Question 1

The number of ways to arrange the letters of the word 'MISSISSIPPI' is:

Accepted Answer

Setup: This is a permutation problem with repeated elements. We use the formula for permutations of n objects where some are identical: n! / (n₁! × n₂! × ... × nₖ!)

Solution:
Step 1: Count the total number of letters in 'MISSISSIPPI': M-I-S-S-I-S-S-I-P-P-I = 11 letters total.
Step 2: Identify the frequency of each letter:
  - M: 1 time
  - I: 4 times
  - S: 4 times
  - P: 2 times
Step 3: Apply the formula for permutations with repetition:
  Number of arrangements = 11! / (1! × 4! × 4! × 2!)
Step 4: Calculate:
  11! = 39,916,800
  4! = 24
  2! = 2
  Denominator = 1 × 24 × 24 × 2 = 1,152
  Result = 39,916,800 / 1,152 = 34,650

Verification: The denominator accounts for identical letters being indistinguishable, reducing the count from 11! to the correct value.

Question 2

From a group of 6 men and 4 women, in how many ways can a committee of 5 be formed such that it contains at least 2 women?

Accepted Answer

Setup: This is a combination problem with a constraint on the composition. We count committees with at least 2 women by considering cases: exactly 2 women, exactly 3 women, or exactly 4 women.

Solution:
Step 1: Total people = 6 men + 4 women = 10. Committee size = 5.
Step 2: 'At least 2 women' means 2, 3, or 4 women (we can't have 5 women since there are only 4).

Case 1 — Exactly 2 women and 3 men:
  Ways = C(4,2) × C(6,3) = 6 × 20 = 120

Case 2 — Exactly 3 women and 2 men:
  Ways = C(4,3) × C(6,2) = 4 × 15 = 60

Case 3 — Exactly 4 women and 1 man:
  Ways = C(4,4) × C(6,1) = 1 × 6 = 6

Step 3: Total = 120 + 60 + 6 = 186.

Verification (using complement): Total committees without restriction = C(10,5) = 252. Committees with 0 women (all men) = C(6,5) = 6. Committees with exactly 1 woman = C(4,1) × C(6,4) = 4 × 15 = 60. Committees with at least 2 women = 252 - 6 - 60 = 186 ✓

Question 3

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

Accepted Answer

Setup: This tests permutation with a constraint on the last digit (even condition).

Solution:
1. For a number to be even, its last digit must be even.
2. From {1, 2, 3, 4, 5}, the even digits are: 2 and 4. So 2 choices for the last digit.
3. Once the last digit is fixed, we have 4 remaining digits to fill the first 3 positions.
4. Number of ways to arrange 4 digits in 3 positions = P(4, 3) = 4!/(4-3)! = 4!/1! = 24.
5. By multiplication principle: Total = 2 × 24 = 48.

Verification: If last digit is 2, first 3 positions use {1, 3, 4, 5}: P(4,3) = 24. If last digit is 4, first 3 positions use {1, 2, 3, 5}: P(4,3) = 24. Total = 48. ✓

Question 4

A committee of 3 members is to be selected from a group of 5 men and 4 women. In how many ways can the committee be formed such that it has at least 1 woman?

Accepted Answer

Setup: This tests combination with a constraint using the complementary counting principle.

Solution:
Method 1 (Complementary Counting — Faster):
1. Total ways to select 3 members from 9 people = C(9,3) = 9!/(3!×6!) = (9×8×7)/(3×2×1) = 84.
2. Ways to select 3 members with NO women (all men) = C(5,3) = 5!/(3!×2!) = (5×4)/(2×1) = 10.
3. Ways with at least 1 woman = Total − All men = 84 − 10 = 74.

Verification (Direct Method):
- 1 woman, 2 men: C(4,1) × C(5,2) = 4 × 10 = 40
- 2 women, 1 man: C(4,2) × C(5,1) = 6 × 5 = 30
- 3 women, 0 men: C(4,3) × C(5,0) = 4 × 1 = 4
- Total = 40 + 30 + 4 = 74 ✓

Question 5

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

Accepted Answer

Setup: This tests permutation with repetition — a fundamental concept where identical objects reduce the total count.

Solution:
1. Count the total letters in 'MISSISSIPPI': M-I-S-S-I-S-S-I-P-P-I = 11 letters.
2. Identify the frequency of each letter:
   - M: 1
   - I: 4
   - S: 4
   - P: 2
3. Formula for permutations with repetition: n! / (n₁! × n₂! × ... × nₖ!)
   where n = total letters, n₁, n₂, ..., nₖ = frequencies of each repeated letter.
4. Arrangements = 11! / (1! × 4! × 4! × 2!)
   = 39,916,800 / (1 × 24 × 24 × 2)
   = 39,916,800 / 1,152
   = 34,650.

Verification of calculation:
- 11! = 39,916,800
- 4! = 24, so 4! × 4! = 576
- 2! = 2
- Denominator = 1 × 576 × 2 = 1,152
- Result = 39,916,800 ÷ 1,152 = 34,650 ✓

Question 6

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

Accepted Answer

Setup: This question tests permutation of objects with repetition — a fundamental concept in combinatorics.

