Question 1

   Question: The number of ways, 5 boys and 4 girls can sit in a row so that either all the boys sit together or no two boys sit together, is.

   Options: (Not explicitly listed in source; Answer is an integer).

   Correct Answer: 17280.

   Year: JEE Main 2025 (Online) 23rd January Evening Shift.

   Solution (as Given in the Source): A : number of ways that all boys sit together $= 5! \times 5!$; B : number of ways if no 2 boys sit together $= 4! \times 5!$; $A \cap B = \phi$. Required no. of ways $= 5! \times 5! + 4! \times 5! = 17280$.

   Step Solution:

    1.  Case 1 (All boys together): Treat the 5 boys as a single unit. This unit plus the 4 girls makes 5 entities to arrange, which is $5! = 120$ ways.

    2.  Internal Arrangement: The 5 boys can be arranged among themselves in $5! = 120$ ways. Total for Case 1 is $120 \times 120 = 14,400$.

    3.  Case 2 (No two boys together): First, arrange the 4 girls in $4! = 24$ ways.

    4.  Gap Method: 4 girls create 5 available gaps ( _ G _ G _ G _ G _ ). Place the 5 boys in these 5 gaps in $5! = 120$ ways.

    5.  Final Calculation: Total for Case 2 is $24 \times 120 = 2,880$. Adding both cases: $14,400 + 2,880 = 17,280$.

   Difficulty Level: Medium.

   Concept Name: Permutations (String Method and Gap Method).

   Short cut solution: Sum the results of the "All Together" formula ($n! \times (N-n+1)!$) and the "None Together" formula (arranging the larger group and placing the smaller group in gaps).

 Question 5

   Question: The number of 6-letter words, with or without meaning, that can be formed using the letters of the word MATHS such that any letter that appears in the word must appear at least twice, is.

   Options: (Not explicitly listed in source; Answer is an integer).

   Correct Answer: 1405.

   Year: JEE Main 2025 (Online) 29th January Morning Shift.

   Solution (as Given in the Source): (i) Single letter used $= 5$; (ii) Two distinct letters used $= ^5C_2 \times (\frac{6!}{2!4!} \times 2 + \frac{6!}{3!3!}) = 500$; (iii) Three distinct letters used $= ^5C_3 \times \frac{6!}{2!2!2!} = 900$. Total no. of words $= 1405$.

   Step Solution:

    1.  Case 1 (1 Distinct Letter): All 6 letters are the same. There are 5 choices (M, A, T, H, or S). Total $= 5$.

    2.  Case 2 (2 Distinct Letters): The frequency can be (4,2) or (3,3). For (4,2): Choose 2 letters ($^5C_2=10$), decide which is used 4 times (2 ways), and arrange: $10 \times 2 \times \frac{6!}{4!2!} = 300$.

    3.  Case 2 continued (3,3): Choose 2 letters ($^5C_2=10$) and arrange: $10 \times \frac{6!}{3!3!} = 200$. Total for Case 2 $= 300 + 200 = 500$.

    4.  Case 3 (3 Distinct Letters): The frequency must be (2,2,2). Choose 3 letters ($^5C_3=10$) and arrange: $10 \times \frac{6!}{2!2!2!} = 900$.

    5.  Total: Summing all cases: $5 + 500 + 900 = 1,405$.

   Difficulty Level: Hard.

   Concept Name: Permutations of Multisets (Case-based counting).

   Short cut solution: Break the problem into integer partitions of 6 where each part $\ge 2$ (i.e., {6}, {4,2}, {3,3}, {2,2,2}) and apply multinomial coefficients.

 Question 10

   Question: In a group of 3 girls and 4 boys, there are two boys $B_1$ and $B_2$. The number of ways, in which these girls and boys can stand in a queue such that all the girls stand together, all the boys stand together, but $B_1$ and $B_2$ are not adjacent to each other, is :.

   Options: A. 120, B. 96, C. 72, D. 144.

   Correct Answer: D.

   Year: JEE Main 2025 (Online) 22nd January Evening Shift.

   Solution (as Given in the Source): There are two blocks (girls and boys) $= 2$ ways. Girls arranged in $3! = 6$ ways. For boys, total arrangements ($4! = 24$) minus adjacent $B_1, B_2$ ($2! \times 3! = 12$) $= 12$ valid arrangements. Total ways $= 2 \times 6 \times 12 = 144$.

   Step Solution:

    1.  Block Arrangement: Since girls must be together and boys must be together, treat them as 2 blocks. Arrange these 2 blocks in $2! = 2$ ways.

    2.  Internal Girl Arrangement: Arrange the 3 girls within their block in $3! = 6$ ways.

    3.  Restricted Boy Arrangement: Calculate valid arrangements for 4 boys where $B_1$ and $B_2$ are not together. Total permutations $= 4! = 24$.

    4.  Subtraction Method: Arrangements where $B_1$ and $B_2$ are together $= (B_1B_2 \text{ as one unit}) \times \text{other 2 boys} = 3! \times 2! = 12$.

    5.  Final Multiplier: Valid boy arrangements $= 24 - 12 = 12$. Total ways $= 2 \times 6 \times 12 = 144$.

   Difficulty Level: Medium.

   Concept Name: String Method with Internal Constraints.

   Short cut solution: Total ways $= (\text{Arrangement of Blocks}) \times (\text{Permutation of Girls}) \times (\text{Permutation of Boys excluding adjacent pair})$. $2 \times 3! \times (4! - 2 \times 3!) = 12 \times 12 = 144$.

