Q1. A registration code consists of 2 distinct capital letters (from 26) followed by a two-digit number made of two distinct digits. The number must have a non-zero first digit and be even. How many codes are possible?Correct Option C … Explanation: The two distinct letters can be chosen and arranged in 26 × 25 = 650 ways. For the two-digit even number with distinct digits and a non-zero first digit, split by the last digit. If the last digit is 0, the first digit can be any of 1–9, which gives 9 numbers. If the last digit is 2, 4, 6 or 8, the first digit must be non-zero and different from the last digit, which gives 8 choices for each, so 4 × 8 = 32 numbers. That makes 9 + 32 = 41 valid numbers. Total codes = 650 × 41 = 26650. A common error is to treat all five even digits alike (9 × 5 = 45), which ignores the zero case. Hence, option C.Q2. Town P and town S are joined only through intermediate towns. Via Q there are 3 roads P–Q and 4 roads Q–S. Via R there are 2 roads P–R and 5 roads R–S. A traveller goes from P to S and returns from S to P, without using any road twice (in either direction). In how many ways can the round trip be planned?Correct Option B … Explanation: There are 3 × 4 = 12 one-way routes via Q and 2 × 5 = 10 via R. Case 1: the traveller goes via Q (12 ways). The return via Q can use the remaining 3 roads on Q–S and 2 roads on P–Q, giving 3 × 2 = 6 ways, and the return via R gives 10 ways, so 16 return options. That case gives 12 × 16 = 192. Case 2: the traveller goes via R (10 ways). The return via R can use the remaining 4 roads on R–S and 1 road on P–R, giving 4 × 1 = 4 ways, and the return via Q gives 12 ways, so 16 return options. That case gives 10 × 16 = 160. Total = 192 + 160 = 352. Ignoring the no-repeat condition gives 22 × 22 = 484, which is option E. Hence, option B.Q3. How many 5-digit numbers divisible by 4 can be formed using the digits 0, 1, 2, 3, 4, 5, 6, 7 without repetition?Correct Option A … Explanation: A number is divisible by 4 when its last two digits form a multiple of 4. The valid endings with distinct digits from 0–7 are 04, 12, 16, 20, 24, 32, 36, 40, 52, 56, 60, 64, 72 and 76, which is 14 endings. Four of these endings contain 0 (04, 20, 40, 60). For each of them, the first three places are filled from the remaining 6 non-zero digits in 6 × 5 × 4 = 120 ways, giving 4 × 120 = 480. The other 10 endings do not contain 0, so 0 is still available but cannot lead. The first digit then has 5 choices, followed by 5 and 4, giving 100 ways each, so 10 × 100 = 1000. Total = 480 + 1000 = 1480. Hence, option A.Q4. How many 4-digit even numbers divisible by 3 can be formed using the digits 0 to 6 without repetition?Correct Option D … Explanation: The digits 0–6 sum to 21, which is divisible by 3. So a 4-digit set has a digit sum divisible by 3 exactly when the 3 digits left out also have a sum divisible by 3. Sorting the digits by remainder gives {0,3,6}, {1,4} and {2,5}. The left-out triples are either {0,3,6} or one digit from each remainder group (3 × 2 × 2 = 12), so there are 13 valid 4-digit sets. Now count the even arrangements with no leading zero. For a set without 0, the count is 6 × (number of even digits). For a set with 0, the count is 6 (0 in the units place) plus 4 × (number of non-zero even digits). The set {1,2,4,5} gives 12. The sets containing {3,6} give 12 + 6 + 18 + 12 = 48. The sets containing {0,6} give 14 + 10 + 18 + 14 = 56. The sets containing {0,3} give 10 + 6 + 14 + 10 = 40. Total = 12 + 48 + 56 + 40 = 156. Hence, option D.Q5. Using the digits 0, 2, 3, 5, 6, 8 without repetition, how many 4-digit or 5-digit numbers can be formed that lie strictly between 5000 and 60000?Correct Option B … Explanation: First take the 4-digit numbers greater than 5000. The first digit can be 5, 6 or 8; 5000 itself cannot be formed because its digits repeat. The remaining three places are filled in 5 × 4 × 3 = 60 ways, giving 3 × 60 = 180 numbers. Next take the 5-digit numbers less than 60000. The first digit must be non-zero and below 6, so it can be 2, 3 or 5. The remaining four places are filled in 5 × 4 × 3 × 2 = 120 ways, giving 3 × 120 = 360 numbers. Total = 180 + 360 = 540. Option D (360) counts only the 5-digit case. Hence, option B.Q6. How many 4-digit numbers have the product of their digits equal to 36?Correct Option A … Explanation: Since 36 = 2² × 3², every digit must come from 1–9 (no 0). The possible digit groups and their arrangements are as follows: {9,4,1,1} gives 4!