Question 1

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 identical red balls and 3 identical blue balls. Step 2: Use the formula for permutations with repetition: n! / (n₁! × n₂!), where n = total objects, n₁ = red balls, n₂ = blue balls. Step 3: Arrangements = 7! / (4! × 3!) = 5040 / (24 × 6) = 5040 / 144 = 35.

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: The number of arrangements of n distinct objects is n!. Step 3: Therefore, arrangements = 5! = 5 × 4 × 3 × 2 × 1 = 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 (committee selection). 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 4

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

Accepted Answer

Step 1: The word 'APPLE' has 5 letters with 'P' repeated twice. Step 2: Formula for arrangements with repetition: n! / (n₁! × n₂! × ... × nₖ!), where n₁, n₂, etc. are frequencies of repeated letters. Step 3: Number of arrangements = 5! / 2! = 120 / 2 = 60.

Question 5

In how many ways can 6 students be divided into 2 groups of 3 each?

Accepted Answer

Step 1: We need to divide 6 students into 2 indistinguishable groups of 3 each. Step 2: If groups were distinguishable, we'd use C(6,3) = 20. Step 3: Since the 2 groups are indistinguishable (Group 1 and Group 2 are not labeled), we divide by 2!. Step 4: Answer = C(6,3) / 2! = 20 / 2 = 10.

Question 6

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

Accepted Answer

Step 1: For divisibility by 5, the last digit must be 5 (since 0 is not available). Step 2: Fix 5 in the units place. Step 3: Fill remaining 3 positions (thousands, hundreds, tens) with 3 digits chosen from {1, 2, 3, 4, 6}. Step 4: Number of ways = P(5,3) = 5 × 4 × 3 = 60.

Question 7

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 8

In a group of 10 people, how many ways can we select a president, a vice-president, and a secretary such that no person holds more than one position?

Accepted Answer

Step 1: This is a permutation problem since order matters (different positions are distinct). Step 2: Select president from 10 people: 10 choices. Step 3: Select vice-president from remaining 9 people: 9 choices. Step 4: Select secretary from remaining 8 people: 8 choices. Step 5: Total ways = 10 × 9 × 8 = 720 or P(10,3) = 10!/(10-3)! = 10!/7! = 720.

Question 9

A committee of 5 members is to be formed from 8 men and 6 women such that it must contain at least 2 women. In how many ways can this be done?

Accepted Answer

Step 1: Total ways to form a 5-member committee from 14 people = C(14,5) = 2002. Step 2: Ways with 0 women (all men) = C(8,5) = 56. Step 3: Ways with 1 woman = C(6,1) × C(8,4) = 6 × 70 = 420. Step 4: Ways with at least 2 women = 2002 - 56 - 420 = 1526.

Question 10

How many 4-digit numbers can be formed using the 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: Find all 2-digit numbers from {1,2,3,4,5,6} (with repetition) 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 11

A password consists of 3 letters followed by 2 digits. Letters are chosen from {A, B, C, D, E} and digits from {0, 1, 2, ..., 9}. If the password must contain at least one vowel (A or E) among the letters and the two digits must be distinct, in how many ways can the password be formed?

Accepted Answer

Step 1: Find ways to choose 3 letters from {A,B,C,D,E} with at least one vowel (A or E). Total ways without restriction = 5³ = 125. Ways with no vowels (only B,C,D) = 3³ = 27. Ways with at least one vowel = 125 - 27 = 98. Step 2: Find ways to choose 2 distinct digits from {0,1,2,...,9}. First digit: 10 choices. Second digit: 9 choices (must differ from first) = 10 × 9 = 90. Step 3: Total = 98 × 90 = 8820.

Question 12

In a circular arrangement of 8 people, 3 specific people (X, Y, Z) must sit together. In how many ways can they be arranged?

Accepted Answer

Step 1: Treat X, Y, Z as a single unit. Now we have 6 units to arrange in a circle (the block of 3 + 5 other people). Circular arrangements of n objects = (n-1)!. So, 6 units = (6-1)! = 5! = 120. Step 2: Within the block, X, Y, Z can be arranged among themselves = 3! = 6 ways. Step 3: Total = 120 × 6 = 720.

Question 13

A team of 6 people is to be selected from 4 engineers and 5 doctors such that the team has more doctors than engineers. How many ways can this be done?

Accepted Answer

Step 1: More doctors than engineers means: doctors > engineers. Since total = 6, possible combinations are (engineers, doctors): (0,6)—impossible (only 5 doctors), (1,5), (2,4). Step 2: Case 1 (1 engineer, 5 doctors): C(4,1) × C(5,5) = 4 × 1 = 4. Step 3: Case 2 (2 engineers, 4 doctors): C(4,2) × C(5,4) = 6 × 5 = 30. Step 4: Total = 4 + 30 = 34.

SBI Clerk Permutation & Combination

SBI Clerk Permutation & Combination — Past Exam Questions

Permutation & Combination— Rules & Concept

Key Points to Remember

Exam-Specific Tips

60-Second Revision — Permutation & Combination