 Question 11

   Question: The number of words, which can be formed using all the letters of the word "DAUGHTER", so that all the vowels never come together, is :

   Options: 

       A. 34000

       B. 37000

       C. 35000

       D. 36000

   Correct Answer: D

   Year: JEE Main 2025 (Online) 23rd January Morning Shift

   Solution (as Given in the Source): DAUGHTER. Total words $= 8!$. Total words in which vowels are together $= 6! \times 3!$. Words in which all vowels are not together $= 8! - 6! \times 3! = 6!  = 720 \times 50 = 36000$.

   Step Solution:

    1.  Total Arrangements: The word DAUGHTER has 8 distinct letters, so total arrangements $= 8! = 40,320$.

    2.  Identify Vowels: The vowels are A, U, and E (3 total).

    3.  Vowels Together: Treat (AUE) as one block. This block plus the 5 consonants gives 6 units to arrange ($6!$).

    4.  Internal Arrangement: The 3 vowels can be arranged within their block in $3!$ ways. Total vowels together $= 6! \times 3! = 720 \times 6 = 4,320$.

    5.  Final Calculation: Vowels never together = (Total) $-$ (Vowels together) $= 40,320 - 4,320 = 36,000$.

   Difficulty Level: Easy.

   Concept Name: Permutations (String Method and Complementary Counting).

   Short cut solution: Total ways $= 6! \times (8 \times 7 - 3!) = 720 \times (56 - 6) = 36,000$.

 Question 15

   Question: Let $P$ be the set of seven digit numbers with sum of their digits equal to 11. If the numbers in $P$ are formed by using the digits 1, 2 and 3 only, then the number of elements in the set $P$ is :

   Options: 

       A. 164 

       B. 158 

       C. 161 

       D. 173

   Correct Answer: C

   Year: JEE Main 2025 (Online) 29th January Morning Shift

   Solution (as Given in the Source): (i) number of numbers created using $111133 = \frac{7!}{5!2!} \Rightarrow 21$; (ii) number of numbers created using $1111223 = \frac{7!}{4!2!} \Rightarrow 105$; (iii) number of numbers created using $1112222 = \frac{7!}{4!3!} \Rightarrow 35$. Total $= 161$.

   Step Solution:

    1.  Identify Cases: We need 7 digits from $\{1, 2, 3\}$ that sum to 11. 

    2.  Case 1 (Five 1s, Two 3s): $5(1) + 2(3) = 11$. Arrangements $= \frac{7!}{5!2!} = 21$.

    3.  Case 2 (Four 1s, Two 2s, One 3): $4(1) + 2(2) + 1(3) = 11$. Arrangements $= \frac{7!}{4!2!1!} = 105$.

    4.  Case 3 (Three 1s, Four 2s): $3(1) + 4(2) = 11$. Arrangements $= \frac{7!}{3!4!} = 35$.

    5.  Final Sum: Total elements in set $P = 21 + 105 + 35 = 161$.

   Difficulty Level: Hard.

   Concept Name: Permutations of Multisets (Partitioning of Sum).

   Short cut solution: List the integer partitions of 11 into seven parts where each part $\in \{1, 2, 3\}$ and sum the multinomial coefficients.

 Question 17

   Question: The number of sequences of ten terms, whose terms are either 0 or 1 or 2 , that contain exactly five 1s and exactly three 2s, is equal to :

   Options: 

       A. 360

       B. 2520

       C. 1820

       D. 45

   Correct Answer: B

   Year: JEE Main 2025 (Online) 2nd April Morning Shift

   Solution (as Given in the Source): Since there are 5 ones and 3 twos, you will need 2 zeros to fill the sequence (because $5 + 3 + 2 = 10$). The number of arrangements is $\frac{10!}{5! \times 3! \times 2!} = 2520$.

   Step Solution:

    1.  Find Remaining Terms: The sequence has 10 positions. If 5 are "1"s and 3 are "2"s, the remaining $10 - (5+3) = 2$ positions must be "0"s.

    2.  Set up Permutation: We are arranging a multiset of $\{1, 1, 1, 1, 1, 2, 2, 2, 0, 0\}$.

    3.  Apply Formula: Number of ways $= \frac{10!}{5! \times 3! \times 2!}$.

    4.  Calculation: $\frac{10 \times 9 \times 8 \times 7 \times 6 \times 5!}{5! \times (3 \times 2 \times 1) \times (2 \times 1)}$.

    5.  Simplify: $\frac{30240}{12} = 2520$.

   Difficulty Level: Easy.

   Concept Name: Permutations of Multisets.

   Short cut solution: Use the selection method: $\binom{10}{5} \times \binom{5}{3} \times \binom{2}{2} = 252 \times 10 \times 1 = 2520$.

 Question 18

   Question: The number of ways, in which the letters A, B, C, D, E can be placed in the 8 boxes of the figure below so that no row remains empty and at most one letter can be placed in a box, is :

   Options: A. 5880, B. 840, C. 960, D. 5760

   Correct Answer: D

   Year: JEE Main 2025 (Online) 2nd April Evening Shift

   Solution (as Given in the Source): Let $x, y, z$ be the number of box which are filled $\Rightarrow 1 \leq x \leq 3, 1 \leq y \leq 3, 1 \leq z \leq 2$. Total way $S = (48)$ to fill boxes Now to arrange $a, b, c, d$ and $e$ Number of ways will be $48 \cdot 5! = 5760$.

   Step Solution:

    1.  Analyze Constraints: There are 8 boxes and 5 distinct letters. Each row must have at least one letter.

    2.  Determine Box Selection: The number of ways to select 5 boxes out of 8 such that each of the three rows (capacities 3, 3, and 2) has at least one box selected is calculated as 48 ways.

    3.  Letter Arrangement: Once 5 boxes are chosen, the 5 distinct letters (A, B, C, D, E) can be arranged in these boxes in $5!$ ways.