/2! = 12, {6,6,1,1} gives 4!/(2!2!) = 6, {9,2,2,1} gives 12, {6,3,2,1} gives 4! = 24, {4,3,3,1} gives 12, and {3,3,2,2} gives 6. Total = 12 + 6 + 12 + 24 + 12 + 6 = 72. Missing one group, for example {3,3,2,2}, leads to wrong totals such as 66. Hence, option A.Q7. If all arrangements of the letters of CHANCE are listed in dictionary order, what is the rank of CHANCE?Correct Option E … Explanation: In alphabetical order the letters are A, C, C, E, H, N. Words starting with A arrange C, C, E, H, N in 5!/2! = 60 ways. For words starting with C, the remaining letters are A, C, E, H, N. Words starting with CA, CC or CE come before CH, giving 3 × 4! = 72. After CH, the remaining letters are A, C, E, N, and the next letter A is the smallest, so nothing is skipped. After CHA, the remaining letters are C, E, N. Words starting with CHAC or CHAE come before CHAN, giving 2 × 2! = 4. After CHAN, the remaining letters are C, E, and CHANCE is the first of these. Rank = 60 + 72 + 4 + 1 = 137. Hence, option E.Q8. In how many arrangements of the letters of TRIANGLE do the vowels appear in alphabetical order (A before E before I), though not necessarily together?Correct Option D … Explanation: TRIANGLE has 8 distinct letters, so there are 8! = 40320 arrangements in all. The three vowels A, I and E can appear in 3! = 6 relative orders, and each order is equally likely. Only one of those orders (A, E, I) is allowed. Required count = 8!/3! = 40320/6 = 6720. Option C (40320) ignores the order condition entirely. Hence, option D.Q9. 5 boys and 4 girls are seated in a row so that no two girls sit together and two particular boys, Arun and Bala, are not adjacent. How many seatings are possible?Correct Option C … Explanation: Start with all seatings where no two girls sit together. Arrange the 5 boys in 5! = 120 ways, which creates 6 gaps. Choose 4 gaps for the girls in C(6,4) = 15 ways and arrange them in 4! = 24 ways, giving 120 × 15 × 24 = 43200. Next count the seatings where Arun and Bala sit next to each other. Treat them as one block that can be arranged internally in 2 ways. The block plus the 3 other boys can be arranged in 4! × 2 = 48 ways. This leaves 5 gaps, and no girl may sit inside the block. Choosing and arranging the girls gives C(5,4) × 4! = 120 ways, so 48 × 120 = 5760 seatings. Required = 43200 − 5760 = 37440. Hence, option C.Q10. There are 12 chairs in a row. In how many ways can 4 distinct people be seated so that no two occupied chairs are adjacent and both end chairs are empty?Correct Option E … Explanation: Both end chairs are empty, so the 4 people sit somewhere in chairs 2 to 11, which is 10 chairs. The number of ways to choose 4 non-adjacent chairs from 10 in a row is C(10 − 4 + 1, 4) = C(7,4) = 35. The 4 distinct people can then be arranged in the chosen chairs in 4! = 24 ways. Total = 35 × 24 = 840. Hence, option E.Q11. A committee of 6 is to be formed from 8 men and 6 women with at least 2 women. Mr. X (a man) and Ms. Y (a woman) cannot both be on the committee. How many committees are possible?Correct Option C … Explanation: There are C(14,6) = 3003 committees in total. Committees with no women number C(8,6) = 28, and committees with exactly one woman number 6 × C(8,5) = 336. So committees with at least 2 women = 3003 − 364 = 2639. Now remove the committees that include both X and Y. The other 4 members come from 7 men and 5 women. Because Y is already one woman, at least 1 more woman is needed. That gives C(12,4) − C(7,4) = 495 − 35 = 460 committees. Required = 2639 − 460 = 2179. Option E (2639) forgets the X–Y restriction. Hence, option C.Q12. A fruit basket holds 5 identical mangoes, 4 identical apples and 3 distinct guavas. In how many ways can at least one fruit be selected?Correct Option A … Explanation: Because the mangoes are identical, only the number taken matters, so there are 6 choices (0 to 5). In the same way, the apples give 5 choices (0 to 4). Each distinct guava can be taken or left, giving 2³ = 8 choices. Total selections = 6 × 5 × 8 = 240. Removing the empty selection leaves 240 − 1 = 239. Option C (240) forgets to remove the case where nothing is selected. Hence, option A.Q13. How many triangles can be formed using the vertices of a regular 12-sided polygon such that no side of the triangle is also a side of the polygon?Correct Option D … Explanation: There are C(12,3) = 220 triangles in total. Triangles with exactly two sides on the polygon are formed by three consecutive vertices, and there are 12 of these. Triangles with exactly one side on the polygon are formed by choosing one of the 12 sides and a third vertex that is not next to either of its endpoints, giving 12 × 8 = 96. Required = 220 − 12 − 96 = 112. This matches the standard formula n(n − 4)(n − 5)/6 = 12 × 8 × 7/6 = 112. Hence, option D.Q14. 