Question 1

In how many ways can the letters of the word 'BOOK' be arranged?

Accepted Answer

Step 1: The word 'BOOK' has 4 letters with O repeated twice. Step 2: Use formula for permutations with repetition: n! / (n₁! × n₂! × ...). Step 3: Number of arrangements = 4! / 2! = 24 / 2 = 12.

Question 2

How many 2-digit numbers can be formed using the digits 3, 4, 5, 6 without repetition?

Accepted Answer

Step 1: We need to form 2-digit numbers from 4 available digits without repetition. Step 2: For the first position, we have 4 choices. For the second position, we have 3 remaining choices. Step 3: Total = 4 × 3 = 12 or P(4,2) = 4!/(4-2)! = 4!/2! = 12.

Question 3

In how many ways can 4 red balls and 3 blue balls be arranged in a line?

Accepted Answer

Step 1: Total balls = 4 + 3 = 7. Step 2: Red balls are identical to each other, and blue balls are identical to each other. Step 3: Use formula: 7! / (4! × 3!) = 5040 / (24 × 6) = 5040 / 144 = 35.

Question 4

A student must answer 5 questions out of 8 questions in an exam. In how many ways can the student choose the questions?

Accepted Answer

Step 1: The student must select 5 questions from 8 available questions. Step 2: Order of selection does not matter, so we use combination. Step 3: C(8,5) = 8! / (5! × 3!) = (8 × 7 × 6) / (3 × 2 × 1) = 336 / 6 = 56.

Question 5

How many ways can a committee of 3 people be selected from a group of 8 people?

Accepted Answer

Step 1: We need to select 3 people from 8, where order does not matter. Step 2: Use combination formula C(n,r) = n! / (r!(n-r)!). Step 3: C(8,3) = 8! / (3! × 5!) = (8 × 7 × 6) / (3 × 2 × 1) = 336 / 6 = 56.

Question 6

In how many ways can 5 different books be arranged on a shelf?

Accepted Answer

Step 1: We need to arrange 5 different books in a line. Step 2: The number of arrangements of n distinct objects is n!. Step 3: Therefore, 5! = 5 × 4 × 3 × 2 × 1 = 120.

Question 7

A password consists of 3 distinct digits followed by 2 distinct letters. How many such passwords can be formed if digits are from 0–9 and letters are from A–Z?

Accepted Answer

Step 1: Select and arrange 3 distinct digits from 10 digits (0–9): P(10,3) = 10 × 9 × 8 = 720. Step 2: Select and arrange 2 distinct letters from 26 letters (A–Z): P(26,2) = 26 × 25 = 650. Step 3: Since the digit part and letter part are independent, total passwords = 720 × 650 = 468,000.

Question 8

In how many ways can the letters of the word 'MISSISSIPPI' be arranged?

Accepted Answer

Step 1: Count total letters: M-I-S-S-I-S-S-I-P-P-I = 11 letters. Step 2: Identify frequencies: M appears 1 time, I appears 4 times, S appears 4 times, P appears 2 times. Step 3: Use the formula for permutations with repetition: n! / (n₁! × n₂! × ... × nₖ!) = 11! / (1! × 4! × 4! × 2!). Step 4: Calculate: 11! = 39,916,800; 4! = 24; 2! = 2. Step 5: Answer = 39,916,800 / (1 × 24 × 24 × 2) = 39,916,800 / 1,152 = 34,650.

Question 9

In how many ways can 5 different books be arranged on a shelf such that two specific books are always together?

Accepted Answer

Step 1: Treat the two specific books as a single unit. Now we have 4 units to arrange (the combined unit + 3 other books). Step 2: These 4 units can be arranged in 4! = 24 ways. Step 3: Within the combined unit, the 2 books can be arranged among themselves in 2! = 2 ways. Step 4: Total arrangements = 24 × 2 = 48.

Question 10

How many 4-digit numbers can be formed using the digits 2, 3, 5, 7, 8, 9 without repetition, such that the number is even?