    4.  Final Calculation: Total ways = (Ways to pick boxes) $\times$ (Ways to arrange letters) $= 48 \times 120$.

    5.  Result: $48 \times 120 = 5760$.

   Difficulty Level: Hard.

   Concept Name: Combinatorial Selection with Row Constraints.

   Short cut solution: Subtract cases where a row is empty from total selections ($^8C_5$) and multiply by $5!$: $(56 - 6 - 2) \times 120 = 48 \times 120 = 5760$.

 Question 26

   Question: The total number of words (with or without meaning) that can be formed out of the letters of the word 'DISTRIBUTION' taken four at a time, is equal to

   Options: (Not explicitly listed in source; Answer is an integer).

   Correct Answer: 3734

   Year: 31-Jan-2024 Shift 1

   Solution (as Given in the Source): We have II, TT, D,S, R, B, U,O, N. (i) Selection (a,a,a,b) $= ^8C_1 \times \frac{4!}{3!} = 32$. (ii) Selection (a,a,b,b) $= \frac{4!}{2!2!} = 6$. (iii) Selection (a,a,b,c) $= ^2C_1 \times ^8C_2 \times \frac{4!}{2!} = 672$. (iv) Selection (a,b,c,d) $= ^9C_4 \times 4! = 3024$. Total $= 3024 + 672 + 6 + 32 = 3734$.

   Step Solution:

    1.  Categorize Letters: Distinct types are {I, T, D, S, R, B, U, O, N}. Count: I(3), T(2), and others are 1 each.

    2.  Case 1 (3 Alike, 1 Diff): Choose 'I' as the triple and 1 from 8 others: $1 \times ^8C_1 \times \frac{4!}{3!} = 32$.

    3.  Case 2 (2 Pairs): Choose both 'I' and 'T' as pairs: $^2C_2 \times \frac{4!}{2!2!} = 6$.

    4.  Case 3 (1 Pair, 2 Diff): Choose 1 pair from {I, T} and 2 from remaining 8 types: $^2C_1 \times ^8C_2 \times \frac{4!}{2!} = 2 \times 28 \times 12 = 672$.

    5.  Case 4 (All Diff): Choose 4 distinct types from 9 available: $^9C_4 \times 4! = 126 \times 24 = 3024$. Total $= 32 + 6 + 672 + 3024 = 3734$.

   Difficulty Level: Hard.

   Concept Name: Permutations of Multisets (Case-based analysis).

   Short cut solution: Systematically list possible groups based on letter frequency ($3+1, 2+2, 2+1+1, 1+1+1+1$) and apply multinomial coefficients.

 Question 47

   Question: The number of words, with or without meaning, that can be formed using all the letters of the word ASSASSINATION so that the vowels occur together, is

   Options: (Not explicitly listed in source; Answer is an integer).

   Correct Answer: 50400

   Year: 1-Feb-2023 Shift 1

   Solution (as Given in the Source): Vowels: A,A,A,I,I,O. Consonants: S,S,S,S,N,N,T. Total number of ways in which vowels come together $= \frac{8!}{4!2!} \times \frac{6!}{3!2!} = 50400$.

   Step Solution:

    1.  Group Vowels: Treat (AAAIIO) as one single block. The consonants are S, S, S, S, N, N, T.

    2.  Total Units: There are 7 consonants + 1 vowel block = 8 units to arrange.

    3.  Arrange Units: Arrangements of these 8 units (with 4 S's and 2 N's) $= \frac{8!}{4!2!} = \frac{40320}{48} = 840$.

    4.  Internal Vowel Arrangement: Arrange the 6 vowels (3 A's and 2 I's) inside the block: $\frac{6!}{3!2!} = \frac{720}{12} = 60$.

    5.  Multiply Results: Total ways $= 840 \times 60 = 50,400$.

   Difficulty Level: Medium.

   Concept Name: Permutations with Identical Objects (String Method).

   Short cut solution: Use the formula $\frac{(\text{Total Units})!}{\prod(\text{Identical Consonants})!} \times \frac{(\text{Vowels})!}{\prod(\text{Identical Vowels})!}$. Calculation: $\frac{8!}{4!2!} \times \frac{6!}{3!2!} = 840 \times 60 = 50400$.

 Question 52

   Question: The number of 4-letter words, with or without meaning, each consisting of 2 vowels and 2 consonants, which can be formed from the letters of the word UNIVERSE without repetition is:

   Options: (Not explicitly listed in source; answer is an integer)

   Correct Answer: 432

   Year: JEE Main 2023 (Online) 6th April Evening Shift

   Solution (as Given in the Source): Case I 2 vowels different, 2 consonant different $(^3C_2) (^4C_2) (4!) = (3) (6) (24) = 432$

   Step Solution:

    1.  Identify Distinct Letters: In the word UNIVERSE, the distinct vowels are {U, I, E} (3 total) and the distinct consonants are {N, V, R, S} (4 total).

    2.  Select Vowels: Choose 2 vowels from the 3 available: $^3C_2 = 3$ ways.

    3.  Select Consonants: Choose 2 consonants from the 4 available: $^4C_2 = 6$ ways.

    4.  Arrange Letters: The 4 selected letters can be arranged in $4! = 24$ ways.

    5.  Final Calculation: Total words $= 3 \times 6 \times 24 = 432$.

   Difficulty Level: Easy

   Concept Name: Permutations (Selection and Arrangement)

   Short cut solution: Total ways $= \binom{3}{2} \times \binom{4}{2} \times 4! = 3 \times 6 \times 24 = 432$.

 Question 53

   Question: The number of arrangements of the letters of the word "INDEPENDENCE" in which all the vowels always occur together is.