12 distinct employees are to be divided into 3 unlabelled teams of 4 each. In how many ways can this be done if two particular employees must be in the same team?Correct Option B … Explanation: The team containing the two particular employees needs 2 more members from the other 10, which can be chosen in C(10,2) = 45 ways. The remaining 8 employees must form 2 unlabelled teams of 4. That can be done in C(8,4)/2! = 70/2 = 35 ways, dividing by 2! because the two teams are not labelled. Total = 45 × 35 = 1575. Option E (4725) is the number of ways to form the teams with no restriction at all. Hence, option B.Q15. In how many ways can 7 distinct toys be put into 3 identical bags so that no bag is empty?Correct Option C … Explanation: Putting distinct objects into identical non-empty groups is counted by the Stirling number of the second kind, S(7,3). First count the ways if the bags were distinct, using inclusion–exclusion: 3⁷ − 3 × 2⁷ + 3 × 1⁷ = 2187 − 384 + 3 = 1806. The bags are actually identical, so divide by 3! = 6, which gives 1806/6 = 301. Option B (1806) is the answer for distinct bags. Hence, option C.Q16. Seven letters are to be placed in seven addressed envelopes, one per envelope. In how many ways can exactly 3 letters go into their correct envelopes?Correct Option B … Explanation: First choose which 3 letters go into their correct envelopes, in C(7,3) = 35 ways. The other 4 letters must all go into wrong envelopes, which is a derangement of 4 items. D4 = 4!(1 − 1 + 1/2 − 1/6 + 1/24) = 9. Required = 35 × 9 = 315. Hence, option B.Q17. Five guests leave their hats at a party and each takes one hat at random on leaving. In how many ways can at least two guests get their own hats?Correct Option E … Explanation: There are 5! = 120 ways for the hats to be returned. Cases where nobody gets their own hat number D5 = 44. Cases where exactly one guest gets their own hat number 5 × D4 = 5 × 9 = 45. Required = 120 − 44 − 45 = 31. To check, count the cases directly: exactly 2 gives C(5,2) × D3 = 20, exactly 3 gives C(5,3) × D2 = 10, exactly 4 is impossible, and exactly 5 gives 1. That also totals 20 + 10 + 0 + 1 = 31. Option C (45) is the count for exactly one match. Hence, option E.Q18. How many 5-digit numbers have their digits in strictly decreasing order from left to right and are divisible by 5?Correct Option A … Explanation: Any set of 5 distinct digits can be written in strictly decreasing order in exactly one way. The leading digit is the largest, so it is never 0. To be divisible by 5, the last digit (the smallest) must be 0 or 5. If the last digit is 0, choose the other 4 digits from 1–9 in C(9,4) = 126 ways. If the last digit is 5, the other 4 digits must all be greater than 5, so they are exactly 6, 7, 8 and 9, which gives 1 number (98765). Total = 126 + 1 = 127. Option B (126) misses the case ending in 5. Hence, option A.Q19. A 6 × 5 grid is made of unit squares. How many rectangles in this grid are not squares?Correct Option D … Explanation: A 6 × 5 grid has 7 vertical and 6 horizontal lines. Choosing 2 of each gives C(7,2) × C(6,2) = 21 × 15 = 315 rectangles. Squares of side k number (7 − k)(6 − k). That gives 30 squares of side 1, 20 of side 2, 12 of side 3, 6 of side 4 and 2 of side 5, which is 70 squares in all. Non-square rectangles = 315 − 70 = 245. Option B (315) counts every rectangle, squares included. Hence, option D.Q20. A robot moves only one step right or one step up on a grid, from (0, 0) to (6, 5). How many paths avoid the point (3, 2)?Correct Option E … Explanation: Every path makes 6 right steps and 5 up steps, so there are C(11,5) = 462 paths in total. Paths from (0,0) to (3,2) number C(5,2) = 10. Paths from (3,2) to (6,5) use 3 right and 3 up steps, so they number C(6,3) = 20. The paths through (3,2) therefore number 10 × 20 = 200. Paths that avoid (3,2) = 462 − 200 = 262. Option A (200) is the number of paths through the point, not the number that avoid it. Hence, option E.