Accepted Answer

Step 1: For a number to be even, the last digit must be even. From {2, 3, 5, 7, 8, 9}, the even digits are 2 and 8 (2 choices). Step 2: Once the last digit is fixed, we have 5 remaining digits to fill the first 3 positions. Step 3: First position: 5 choices, Second position: 4 choices, Third position: 3 choices. Step 4: Total = 2 × 5 × 4 × 3 = 120.

Question 11

A committee of 4 people is to be selected from 6 men and 5 women. In how many ways can this be done if the committee must have at least 2 women?

Accepted Answer

Step 1: 'At least 2 women' means 2 women or 3 women or 4 women. Step 2: Case 1 (2 women, 2 men): C(5,2) × C(6,2) = 10 × 15 = 150. Step 3: Case 2 (3 women, 1 man): C(5,3) × C(6,1) = 10 × 6 = 60. Step 4: Case 3 (4 women, 0 men): C(5,4) × C(6,0) = 5 × 1 = 5. Step 5: Total = 150 + 60 + 5 = 215.

Question 12

In how many ways can the letters of the word 'GARDEN' be arranged so that the vowels always occupy the odd positions?

Accepted Answer

Step 1: Identify the letters in 'GARDEN' — G, A, R, D, E, N (6 letters total). Vowels: A, E (2 vowels). Consonants: G, R, D, N (4 consonants). Step 2: Odd positions in a 6-letter word are positions 1, 3, 5 (3 positions). We need to place 2 vowels in these 3 odd positions. Number of ways = P(3,2) = 3×2 = 6. Step 3: The remaining 4 consonants fill the remaining 4 positions (2 even + 1 odd). Number of ways = 4! = 24. Step 4: Total arrangements = 6 × 24 = 144. Therefore, answer is 144.

Question 13

From a group of 8 people, in how many ways can a president, a vice-president, and a treasurer be selected if no person can hold more than one position?

Accepted Answer

Step 1: This is a permutation problem because the positions are distinct (order matters). Step 2: For president: 8 choices. Step 3: For vice-president: 7 remaining choices (one person already selected). Step 4: For treasurer: 6 remaining choices. Step 5: Total = 8 × 7 × 6 = 336. Alternatively, P(8,3) = 8! / (8-3)! = 8! / 5! = 336.

Question 14

A committee of 6 people is to be formed from 8 men and 5 women. In how many ways can this be done if the committee must have at least 2 women and at most 4 women?

Accepted Answer

Step 1: Valid compositions are: (4M, 2W), (3M, 3W), (2M, 4W). Step 2: For 4M, 2W: C(8,4) × C(5,2) = 70 × 10 = 700. Step 3: For 3M, 3W: C(8,3) × C(5,3) = 56 × 10 = 560. Step 4: For 2M, 4W: C(8,2) × C(5,4) = 28 × 5 = 140. Step 5: Total = 700 + 560 + 140 = 1,400.

Question 15

How many 4-digit numbers can be formed using digits 1, 2, 3, 4, 5, 6 (with repetition allowed) such that the number is divisible by 4? (A number is divisible by 4 if its last two digits form a number divisible by 4.)

Accepted Answer

Step 1: For divisibility by 4, the last two digits must form a number divisible by 4. Step 2: List all 2-digit numbers from {1,2,3,4,5,6} divisible by 4: 12, 16, 24, 32, 36, 44, 52, 56, 64. That's 9 valid endings. Step 3: For each valid ending, the first two digits can be any of 6 choices each: 6 × 6 = 36 ways. Step 4: Total = 9 × 36 = 324.

Question 16

In a group of 10 people, a team of 4 is to be selected. In how many ways can this be done if two specific people (A and B) must not both be selected together?

Accepted Answer

Step 1: Total ways to select 4 from 10 without restriction: C(10,4) = 210. Step 2: Ways where both A and B are selected: Select A and B (fixed), then choose 2 more from remaining 8: C(8,2) = 28. Step 3: Ways where A and B are NOT both selected = Total - (Both selected) = 210 - 28 = 182.