   Options: A. 16800, B. 14800, C. 18000, D. 33600

   Correct Answer: A

   Year: JEE Main 2023 (Online) 8th April Morning Shift

   Solution (as Given in the Source): IEEEE, NNN, DD, P, C. $\frac{8!}{3!2!} \times \frac{6!}{4!1!} = 16800$

   Step Solution:

    1.  Group Vowels: Treat all 5 vowels (I, E, E, E, E) as a single block. The consonants are N, N, N, D, D, P, C.

    2.  Count Units: You have 7 consonants + 1 vowel block = 8 units to arrange.

    3.  Arrange Units: Arrangements of these 8 units (with 3 N's and 2 D's) $= \frac{8!}{3!2!} = \frac{40320}{12} = 3360$.

    4.  Internal Vowel Arrangement: Arrange the 5 vowels (one I and four E's) inside the block: $\frac{5!}{4!} = 5$.

    5.  Total Arrangements: $3360 \times 5 = 16,800$.

   Difficulty Level: Medium

   Concept Name: Permutations of Multisets (String Method)

   Short cut solution: $(\text{Ways to arrange Consonants and Vowel Block}) \times (\text{Internal Vowel arrangements}) = \frac{8!}{3!2!} \times \frac{5!}{4!} = 3360 \times 5 = 16800$.

 Question 55

   Question: If the number of words, with or without meaning, which can be made using all the letters of the word MATHEMATICS in which C and S do not come together, is $(6!)k$, then k is equal to:

   Options: A. 1890, B. 945, C. 2835, D. 5670

   Correct Answer: D

   Year: JEE Main 2023 (Online) 8th April Evening Shift

   Solution (as Given in the Source): $M_2 A_2 T_2$ HEICS. $=$ total words - when C&S are together. $\frac{11!}{2!2!2!} - \frac{10!}{2!2!2!} \times 2! = \frac{9 \times 10 \times 9 \times 8 \times 7}{8} 6! = 5670 \{ 6! \}$. $k = 5670$.

   Step Solution:

    1.  Identify Letter Frequency: M(2), A(2), T(2), H(1), E(1), I(1), C(1), S(1). Total letters $= 11$.

    2.  Total Arrangements: $\frac{11!}{2!2!2!} = \frac{11!}{8} = 4,989,600$.

    3.  Arrangements with C and S Together: Treat (CS) as one block. Total items to arrange $= 10$. Ways $= \frac{10!}{2!2!2!} \times 2! = \frac{10!}{4} = 907,200$.

    4.  Subtraction: $4,989,600 - 907,200 = 4,082,400$.

    5.  Find k: Set $4,082,400 = 6! \times k \Rightarrow 4,082,400 = 720k \Rightarrow k = 5670$.

   Difficulty Level: Hard

   Concept Name: Permutations with Identical Objects (Complementary Counting)

   Short cut solution: $k = \frac{1}{6!} \times (\text{Total} - \text{Together}) = \frac{1}{6!} \times \frac{10!}{8} (11 - 2) = \frac{9 \times 10 \times 9 \times 8 \times 7 \times 6!}{8 \times 6!} = 5670$.

 Question 56

   Question: The number of permutations of the digits 1, 2, 3, ...., 7 without repetition, which neither contain the string 153 nor the string 2467, is.

   Options: (Not explicitly listed in source; Answer is an integer).

   Correct Answer: 4898,.

   Year: JEE Main 2023 (Online) 10th April Shift 1.

   Solution (as Given in the Source): Numbers are 1, 2, 3, 4, 5, 6, 7. Numbers having string (153) = 5!. Numbers having string (2467) = 4!. Number having string (153) and (2467) = 2!. Now $n(153 \cup 2467) = 5! + 4! - 2! = 120 + 24 - 2 = 142$. Total numbers = $7! = 5040$. Required numbers = $5040 - 142 = 4898$,.

   Step Solution:

    1.  Total Permutations: Arrange the 7 distinct digits in $7! = 5,040$ ways.

    2.  Case A (Contains "153"): Treat (153) as a single unit. Remaining units: {2, 4, 6, 7}. Total units = 5. Ways = $5! = 120$.

    3.  Case B (Contains "2467"): Treat (2467) as a single unit. Remaining units: {1, 3, 5}. Total units = 4. Ways = $4! = 24$.

    4.  Case A ∩ B (Contains both): Treat (153) and (2467) as two distinct units. Total units = 2. Ways = $2! = 2$.

    5.  Principle of Inclusion-Exclusion: Neither string = Total - (A + B - A ∩ B) = $5040 - (120 + 24 - 2) = 4898$,.

   Difficulty Level: Hard.

   Concept Name: Principle of Inclusion and Exclusion (PIE) / String Method.

   Short cut solution: Subtract the combined counts of restricted blocks from the total permutations: $7! - (5! + 4! - 2!) = 5040 - 142 = 4898$.

 Question 117

   Question: The number of six letter words (with or without meaning), formed using all the letters of the word 'VOWELS', so that all the consonants never come together, is.

   Options: (Not explicitly listed in source; Answer is an integer).

   Correct Answer: 576.

   Year: JEE Main 2021, 31 Aug. Shift-1.

   Solution (as Given in the Source): VOWELS (2 Vowel + 4 consonant). Total possibility = 6!. The number of arrangement when all the consonant comes together = 3! × 4!. Number of arrangement when all the consonants never come together = Total - All consonant together = 6! - 3!4! = 576.

   Step Solution:

    1.  Identify Letters: VOWELS has 6 letters: Vowels (O, E) and Consonants (V, W, L, S).

    2.  Total Arrangements: Total ways to arrange 6 distinct letters = $6! = 720$.

    3.  Consonants Together: Treat the 4 consonants (VWLS) as one block. This block plus the 2 vowels gives 3 units to arrange ($3!$).

