter 6 Permutations and Combinations 6.4 Solutions

Question 1. Find the number of permutations of letters in each of the following words:
(i) DIVYA
(ii) SHANTARAM
(iii) REPRESENT
(iv) COMBINE
Answer:
Solution:
(i) There are 5 letters in the word DIVYA which can be arranged in \(5!\) Way = \(120\) ways

(ii) There are 9 letters in the word SHANTARAM in which 'A' repeats 3 times.
\(\therefore\) Number of permutations of the letters of the word SHANTARAM = \( \frac{9!}{3!} \)
= \( 9 \times 8 \times 7 \times 6 \times 5 \times 4 \)
= \(60480\)

(iii) There are 9 letters in the word REPRESENT in which 'E' repeats 3 times and 'R' repeats 2 times.
\(\therefore\) Number of permutations of the letters of the word REPRESENT = \( \frac{9!}{3!2!} \)
= \( \frac{9 \times 8 \times 7 \times 6 \times 5 \times 4}{2} \)
= \(30240\)

(iv) There are 7 distinct letters in the word COMBINE which can be arranged among themselves in \(7!\) = \(5040\) ways
In simple words: This question asks to calculate the number of unique ways to arrange the letters in given words, considering repeated letters reduce the total number of distinct arrangements.

🎯 Exam Tip: Pay attention to repeated letters in a word. If a letter repeats 'n' times, divide the total factorial of letters by 'n!'.

Question 2. You have 2 identical books on English, 3 identical books on Hindi and 4 identical books on Mathematics. Find the number of distinct ways of arranging them on a shelf.
Answer:
Solution:
There are total 9 books to be arranged on a shelf.
Out of these 9 books, 2 books on English, 3 books on Hindi and 4 books on Mathematics are identical.
\(\therefore\) Total number of arrangements = \( \frac{9!}{2!3!4!} \)
= \( \frac{9 \times 8 \times 7 \times 6 \times 5 \times 4!}{2 \times 3 \times 2 \times 4!} \)
= \( 9 \times 4 \times 7 \times 5 \)
= \(1260\)
\(\therefore\) In \(1260\) distinct ways the books can be arranged on a shelf.
In simple words: This problem involves finding the number of distinct arrangements for a set of items where some items are identical, using the formula for permutations with repetitions.

🎯 Exam Tip: Clearly identify the total number of items and the count of each type of identical item to correctly apply the permutation formula.

Question 3. A coin is tossed 8 times. In how many ways can we obtain
(i) 4 heads and 4 tails?
(ii) at least 6 heads?
Answer:
Solution:
A coin is tossed 8 times. All heads are identical and all tails are identical.
(i) We can obtain 4 heads and 4 tails in \( \frac{8!}{4!4!} \)
= \( \frac{8 \times 7 \times 6 \times 5}{4 \times 3 \times 2} \)
= \(70\) ways
\(\therefore\) In \(70\) different ways we can obtain 4 heads and 4 tails.

(ii) When at least 6 heads are to be obtained
\(\therefore\) Outcome can be (6 heads and 2 tails) or (7 heads and 1 tail) or (8 heads)
\(\therefore\) Number of ways in which it can be obtained = \( \frac{8!}{6!2!} + \frac{8!}{7!1!} + \frac{8!}{8!} \)
= \( \frac{8 \times 7}{2} + 8 + 1 \)
= \(28 + 8 + 1\)
= \(37\)
\(\therefore\) In \(37\) different ways we can obtain at least 6 heads.
In simple words: This question applies permutation principles to coin tosses, first for an exact number of heads/tails, then for a condition of "at least" a certain number of heads, which requires summing up probabilities for multiple outcomes.

🎯 Exam Tip: For "at least" problems, break down the condition into mutually exclusive cases (e.g., exactly 6, exactly 7, exactly 8) and sum their individual permutations.

Question 4. A bag has 5 red, 4 blue, and 4 green marbles. If all are drawn one by one and their colours are recorded, how many different arrangements can be found?
Answer:
Solution:
There is a total of 13 marbles in a bag.
Out of these 5 are Red, 4 Blue, and 4 are Green marbles.
All balls of the same colour are taken to be identical.
\(\therefore\) Required number of arrangements = \( \frac{13!}{5!4!4!} \)
In simple words: This problem asks for the number of distinct sequences when drawing colored marbles, where marbles of the same color are indistinguishable, using the formula for permutations with repetitions.

🎯 Exam Tip: Similar to arranging letters in a word, identify the total number of items and the count of each identical type.