Question 17

From a deck of 52 playing cards, how many ways can 5 cards be selected such that there is at least one card from each suit (hearts, diamonds, clubs, spades)?

Accepted Answer

Step 1: With 4 suits and 5 cards, by pigeonhole principle, exactly one suit has 2 cards and the other three suits have 1 card each. Step 2: Choose which suit contributes 2 cards: C(4,1) = 4 ways. Step 3: Choose 2 cards from the selected suit (13 cards): C(13,2) = 78 ways. Step 4: Choose 1 card from each of the remaining 3 suits: C(13,1) × C(13,1) × C(13,1) = 13 × 13 × 13 = 2,197 ways. Step 5: Total = 4 × 78 × 2,197 = 685,416.

Question 18

In how many ways can 5 men and 4 women be arranged in a row such that no two women sit adjacent to each other?

Accepted Answer

Step 1: First arrange 5 men in a row: 5! = 120 ways. Step 2: This creates 6 possible positions for women (before first man, between each pair of men, and after last man): _M_M_M_M_M_. Step 3: Choose 4 of these 6 positions for women: C(6,4) = 15 ways. Step 4: Arrange 4 women in chosen positions: 4! = 24 ways. Step 5: Total = 5! × C(6,4) × 4! = 120 × 15 × 24 = 43,200.

Question 19

In how many ways can 5 different books be arranged on a shelf?

Accepted Answer

Step 1: We need to arrange 5 different books in a line. Step 2: This is a permutation problem where all 5 objects are used. Step 3: Number of arrangements = 5! = 5 × 4 × 3 × 2 × 1 = 120.

Question 20

In how many ways can the letters of the word 'BOOK' be arranged?

Accepted Answer

Step 1: The word 'BOOK' has 4 letters with 'O' repeated twice. Step 2: When objects are repeated, we use the formula: n!/(n₁! × n₂! × ... × nₖ!). Step 3: Number of arrangements = 4!/2! = (4 × 3 × 2 × 1)/2 = 24/2 = 12.

Question 21

How many 3-digit numbers can be formed using the digits 2, 3, 4, 5, and 6 without repetition?

Accepted Answer

Step 1: We need to form 3-digit numbers using 5 different digits without repetition. Step 2: This is a permutation problem: P(5,3). Step 3: P(5,3) = 5!/(5-3)! = 5!/2! = (5 × 4 × 3) = 60.

Question 22

In how many ways can 4 red balls and 3 blue balls be arranged in a line?

Accepted Answer

Step 1: We have 7 balls total: 4 red (identical) and 3 blue (identical). Step 2: This is a permutation with repetition problem. Step 3: Number of arrangements = 7!/(4! × 3!) = (7 × 6 × 5 × 4!)/(4! × 3 × 2 × 1) = (7 × 6 × 5)/6 = 35.

Question 23

How many ways can a committee of 3 people be selected from a group of 8 people?

Accepted Answer

Step 1: We need to select 3 people from 8 where order does not matter. Step 2: This is a combination problem: C(8,3) = 8!/(3! × 5!). Step 3: C(8,3) = (8 × 7 × 6)/(3 × 2 × 1) = 336/6 = 56.

Question 24

How many ways can 6 students be divided into two groups of 3 each?

Accepted Answer

Step 1: We need to divide 6 students into two indistinguishable groups of 3 each. Step 2: If we select 3 students for one group, the remaining 3 automatically form the other group. Step 3: Number of ways = C(6,3)/2 = 20/2 = 10. (We divide by 2 because the two groups are indistinguishable.)

Question 25

A student must answer 4 questions out of 7 questions on an exam. In how many ways can the student choose which questions to answer?

Accepted Answer

Step 1: The student must choose 4 questions from 7 available questions. Step 2: Order of selection does not matter, so use combination: C(7,4). Step 3: C(7,4) = 7! / [4! × 3!] = (7 × 6 × 5) / (3 × 2 × 1) = 210 / 6 = 35.

