Question 1

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: The order of selection does not matter, so this is a combination problem. Step 3: Number of ways = C(8,5) = 8!/(5! × 3!) = (8 × 7 × 6)/(3 × 2 × 1) = 336/6 = 56. Therefore, the answer is 56.

Question 2

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. Therefore, the answer is 120.

Question 3

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. Step 3: Number of ways = C(8,3) = 8!/(3! × 5!) = (8 × 7 × 6)/(3 × 2 × 1) = 336/6 = 56. Therefore, the answer is 56.

Question 4

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

Accepted Answer

Step 1: The word 'APPLE' has 5 letters total. Step 2: The letter 'P' appears 2 times (repeated), while A, L, E appear once each. Step 3: Number of arrangements = 5!/(2!) = (5 × 4 × 3 × 2 × 1)/2 = 120/2 = 60. Therefore, the answer is 60.

Question 5

How many 2-digit numbers can be formed using the digits 3, 5, 7, and 9 without repetition?

Accepted Answer

Step 1: We need to form 2-digit numbers using 4 distinct digits without repetition. Step 2: For the first position (tens place), we have 4 choices. Step 3: For the second position (units place), we have 3 remaining choices (since repetition is not allowed). Step 4: Total numbers = 4 × 3 = 12. Alternatively, this is P(4,2) = 4!/(4-2)! = 4!/2! = 12. Therefore, the answer is 12.

Question 6

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: We have 4 identical red balls and 3 identical blue balls. Step 3: Number of arrangements = 7!/(4! × 3!) = (7 × 6 × 5 × 4!)/(4! × 3 × 2 × 1) = (7 × 6 × 5)/6 = 210/6 = 35. Therefore, the answer is 35.

Question 7

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 8

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

Accepted Answer

Step 1: Count the total letters in MISSISSIPPI: M(1), I(4), S(4), P(2) = 11 letters total. Step 2: When letters repeat, the formula is n! / (n₁! × n₂! × ... × nₖ!), where n is total letters and n₁, n₂, etc. are frequencies of each repeated letter. Step 3: Arrangements = 11! / (1! × 4! × 4! × 2!) = 39,916,800 / (1 × 24 × 24 × 2) = 39,916,800 / 1,152 = 34,650.

Question 9

From a group of 8 students, in how many ways can a President, Vice-President, and Secretary be selected such that no student holds more than one position?

Accepted Answer

Step 1: This is a permutation problem because the positions are distinct (order matters). Step 2: We need to select 3 students from 8 for 3 different positions. Step 3: President can be chosen in 8 ways, Vice-President in 7 ways (from remaining), Secretary in 6 ways (from remaining). Step 4: Total = 8 × 7 × 6 = 336. Alternatively, P(8,3) = 8!/(8-3)! = 8!/5! = 336.

Question 10

A password consists of 2 letters followed by 3 digits. If letters can be repeated but digits cannot, how many different passwords are possible? (Consider 26 letters and 10 digits: 0–9)

Accepted Answer

Step 1: The password has 2 letter positions and 3 digit positions. Step 2: For the 2 letters (repetition allowed): First letter has 26 choices, second letter has 26 choices = 26 × 26 = 676. Step 3: For the 3 digits (no repetition): First digit has 10 choices, second digit has 9 choices (one used), third digit has 8 choices (two used) = 10 × 9 × 8 = 720. Step 4: Total passwords = 676 × 720 = 486,720.

Question 11

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. Step 3: Now we need to fill the remaining 3 positions (thousands, hundreds, tens) with 5 remaining digits. Step 4: This can be done in P(5,3) = 5!/(5-3)! = 5!/2! = 60 ways. Step 5: Total = 2 × 60 = 120.

Question 12

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. 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, books A and B can be arranged in 2! = 2 ways. Step 4: Total arrangements = 4! × 2! = 24 × 2 = 48.

Question 13

A committee of 6 members is to be formed from 8 engineers and 5 doctors such that at least 2 engineers and at least 2 doctors are included. In how many ways can this be done?

Accepted Answer

Step 1: Total committee size = 6. Minimum 2 engineers and 2 doctors means possible distributions are: (2E, 4D), (3E, 3D), (4E, 2D). Step 2: Case 1 (2E, 4D): C(8,2) × C(5,4) = 28 × 5 = 140. Step 3: Case 2 (3E, 3D): C(8,3) × C(5,3) = 56 × 10 = 560. Step 4: Case 3 (4E, 2D): C(8,4) × C(5,2) = 70 × 10 = 700. Step 5: Total = 140 + 560 + 700 = 1,400.

Question 14

From a group of 7 boys and 6 girls, a team of 5 is to be selected. In how many ways can this be done such that the team has more girls than boys?

Accepted Answer

Step 1: For more girls than boys in a 5-person team, valid distributions are: (4G, 1B) and (5G, 0B). Step 2: Case 1 (4G, 1B): C(6,4) × C(7,1) = 15 × 7 = 105. Step 3: Case 2 (5G, 0B): C(6,5) × C(7,0) = 6 × 1 = 6. Step 4: Total = 105 + 6 = 111.

Question 15

How many 4-digit numbers can be formed using digits 1, 2, 3, 4, 5, 6 (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 digits {1,2,3,4,5,6} divisible by 4: 12, 16, 24, 32, 36, 44, 52, 56, 64. That is 9 valid endings. Step 3: First two digits can be any of 6 digits each: 6 × 6 = 36 ways. Step 4: Total = 36 × 9 = 324.

Question 16

In how many ways can the letters of the word 'MISSISSIPPI' be arranged such that all I's are together and all S's are together?

Accepted Answer

Step 1: Word MISSISSIPPI has 11 letters: M(1), I(4), S(4), P(2). Step 2: Treat all 4 I's as one block and all 4 S's as one block. Step 3: Now we have units: [IIII], [SSSS], M, P, P = 5 units. Step 4: Arrange these 5 units: 5!/2! = 60 (divide by 2! for two identical P's). Step 5: Within [IIII] block: 1 way (all identical). Within [SSSS] block: 1 way (all identical). Step 6: Total = 60 × 1 × 1 = 60.

Question 17

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 18

A password consists of 3 letters followed by 2 digits. Letters can be repeated but must be from {A, B, C, D, E}, and digits must be distinct and from {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}. How many such passwords are possible?

Accepted Answer

Step 1: First 3 positions are letters from {A, B, C, D, E} with repetition allowed: 5 × 5 × 5 = 125 ways. Step 2: Last 2 positions are distinct digits from {0, 1, 2, ..., 9}: First digit has 10 choices, second digit has 9 choices (must be different): 10 × 9 = 90 ways. Step 3: Total = 125 × 90 = 11,250.

SSC MTS Permutation & Combination

SSC MTS Permutation & Combination — Past Exam Questions

Permutation & Combination— Rules & Concept

Key Points to Remember

Exam-Specific Tips

60-Second Revision — Permutation & Combination