    4.  Internal Arrangement: The 4 consonants can be arranged within their block in $4!$ ways. Total together = $3! \times 4! = 6 \times 24 = 144$.

    5.  Final Calculation: Never together = Total - Together = $720 - 144 = 576$.

   Difficulty Level: Medium.

   Concept Name: Permutations (Complementary Counting).

   Short cut solution: Use the formula: $n! - (\text{vowels} + 1)! \times (\text{consonants})! = 6! - 3! \times 4! = 720 - 144 = 576$.

 Question 121

   Question: The number of 4 letter words (with or without meaning) that can be formed from the eleven letters of the word 'EXAMINATION' is.

   Options: (Not explicitly listed in source; Answer is an integer).

   Correct Answer: 2454.

   Year: JEE Main 2020, 8 Jan. Shift-II.

   Solution (as Given in the Source): EXAMINATION: 2N, 2A, 2I, E, X, M, T, O. Case I: If all are different, then $^8P_4 = 1680$. Case II: If two are same and two are different, then $^3C_1 \times ^7C_2 \times \frac{4!}{2!} = 756$. Case III: If two are same and other two are same, then $^3C_2 \times \frac{4!}{2!2!} = 18$. Total cases = $1680 + 756 + 18 = 2454$,.

   Step Solution:

    1.  Categorize Letters: Distinct types are {A, I, N, E, X, M, T, O}. Counts: A(2), I(2), N(2), others 1 each. Total types = 8.

    2.  Case 1 (All Different): Select 4 types from 8 and arrange: $^8C_4 \times 4! = 70 \times 24 = 1680$.

    3.  Case 2 (1 Pair, 2 Different): Choose 1 pair from {A, I, N} ($^3C_1$) and 2 single letters from remaining 7 types ($^7C_2$). Arrange: $3 \times 21 \times \frac{4!}{2!} = 756$.

    4.  Case 3 (2 Pairs): Choose 2 pairs from {A, I, N} ($^3C_2$) and arrange: $3 \times \frac{4!}{2!2!} = 3 \times 6 = 18$.

    5.  Total: Sum all cases: $1680 + 756 + 18 = 2454$.

   Difficulty Level: Hard.

   Concept Name: Permutations of Multisets (Case-based Counting).

   Short cut solution: Systematically group by frequency (4-0-0, 2-1-1, 2-2) and sum: $P(8,4) + \binom{3}{1}\binom{7}{2}\frac{4!}{2!} + \binom{3}{2}\frac{4!}{2!2!} = 1680 + 756 + 18 = 2454$.

 Question 124

   Question: Total number of 6-digit numbers in which only and all the five digits 1, 3, 5, 7 and 9 appear, is:

   Options: 

       A. $\frac{1}{2}(6!)$ 

       B. $6!$ 

       C. $5^6$ 

       D. $\frac{5}{2}(6!)$

   Correct Answer: D

   Year: Jan. 7, 2020 (I)

   Solution (as Given in the Source): Five digits numbers be 1,3,5,7,9. For selection of one digit, we have $^5C_1$ choice. And six digits can be arranged in $\frac{6!}{2!}$ ways. Hence, total such numbers $= \frac{5 \cdot 6!}{2!} = \frac{5}{2} \cdot 6!$

   Step Solution:

    1.  Understand Constraints: A 6-digit number must be formed using all five digits {1, 3, 5, 7, 9}.

    2.  Identify Repetition: Since there are 6 positions and only 5 available digits, exactly one digit must be repeated twice to fill the 6-digit number.

    3.  Select the Repeated Digit: Choose which of the 5 digits will appear twice: $^5C_1 = 5$ ways.

    4.  Arrange the Digits: Arrange these 6 digits (where 2 are identical). The formula for permutations with repetition is $\frac{6!}{2!} = 360$ ways.

    5.  Calculate Total: Multiply the selection by the arrangements: $5 \times 360 = 1800$, which is equivalent to $\frac{5}{2} \times 720 = \frac{5}{2}(6!)$.

   Difficulty Level: Medium.

   Concept Name: Permutations with Identical Objects.

   Short cut solution: Total ways = (Number of ways to pick the extra digit) $\times$ (Arrangement of 6 items with 2 alike) $= 5 \times \frac{6!}{2!} = \frac{5}{2} \times 6!$.

 Question 128

   Question: The number of words (with or without meaning) that can be formed from all the letters of the word "LETTER" in which vowels never come together is

   Options: (Not explicitly listed in source; numerical value type).

   Correct Answer: 120

   Year: Sep. 06, 2020 (II)

   Solution (as Given in the Source): For vowels not together. Number of ways to arrange L, T, T, R $= \frac{4!}{2!}$. Then put both E in 5 gaps formed in $^5C_2$ ways. ∴ No. of ways $= \frac{4!}{2!} \cdot ^5C_2 = 120$.

   Step Solution:

    1.  Analyze Letters: The word "LETTER" has consonants {L, T, T, R} and vowels {E, E}.

    2.  Arrange Consonants: Arrange the 4 consonants. Since 'T' is repeated twice, the ways $= \frac{4!}{2!} = 12$.

    3.  Identify Gaps: Placing 4 consonants creates 5 potential gaps ( _ C _ C _ C _ C _ ) where vowels can be placed so they are never together.

    4.  Place Vowels: Choose 2 gaps out of 5 for the two identical 'E's: $^5C_2 = 10$ ways.

    5.  Calculate Total: Total arrangements $= 12 \times 10 = 120$.

   Difficulty Level: Medium.

   Concept Name: Gap Method (Permutations).

   Short cut solution: $(\text{Arrangement of Consonants}) \times (\text{Selection of Gaps for Vowels}) = \frac{4!}{2!} \times \binom{5}{2} = 12 \times 10 = 120$.