Question 26

A committee of 4 people is to be selected from a group of 6 men and 5 women. In how many ways can this be done if the committee must have at least 2 women?

Accepted Answer

Step 1: 'At least 2 women' means 2 women or 3 women or 4 women. Step 2: Case 1 (2 women, 2 men): C(5,2) × C(6,2) = 10 × 15 = 150. Step 3: Case 2 (3 women, 1 man): C(5,3) × C(6,1) = 10 × 6 = 60. Step 4: Case 3 (4 women, 0 men): C(5,4) × C(6,0) = 5 × 1 = 5. Step 5: Total = 150 + 60 + 5 = 215.

Question 27

In how many ways can 5 different books be arranged on a shelf such that two specific books (A and B) are always together?

Accepted Answer

Step 1: Treat the two specific books (A and B) as a single unit. So we have 4 units to arrange: (AB), C, D, E. Step 2: These 4 units can be arranged in 4! = 24 ways. Step 3: Within the unit (AB), books A and B can be arranged among themselves in 2! = 2 ways. Step 4: Total arrangements = 4! × 2! = 24 × 2 = 48.

Question 28

In how many ways can the letters of the word 'MISSISSIPPI' be arranged?

Accepted Answer

Step 1: Count total letters in MISSISSIPPI: 11 letters. Step 2: Identify repeated letters: M appears 1 time, I appears 4 times, S appears 4 times, P appears 2 times. Step 3: Apply formula for permutations with repetition: n! / (n₁! × n₂! × ... × nₖ!) = 11! / (1! × 4! × 4! × 2!). Step 4: Calculate: 11! = 39,916,800; 4! = 24; 2! = 2. Step 5: Result = 39,916,800 / (1 × 24 × 24 × 2) = 39,916,800 / 1,152 = 34,650.

Question 29

How many 4-digit numbers can be formed using the digits 2, 3, 5, 7, 8, 9 without repetition, such that the number is even?

Accepted Answer

Step 1: For a number to be even, it must end in an even digit. From {2, 3, 5, 7, 8, 9}, the even digits are 2 and 8 (2 choices). Step 2: Fix one even digit at the units place (1 way). Step 3: Fill the remaining 3 positions (thousands, hundreds, tens) with 3 digits chosen from the remaining 5 digits. Step 4: Number of ways = 5 × 4 × 3 = 60 for each choice of units digit. Step 5: Total = 2 × 60 = 120.

Question 30

From a deck of 52 playing cards, in how many ways can 5 cards be selected such that there are exactly 2 cards of one suit and exactly 3 cards of another suit?

Accepted Answer

Step 1: Choose 2 suits from 4 suits: C(4,2) = 6 ways. Step 2: Decide which suit gets 2 cards and which gets 3 cards: 2 ways (suit A gets 2 and suit B gets 3, or vice versa). Step 3: Select 2 cards from the first suit (13 cards): C(13,2) = 78. Step 4: Select 3 cards from the second suit (13 cards): C(13,3) = 286. Step 5: Total = 6 × 2 × 78 × 286 = 268,512. Alternatively: C(4,2) × 2 × C(13,2) × C(13,3) = 6 × 2 × 78 × 286 = 268,512.

Question 31

A committee of 4 members is to be formed from 6 men and 5 women. In how many ways can this be done if the committee must have at least 2 women?

Accepted Answer

Step 1: 'At least 2 women' means 2 women + 2 men, OR 3 women + 1 man, OR 4 women + 0 men. Step 2: Case 1 (2W, 2M): C(5,2) × C(6,2) = 10 × 15 = 150. Step 3: Case 2 (3W, 1M): C(5,3) × C(6,1) = 10 × 6 = 60. Step 4: Case 3 (4W, 0M): C(5,4) × C(6,0) = 5 × 1 = 5. Step 5: Total = 150 + 60 + 5 = 215.

Question 32

In how many ways can 5 different books be arranged on a shelf such that two specific books are always together?