Question 5. Find the number of ways of arranging letters of the word MATHEMATICAL. How many of these arrangements have all vowels together?
Answer:
Solution:
There are 12 letters in the word MATHEMATICAL in which 'M' repeats 2 times, 'A' repeats 3 times, and 'T' repeats 2 times.
\(\therefore\) Total number of arrangements = \( \frac{12!}{2!3!2!} \)
When all the vowels
i.e., 'A', 'A', 'A', 'E', T are to be kept together
Number of arrangements of these vowels = \( \frac{5!}{3!} \) ways.
Let us consider these vowels together as one unit.
This unit is to be arranged with 7 other letters in which ‘M' and 'T' repeated 2 times each.
\(\therefore\) Number of arrangements = \( \frac{8!}{2!2!} \)
\(\therefore\) Total number of arrangements = \( \frac{8! \times 5!}{2!2!3!} \)
In simple words: This problem involves finding total permutations of a word with repeated letters, then a specific case where all vowels must stay together, treating the block of vowels as a single unit.

🎯 Exam Tip: When a group of letters must stay together, treat them as one single entity. Calculate permutations within that entity and then permutations of the entities themselves.

Question 6. Find the number of different arrangements of letters in the word MAHARASHTRA. How many of these arrangements have
(i) letters M and T never together?
(ii) all vowels together?
Answer:
Solution:
There are 11 letters in the word MAHARASHTRA in which 'A' is repeated 4 times, 'H' repeated 2 times, and 'R' repeated 2 times.
\(\therefore\) Total number of arrangements is \( \frac{11!}{4!2!2!} \)
\(\therefore\) \( \frac{11!}{4!2!2!} \) different words can be formed from the letters of the word MAHARASHTRA.

(i) Other than M and T. there are 9 letters in which A repeats 4 times, H repeats twice, R repeats twice
The number of arrangements of the a letter = \( \frac{9!}{4!2!2!} \)
These 9 letters create 10 gaps in which M and T are to be arranged
The number of arrangements of M and T = \( ^{10}P_2 \)
\(\therefore\) Total number arrangement having M and T never together = \( \frac{9! \times ^{10}P_2}{4!2!2!} \)

(ii) When all vowels are together.
There are 4 vowels in the word MAHARASHTRA i.e., A, A, A, A
Let us consider these 4 vowels as one unit, they themselves can be arranged in \( \frac{4!}{4!} = 1 \) way.
This unit is to be arranged with 7 other letters which can be done in \(8!\) ways
\(\therefore\) Total number of arrangements = \( \frac{8!}{2!2!} \)
\(\therefore\) \( \frac{8!}{2!2!} \) different words can be formed if vowels are always together.
In simple words: This question explores permutations of a word with repeated letters, first for the total arrangements, then for specific conditions: when two particular letters are never together (calculated by total - together), and when all vowels are grouped.

🎯 Exam Tip: For "never together" problems, it's often easier to calculate "total arrangements" minus "arrangements where they are together". For "all vowels together", treat the vowel group as a single unit.

Question 7. How many different words are formed if the letter R is used thrice and letters S and T are used twice each?
Answer:
Solution:
When 'R' is used thrice, 'S' is used twice and 'T' is used twice,
\(\therefore\) Total number of letters available = 7, of which 'S' and 'T' repeat 2 times each, 'R' repeats 3 times.
\(\therefore\) Required number of arrangements = \( \frac{7!}{2!2!3!} \)
= \( \frac{7 \times 6 \times 5 \times 4 \times 3!}{2 \times 1 \times 2 \times 1 \times 3!} \)
= \( 7 \times 6 \times 5 \)
= \(210\)
\(\therefore\) 210 different words can be formed with the letter R is used thrice and letters S and T are used twice each.
In simple words: This problem asks for the number of distinct words that can be formed using a given set of letters, where some letters have specified repetition counts.

🎯 Exam Tip: Identify the total count of letters and the exact repetition count for each distinct letter to correctly apply the permutation formula for identical items.

Question 8. Find the number of arrangements of letters in the word MUMBAI so that the letter B is always next to A.
Answer:
Solution:
There are 6 letters in the word MUMBAI.
These letters are to be arranged in such a way that 'B' is always next to 'A'.
Let us consider AB as one unit. This unit with other 4 letters in which 'M' repeats twice, is to be arranged.
\(\therefore\) Total number of arrangements when B is always next to A = \( \frac{5!}{2!} \)
= \( \frac{5 \times 4 \times 3 \times 2!}{2!} \)
= \(60\)
In simple words: This problem requires arranging letters where two specific letters (AB) must always appear together, treating them as a single combined unit.