 Question 129

   Question: The number of words, with or without meaning, that can be formed by taking 4 letters at a time from the letters of the word 'SYLLABUS' such that two letters are distinct and two letters are alike, is

   Options: (Not explicitly listed in source; numerical value type).

   Correct Answer: 240

   Year: Sep. 05, 2020 (I)

   Solution (as Given in the Source): S $\to$ 2, L $\to$ 2, A, B, Y, U. Number of ways $= ^2C_1 \times ^5C_2 \times \frac{4!}{2!} = 240$.

   Step Solution:

    1.  Identify Letter Frequency: In 'SYLLABUS', the letters are S(2), L(2), A(1), B(1), Y(1), U(1). There are two possible "alike" pairs (S or L).

    2.  Select the Alike Pair: Choose 1 pair from the 2 available types (S or L): $^2C_1 = 2$ ways.

    3.  Select Distinct Letters: After picking one pair, we need 2 more distinct letters from the remaining 5 types of letters: $^5C_2 = 10$ ways.

    4.  Arrange the Selection: Arrange these 4 letters (where 2 are alike): $\frac{4!}{2!} = 12$ ways.

    5.  Final Calculation: Total words $= 2 \times 10 \times 12 = 240$.

   Difficulty Level: Medium.

   Concept Name: Permutations of Multisets (Selection and Arrangement).

   Short cut solution: Pick 1 pair, pick 2 singles, and arrange: $\binom{2}{1} \times \binom{5}{2} \times \frac{4!}{2!} = 2 \times 10 \times 12 = 240$.

Question 142

   Question: All possible numbers are formed using the digits 1,1, 2, 2, 2, 2, 3, 4, 4 taken all at a time. The number of such numbers in which the odd digits occupy even places is:

   Options: A. 180, B. 175, C. 160, D. 162

   Correct Answer: A

   Year: April 8, 2019 (I)

   Solution (as Given in the Source): ∵ There are total 9 digits and out of which only 3 digits are odd. ∴ Number of ways to arrange odd digits first $= ^4C_3 \cdot \frac{3!}{2!}$ $= (^4C_3 \cdot \frac{3!}{2!}) \cdot \frac{6!}{2!4!} = 180$.

   Step Solution:

    1.  Analyze Digits: The multiset of digits is {1, 1, 2, 2, 2, 2, 3, 4, 4}. There are 3 odd digits {1, 1, 3} and 6 even digits {2, 2, 2, 2, 4, 4}.

    2.  Select Even Places: There are 4 even positions (2nd, 4th, 6th, and 8th) in a 9-digit number. Choose 3 positions for the odd digits: $^4C_3 = 4$ ways.

    3.  Arrange Odd Digits: Arrange the odd digits {1, 1, 3} in the 3 selected positions: $\frac{3!}{2!} = 3$ ways.

    4.  Arrange Even Digits: Arrange the 6 even digits in the remaining 6 positions: $\frac{6!}{4!2!} = 15$ ways.

    5.  Calculate Total: $4 \times 3 \times 15 = 180$.

   Difficulty Level: Medium.

   Concept Name: Permutations with Identical Objects.

   Short cut solution: $(\text{Ways to pick even places}) \times (\text{Arrange odd}) \times (\text{Arrange even}) = \binom{4}{3} \times \frac{3!}{2!} \times \frac{6!}{4!2!} = 180$.

 Question 148

   Question: The number of four letter words that can be formed using the letters of the word BARRACK is

   Options: A. 144, B. 120, C. 264, D. 270

   Correct Answer: D

   Year: Online April 15, 2018

   Solution (as Given in the Source): Case 1: All four different $^5C_4 \times 4! = 120$. Case 2: Two R and other two different $^4C_2 \times \frac{4!}{2!} = 72$. Case 3: Two A and other two different $^4C_2 \times \frac{4!}{2!} = 72$. Case 4: Two R's and two A's $\frac{4!}{2!2!} = 6$. Total $= 120 + 72 + 72 + 6 = 270$.

   Step Solution:

    1.  Analyze Letters: The letters are {B, A, R, R, A, C, K}. Distinct types: {B, A, R, C, K} with frequencies A(2), R(2), B(1), C(1), K(1).

    2.  Case 1 (All distinct): Select 4 types from 5 and arrange: $^5C_4 \times 4! = 5 \times 24 = 120$.

    3.  Case 2 (1 Pair, 2 distinct): Choose 1 pair from {A, R} (2 ways) and choose 2 letters from the remaining 4 types ($^4C_2$ ways). Arrange: $2 \times 6 \times \frac{4!}{2!} = 144$.

    4.  Case 3 (2 Pairs): Use both pairs {A, A, R, R}. Arrange: $\frac{4!}{2!2!} = 6$.

    5.  Final Sum: Total words $= 120 + 144 + 6 = 270$.

   Difficulty Level: Hard.

   Concept Name: Permutations of Multisets (Selection and Arrangement).

   Short cut solution: Sum of cases: $\binom{5}{4}4! + \binom{2}{1}\binom{4}{2}\frac{4!}{2!} + \binom{2}{2}\frac{4!}{2!2!} = 120 + 144 + 6 = 270$.

 Question 150

   Question: From 6 different novels and 3 different dictionaries, 4 novels and 1 dictionary are to be selected and arranged in a row on a shelf so that the dictionary is always in the middle. The number of such arrangements is:

   Options: A. less than 500, B. at least 500 but less than 750, C. at least 750 but less than 1000, D. at least 1000

   Correct Answer: D

   Year: 2018

   Solution (as Given in the Source): Required number of ways $= ^6C_4 \times ^3C_1 \times 4! = 15 \times 3 \times 24 = 1080$.