Accepted Answer

Step 1: Treat the two specific books as a single unit. Now we have 4 units to arrange (the pair + 3 other books). Step 2: These 4 units can be arranged in 4! = 24 ways. Step 3: Within the pair, the 2 books can be arranged in 2! = 2 ways. Step 4: Total arrangements = 4! × 2! = 24 × 2 = 48.

Question 33

How many 4-digit numbers can be formed using the digits 1, 2, 3, 4, 5, 6 without repetition, such that the number is divisible by 2?

Accepted Answer

Step 1: For divisibility by 2, the last digit must be even (2, 4, or 6). Step 2: Choose the last digit: 3 choices (2, 4, or 6). Step 3: Fill the first three positions with remaining 5 digits in order: P(5,3) = 5 × 4 × 3 = 60. Step 4: Total = 3 × 60 = 180.

Question 34

In a group of 8 people, how many ways can we select a president, a vice-president, and a treasurer if no person can hold more than one position?

Accepted Answer

Step 1: This is a permutation problem because the positions are distinct (order matters). Step 2: Choose a president from 8 people: 8 choices. Step 3: Choose a vice-president from remaining 7 people: 7 choices. Step 4: Choose a treasurer from remaining 6 people: 6 choices. Step 5: Total = 8 × 7 × 6 = 336 or P(8,3) = 8!/(8−3)! = 8!/5! = 336.

Question 35

In a group of 10 people, how many ways can we select a president, a vice-president, and a secretary such that the president and vice-president are not the same person, and the secretary is different from both?

Accepted Answer

Step 1: Select president: 10 choices. Step 2: Select vice-president (must be different from president): 9 choices. Step 3: Select secretary (must be different from both president and vice-president): 8 choices. Step 4: Total = 10 × 9 × 8 = 720.

Question 36

From a deck of 52 playing cards, how many ways can we select 5 cards such that we have exactly 2 cards of one suit and exactly 3 cards of another suit?

Accepted Answer

Step 1: Choose 2 suits from 4 suits for the 2-card and 3-card groups: C(4,2) = 6 ways. Step 2: Decide which of the 2 chosen suits gets 2 cards and which gets 3 cards: 2 ways. Step 3: Select 2 cards from 13 cards of the first suit: C(13,2) = 78 ways. Step 4: Select 3 cards from 13 cards of the second suit: C(13,3) = 286 ways. Step 5: Total = 6 × 2 × 78 × 286 = 268,512.

Question 37

A committee of 6 members is to be formed from 8 engineers and 5 doctors such that the committee contains at least 2 doctors and at most 4 doctors. How many such committees are possible?

Accepted Answer

Step 1: Identify valid cases: 2 doctors + 4 engineers, 3 doctors + 3 engineers, or 4 doctors + 2 engineers. Step 2: Case 1 (2D, 4E): C(5,2) × C(8,4) = 10 × 70 = 700. Step 3: Case 2 (3D, 3E): C(5,3) × C(8,3) = 10 × 56 = 560. Step 4: Case 3 (4D, 2E): C(5,4) × C(8,2) = 5 × 28 = 140. Step 5: Total = 700 + 560 + 140 = 1,400.

Question 38

How many 4-digit numbers can be formed using digits 1, 2, 3, 4, 5, 6, 7 such that the number is divisible by 5 and no digit is repeated?

Accepted Answer

Step 1: For divisibility by 5, the last digit must be 5 (since 0 is not available). Step 2: Last digit is fixed as 5. Step 3: First three positions must be filled with 3 different digits from remaining 6 digits {1,2,3,4,6,7}. Step 4: Number of ways = P(6,3) = 6 × 5 × 4 = 120.

RRB NTPC Permutation & Combination

RRB NTPC Permutation & Combination — Past Exam Questions

Permutation & Combination— Rules & Concept

Key Points to Remember

Exam-Specific Tips

Permutation & Combination — Practice Questions

60-Second Revision — Permutation & Combination