🎯 Exam Tip: When elements must stay together, group them into a single unit. Remember to also consider permutations within that grouped unit if the order matters (though in this case, "B next to A" implies a specific order).

Question 9. Find the number of arrangements of letters in the word CONSTITUTION that begin and end with N.
Answer:
Solution:
There are 12 letters in the word CONSTITUTION, in which 'O', 'N', T repeat two times each, 'T' repeats 3 times.
The arrangement starts and ends with 'N', 10 letters other than N can be arranged between two N, in which 'O' and 'l' repeat twice each and 'T' repeats 3 times.
\(\therefore\) Total number of arrangements with the letter N at the beginning and at the end = \( \frac{10!}{2!2!3!} \)
In simple words: This question asks for permutations of a word with fixed letters at the start and end, and repetitions among the remaining letters.

🎯 Exam Tip: When letters are fixed in certain positions, remove them from the total pool, and calculate permutations for the remaining letters in the remaining positions, accounting for any repetitions.

Question 10. Find the number of different ways of arranging letters in the word ARRANGE. How many of these arrangements have two R's and two A's not together?
Answer:
Solution:
(i) There are 7 letters in the word ARRANGE in which A is repeated 2 times and R is repeated 2 times
\(\therefore\) The number of arrangements = \( \frac{7!}{2!2!} = 1260 \)

(ii) A: set of words having 2A together
B: set of words having 2R together
Number of words having both A and both R not together
= \(1260 - n(A \cup B)\)
= \(1260 - [n(A) + n(B) - n(A \cap B)]\) ......(i)
n(A) = number of ways in which (AA) R, R, N, G, E are to be arranged
\(\therefore\) n(A) = \( \frac{6!}{2!} = 360 \)
n(B) = number of ways in which (RR), A, A, N, G, E are to be arranged
\(\therefore\) n(B) = \( \frac{6!}{2!} = 360 \)
n(A \( \cap \) B) = number of ways in which (AA), (RR), N, G, E are to be arranged
\(\therefore\) n(A \( \cap \) B) = \(5! = 120\)
Substituting n(A), n(B), n(A \( \cap \) B) in (i), we get
Number of words having both A and both R not together
= \(1260 - [360 + 360 - 120]\)
= \(1260 - 600\)
= \(660\)
In simple words: This problem involves finding arrangements where two pairs of identical letters are *not* together, using the Principle of Inclusion-Exclusion by subtracting cases where A's are together or R's are together (and adding back the overlap).

🎯 Exam Tip: For "not together" conditions involving multiple groups, apply the Principle of Inclusion-Exclusion: Total - (Group 1 together) - (Group 2 together) + (Both Groups together).

Question 11. How many distinct 5 digit numbers can be formed using the digits 3, 2, 3, 2, 4, 5.
Answer:
Solution:
5 digit numbers are to be formed from 2, 3, 2, 3, 4, 5.
Case I: Numbers formed from 2, 2, 3, 4, 5 OR 2, 3, 3, 4, 5
Number of such numbers = \( \frac{5!}{2!} \times 2 \)
= \(5!\)
= \(120\)
Case II: Numbers formed from 2, 2, 3, 3 and any one of 4 or 5
Number of such numbers = \( \frac{5!}{2!2!} \times 2 = 60 \)
Required number = \(120 + 60 = 180\)
\(\therefore\) 180 distinct 5 digit numbers can be formed using the digit 3, 2, 3, 2, 4, 5.
In simple words: This problem requires forming 5-digit numbers from a given set of 6 digits with repetitions, considering different combinations of repeated digits.

🎯 Exam Tip: Break down problems involving selecting from a pool with repetitions into cases based on which digits are chosen and how many times they repeat.

Question 12. Find the number of distinct numbers formed using the digits 3, 4, 5, 6, 7, 8, 9, so that odd positions are occupied by odd digits.
Answer:
Solution:
A number is to be formed with digits 3, 4, 5, 6, 7, 8, 9 such that odd digits always occupy the odd places.
There are 4 odd digits i.e. 3, 5, 7, 9.
They can be arranged at 4 odd places among themselves in \(4!\) ways = \(24\) ways
3 even places of the number are occupied by even digits (i.e. 4, 6, 8).
\(\therefore\) They can be arranged in \(3!\) ways = \(6\) ways
\(\therefore\) Total number of arrangements = \( 24 \times 6 = 144 \)
\(\therefore\) 144 numbers can be formed so that odd digits always occupy the odd positions.
In simple words: This problem involves arranging digits with a constraint: odd digits must be in odd positions and even digits in even positions. The arrangements are calculated independently for odd and even positions and then multiplied.