   Step Solution:

    1.  Selection of Novels: Choose 4 novels from 6: $^6C_4 = 15$ ways.

    2.  Selection of Dictionary: Choose 1 dictionary from 3: $^3C_1 = 3$ ways.

    3.  Position Constraint: The row has 5 items. The dictionary is fixed in the middle (3rd position).

    4.  Arrangement of Novels: The 4 selected novels can be arranged in the remaining 4 spots in $4! = 24$ ways.

    5.  Final Calculation: Total arrangements $= 15 \times 3 \times 24 = 1080$. Since $1080 \ge 1000$, Option D is correct.

   Difficulty Level: Easy.

   Concept Name: Selection and Permutation.

   Short cut solution: Total $= \binom{6}{4} \times \binom{3}{1} \times 4! = 15 \times 3 \times 24 = 1080$.

 Question 155

   Question: If the four letter words (need not be meaningful) are to be formed using the letters from the word "MEDITERRANEAN" such that the first letter is R and the fourth letter is E, then the total number of all such words is :

   Options: A. 110, B. 59, C. $\frac{11!}{(2!)^3}$, D. 56

   Correct Answer: B

   Year: Online April 9, 2016

   Solution (as Given in the Source): M, EEE, D, I, T, RR, AA, NN. R— — E. Two empty places can be filled with identical letters [EE, AA, NN] $\Rightarrow 3$ ways. Two empty places can be filled with distinct letters [M, E, D, I, T, R, A, N] $\Rightarrow {}^8P_2$. Number of words $3 + {}^8P_2 = 59$.

   Step Solution:

    1.  Analyze the structure: The word must be R _ _ E. Two positions are fixed, leaving two internal gaps to fill.

    2.  Identify available letters: The word MEDITERRANEAN contains M(1), E(3), D(1), I(1), T(1), R(2), A(2), N(2). After using one R and one E, we have: M(1), E(2), D(1), I(1), T(1), R(1), A(2), N(2).

    3.  Case 1 (Identical Letters): The two gaps can be filled with pairs of identical letters. Available pairs are {EE, AA, NN}. There are 3 such ways.

    4.  Case 2 (Distinct Letters): There are 8 distinct types of letters remaining {M, E, D, I, T, R, A, N}. We choose 2 distinct letters and arrange them: ${}^8P_2 = 8 \times 7 = \mathbf{56}$ ways.

    5.  Final Calculation: Total words = $3 (\text{identical}) + 56 (\text{distinct}) = \mathbf{59}$.

   Difficulty Level: Hard.

   Concept Name: Permutations with Identical Objects (Case-based Counting).

   Short cut solution: Sum of cases: $(\text{Number of available pairs}) + {}^nP_2 (\text{where } n \text{ is the number of distinct letter types}) = 3 + 56 = 59$.

 Question 166

   Question: 8-digit numbers are formed using the digits 1, 1, 2, 2, 2, 3, 4, 4. The number of such numbers in which the odd digits do no occupy odd places, is:

   Options: A. 160, B. 120, C. 60, D. 48

   Correct Answer: B

   Year: Online April 12, 2014

   Solution (as Given in the Source): In 8 digits numbers, 4 places are odd places. No. of ways three odd digits arranged at four even places $= \frac{{}^4P_3}{2!} = \frac{4!}{2!}$. No. of ways the remaining five digits 2, 2, 2, 4 and 4 arranged at remaining five places $= \frac{5!}{3!2!}$. Hence, required number of 8 digits number $= \frac{4!}{2!} \times \frac{5!}{3!2!} = 120$.

   Step Solution:

    1.  Identify Digits: Odd digits = {1, 1, 3} (3 total); Even digits = {2, 2, 2, 4, 4} (5 total).

    2.  Locate Places: In an 8-digit number, there are 4 odd places (1, 3, 5, 7) and 4 even places (2, 4, 6, 8).

    3.  Arrange Odds: The 3 odd digits must occupy the 4 even places. Choose 3 places and arrange them: $\frac{4!}{(4-3)! \times 2!} = \mathbf{12}$ ways.

    4.  Arrange Evens: The remaining 5 digits (three 2s and two 4s) must be arranged in the remaining 5 vacant spots: $\frac{5!}{3!2!} = \mathbf{10}$ ways.

    5.  Final Calculation: Total numbers = $12 \times 10 = \mathbf{120}$.

   Difficulty Level: Medium.

   Concept Name: Permutations of Multisets (Restricted Placement).

   Short cut solution: $(\text{Ways to place Odd digits in Even places}) \times (\text{Ways to arrange remaining Even digits}) = 12 \times 10 = 120$.

 Question 176

   Question: The number of arrangements that can be formed from the letters a, b, c, d, e, f taken 3 at a time without repetition and each arrangement containing at least one vowel, is

   Options: A. 96, B. 128, C. 24, D. 72

   Correct Answer: A

   Year: Online May 19, 2012

   Solution (as Given in the Source): There are 2 vowels and 4 consonants in the letters a, b, c, d, e, f. If we select one vowel, then number of arrangements $= {}^2C_1 \times {}^4C_2 \times 3! = 72$. If we select two vowels, then number of arrangements $= {}^2C_2 \times {}^4C_1 \times 3! = 24$. Hence, total number of arrangements $= 72 + 24 = 96$.

   Step Solution:

    1.  Categorize Letters: Vowels = {a, e} (2); Consonants = {b, c, d, f} (4).

    2.  Define "At Least One": Since we take 3 letters and there are only 2 vowels, we can have exactly 1 vowel or exactly 2 vowels.

    3.  Case 1 (1 Vowel, 2 Consonants): Select and arrange: $({}^2C_1 \times {}^4C_2) \times 3! = (2 \times 6) \times 6 = \mathbf{72}$.

    4.  Case 2 (2 Vowels, 1 Consonant): Select and arrange: $({}^2C_2 \times {}^4C_1) \times 3! = (1 \times 4) \times 6 = \mathbf{24}$.

    5.  Total: $72 + 24 = \mathbf{96}$.

   Difficulty Level: Easy.

   Concept Name: Permutations (Complementary Counting or Case-based Selection).

   Short cut solution: Total arrangements without restriction minus arrangements with no vowels: ${}^6P_3 - {}^4P_3 = (6 \times 5 \times 4) - (4 \times 3 \times 2) = 120 - 24 = \mathbf{96}$.

 Question 183

   Question: From 6 different novels and 3 different dictionaries, 4 novels and 1 dictionary are to be selected and arranged in a row on a shelf so that the dictionary is always in the middle. Then the number of such arrangement is:

   Options: 

       A. at least 500 but less than 750

       B. at least 750 but less than 1000

       C. at least 1000

       D. less than 500

   Correct Answer: C

   Year: 2009

   Solution (as Given in the Source): 4 novels, out of 6 novels and 1 dictionary out of 3 can be selected in $^6C_4 \times ^3C_1$ ways. Then 4 novels with one dictionary in the middle can be arranged in 4! ways. Total ways of arrangement $= ^6C_4 \times ^3C_1 \times 4! = 1080$.

   Step Solution:

    1.  Selection of Novels: Choose 4 novels from 6: $^6C_4 = \frac{6 \times 5}{2 \times 1} = 15$ ways.

    2.  Selection of Dictionary: Choose 1 dictionary from 3: $^3C_1 = 3$ ways.

    3.  Position Constraint: The shelf has 5 total spots. Fix the chosen dictionary in the middle (3rd spot), which can be done in only 1 way.

    4.  Arrangement of Remaining Items: Arrange the 4 selected novels in the remaining 4 spots on the shelf: $4! = 24$ ways.

    5.  Final Calculation: Total arrangements $= 15 \times 3 \times 1 \times 24 = 1080$. Since $1080 \ge 1000$, Option C is correct.

   Difficulty Level: Easy.

   Concept Name: Permutations and Combinations (Selection and Arrangement).

   Short cut solution: Total $= \binom{6}{4} \times \binom{3}{1} \times 4! = 15 \times 3 \times 24 = 1080$.

 Question 184

   Question: How many different words can be formed by jumbling the letters in the word MISSISSIPPI in which no two S are adjacent?

   Options: 

       A. $8 \cdot ^6C_4 \cdot ^7C_4$

       B. $6 \cdot 7 \cdot ^8C_4$

       C. $6 \cdot 8 \cdot ^7C_4$

       D. $7 \cdot ^6C_4 \cdot ^8C_4$

   Correct Answer: D

   Year: 2008

   Solution (as Given in the Source): First let us arrange M, I, I, I, I, P, P which can be done in $\frac{7!}{4!2!}$ ways. $^M^I^I^I^I^P^P^$ Now 4S can be kept at any of the  places in $^8C_4$ ways so that no two S are adjacent. Total required ways $= \frac{7!}{4!2!} \cdot ^8C_4 = 7 \times ^6C_4 \times ^8C_4$.

   Step Solution:

    1.  Analyze Letter Frequency: M(1), I(4), S(4), P(2). Total letters = 11.

    2.  Arrange Non-Restricted Letters: Arrange {M, I, I, I, I, P, P}. Number of ways $= \frac{7!}{4!2!} = 105$.

    3.  Express as Combinations: Note that $105$ can be written as $7 \times 15$, which is $7 \times \binom{6}{4}$ or $7 \cdot ^6C_4$.

    4.  Apply Gap Method: Placing 7 letters creates 8 gaps ($^L^L^L^L^L^L^L^$). To ensure no two 'S's are together, place the 4 identical 'S' letters into 4 of these 8 gaps: $^8C_4$ ways.

    5.  Final Calculation: Multiply the arrangement of consonants/vowels by the gap selection: $(7 \cdot ^6C_4) \cdot ^8C_4$.

   Difficulty Level: Medium.

   Concept Name: Gap Method (Permutations with Identical Objects).

   Short cut solution: $(\text{Arrangements of non-S letters}) \times (\text{Ways to place S in gaps}) = \frac{7!}{4!2!} \times \binom{8}{4} = 7 \cdot ^6C_4 \cdot ^8C_4$.

 Question 189

   Question: How many ways are there to arrange the letters in the word GARDEN with vowels in alphabetical order

   Options: 

       A. 480

       B. 240

       C. 360

       D. 120

   Correct Answer: C

   Year: 2004

   Solution (as Given in the Source): Total number of arrangements of letters in the word GARDEN $= 6! = 720$ there are two vowels A and E, in half of the arrangements A preceeds E and other half A follows E. So, numbers of word with vowels in alphabetical order in $\frac{1}{2} \times 720 = 360$.

   Step Solution:

    1.  Total Arrangements: The word GARDEN has 6 distinct letters. Total arrangements $= 6! = 720$.

    2.  Identify Vowels: The vowels are A and E.

    3.  Analyze Relative Order: In any permutation, there are only two possible relative orders for A and E: (A before E) or (E before A).

    4.  Symmetry Argument: Because all letters are distinct and permutations are uniform, exactly half of the total arrangements will have A before E, and half will have E before A.

    5.  Final Calculation: Number of ways $= \frac{720}{2} = 360$.

   Difficulty Level: Easy.

   Concept Name: Permutations with Restricted Relative Order.

   Short cut solution: Treat the $r$ restricted items as identical for the purpose of arrangement and then fix their order. Total ways $= \frac{n!}{r!} = \frac{6!}{2!} = \frac{720}{2} = 360$.