🎯 Exam Tip: When there are separate constraints for different positions (like odd/even positions), calculate the arrangements for each set of positions independently and then multiply the results.

Question 13. How many different 6-digit numbers can be formed using digits in the number 659942? How many of them are divisible by 2?
Answer:
Solution:
A 6-digit number is to be formed using digits of 659942, in which 9 repeats twice.
\(\therefore\) Total number of arrangements = \( \frac{6!}{2!} \)
= \( \frac{6 \times 5 \times 4 \times 3 \times 2!}{2!} \)
= \(360\)
\(\therefore\) 360 different 6-digit numbers can be formed.
For a number to be divisible by 2,
Last digits can be selected in 3 ways
Remaining 5 digit in which, 9 appears twice are arranged in \( \frac{5!}{2!} \) ways
\(\therefore\) Total number of arrangements = \( \frac{5!}{2!} \times 3 = 180 \)
\(\therefore\) 180 numbers are divisible by 2.
In simple words: This problem deals with forming 6-digit numbers from a set with repetitions, first finding total permutations, then finding permutations that are divisible by 2 (meaning the last digit must be even).

🎯 Exam Tip: For divisibility rules, focus on the constraint on the last digit. Fix the last digit to meet the condition, then permute the remaining digits, accounting for repetitions.

Question 14. Find the number of distinct words formed from letters in the word INDIAN. How many of them have the two N's together?
Answer:
Solution:
There are 6 letters in the word INDIAN in which I and N repeat twice.
Number of different words that can be formed using the letters of the word INDIAN = \( \frac{6!}{2!2!} \)
= \( \frac{6 \times 5 \times 4 \times 3 \times 2!}{2 \times 2!} \)
= \(180\)
\(\therefore\) 180 different words can be formed with the letters of the word INDIAN.
When two N's are together.
Let us consider the two N's as one unit.
They can be arranged with 4 other letters in \( \frac{5!}{2!} \) ways
= \( \frac{5 \times 4 \times 3 \times 2!}{2!} \)
= \(60\) ways.
\(\therefore\) 2N can be arranged in 1 way
\(\therefore\) Total number of arrangements = \( 60 \times 1 = 60 \) ways
\(\therefore\) 60 words are such that two N's are together.
In simple words: This problem first calculates the total distinct permutations of the word INDIAN. Then, it determines how many of these arrangements have the two 'N's adjacent by treating them as a single block.

🎯 Exam Tip: When specific letters must be together, treat them as a single combined unit. Then, find the permutations of these units, including any internal permutations of the combined unit (if applicable, in this case 2 N's are identical, so 1 way).

Question 15. Find the number of different ways of arranging letters in the word PLATOON if
(i) the two O's are never together.
(ii) consonants and vowels occupy alternate positions.
Answer:
Solution:
(i) When the two O's are never together:
Let us arrange the other 5 letters first, which can be done in \(5! = 120\) ways.
The letters P, L, A, T, N create 6 gaps, in which O's are arranged.
\(\therefore\) Two O's in 6 gaps can be arranged in \( \frac{^{6}P_2}{2!} \) ways
= \( \frac{6!}{(6-2)!2!} \) ways
= \( \frac{6 \times 5 \times 4!}{4! \times 2 \times 1} \) ways
= \( 3 \times 5 \) ways
= \(15\) ways
\(\therefore\) Total number of arrangements if the two O's are never together = \( 120 \times 15 = 1800 \)

(ii) When consonants and vowels occupy alternate positions:
There are 4 consonants and 3 vowels in the word PLATOON.
\(\therefore\) At odd places consonants occur and at even places vowels occur.
4 consonants can be arranged among themselves in \(4!\) ways
3 vowels in which O occurs twice and A occurs once.
\(\therefore\) They can be arranged in \( \frac{3!}{2!} \) ways
\(\therefore\) Required number of arrangements if the consonants and vowels occupy alternate positions = \( 4! \times \frac{3!}{2!} \)
= \( 4 \times 3 \times 2 \times \frac{3!}{2!} \)
= \(72\)
In simple words: This problem involves arranging letters of a word with two scenarios: first, where two specific letters are never adjacent (calculated by placing other letters and then finding gaps for the non-adjacent ones); second, where consonants and vowels must alternate positions.

🎯 Exam Tip: For "never together" cases, arrange the other letters first, create gaps, and then place the "never together" items into those gaps. For "alternate positions", arrange consonants in consonant slots and vowels in vowel slots separately, then multiply the results.