Solution:
Step 1: Count the total number of letters in 'MISSISSIPPI'.
M-I-S-S-I-S-S-I-P-P-I
Total letters = 11

Step 2: Count the frequency of each distinct letter.
- M: 1
- I: 4
- S: 4
- P: 2

Step 3: Apply the formula for permutations with repetition:
Number of arrangements = n! / (n₁! × n₂! × ... × nₖ!)
where n is the total number of objects and n₁, n₂, ..., nₖ are the frequencies of each repeated object.

Step 4: Calculate:
Number of arrangements = 11! / (1! × 4! × 4! × 2!)
= 39,916,800 / (1 × 24 × 24 × 2)
= 39,916,800 / 1,152
= 34,650

Trap Explanation: Students commonly forget to account for all repeated letters, or they miscalculate the frequency of letters. Some may use 10! instead of 11!, or forget to divide by the factorials of repeated letters entirely.

Question 7

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

Accepted Answer

Setup: This tests combination with a logical constraint ('more boys than girls') requiring case-by-case enumeration.

Solution:
For a team of 5 with more boys than girls:
- Boys > Girls
- Boys + Girls = 5
- Possible cases: (Boys, Girls) = (3,2), (4,1), (5,0)

Case 1: 3 boys and 2 girls
- Ways = C(6,3) × C(5,2) = 20 × 10 = 200

Case 2: 4 boys and 1 girl
- Ways = C(6,4) × C(5,1) = 15 × 5 = 75

Case 3: 5 boys and 0 girls
- Ways = C(6,5) × C(5,0) = 6 × 1 = 6

Total = 200 + 75 + 6 = 281

Verification:
- C(6,3) = 6!/(3!×3!) = 20 ✓
- C(5,2) = 5!/(2!×3!) = 10 ✓
- C(6,4) = C(6,2) = 15 ✓
- C(5,1) = 5 ✓
- C(6,5) = 6 ✓

Question 8

A team of 5 players is to be selected from 6 men and 4 women such that the team has at least 2 women. In how many ways can this be done?

Accepted Answer

Setup: This question tests combination with a constraint on the composition of the selection (at least 2 women).

Solution:
Step 1: Identify the constraint — The team must have at least 2 women.
This means the team can have: 2 women and 3 men, OR 3 women and 2 men, OR 4 women and 1 man.

Step 2: Calculate each case using the combination formula C(n,r) = n! / (r! × (n−r)!).

Case 1: 2 women and 3 men
- Ways to select 2 women from 4: C(4,2) = 4! / (2! × 2!) = 6
- Ways to select 3 men from 6: C(6,3) = 6! / (3! × 3!) = 20
- Total for Case 1 = 6 × 20 = 120

Case 2: 3 women and 2 men
- Ways to select 3 women from 4: C(4,3) = 4! / (3! × 1!) = 4
- Ways to select 2 men from 6: C(6,2) = 6! / (2! × 4!) = 15
- Total for Case 2 = 4 × 15 = 60

Case 3: 4 women and 1 man
- Ways to select 4 women from 4: C(4,4) = 1
- Ways to select 1 man from 6: C(6,1) = 6
- Total for Case 3 = 1 × 6 = 6

Step 3: Apply the addition principle (sum all cases):
Total ways = 120 + 60 + 6 = 186

Trap Explanation: Students commonly forget to enumerate all valid cases, or they calculate only one case (usually 2 women and 3 men). Some may also use the complement method incorrectly, calculating C(10,5) − [C(6,5) + C(6,4)×C(4,1)] and making arithmetic errors.

Question 9

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

Accepted Answer

Setup: We need to count 4-digit permutations from 6 digits where the last digit (units place) must be even to ensure the number is even.

Solution:
Step 1: For a number to be even, its units digit must be even. From {1, 2, 3, 4, 5, 6}, the even digits are {2, 4, 6}. So there are 3 choices for the units place.
Step 2: Once the units digit is chosen, we have 5 remaining digits to fill the remaining 3 positions (thousands, hundreds, tens).
Step 3: The number of ways to arrange 5 digits taken 3 at a time is P(5,3) = 5!/(5-3)! = 5!/2! = 120/2 = 60.
Step 4: By the multiplication principle: Total = 3 × 60 = 180.

Alternative verification: Total 4-digit numbers without restriction = P(6,4) = 6!/2! = 360. Exactly half should be even (since 3 out of 6 digits are even), giving 360/2 = 180 ✓

Question 10

In how many ways can 5 distinct books be arranged on a shelf such that a specific book (say Book A) is always at one of the two ends?

Accepted Answer

Setup: This question tests the application of the multiplication principle combined with permutation of remaining objects.

Solution:
Step 1: Identify the constraint — Book A must be at one of the two ends (leftmost or rightmost position).

Step 2: Apply the multiplication principle.
- Number of ways to place Book A at an end = 2 (either left end or right end)
- After placing Book A, we have 4 remaining distinct books to arrange in the remaining 4 positions
- Number of ways to arrange 4 distinct books = 4! = 4 × 3 × 2 × 1 = 24

Step 3: By the multiplication principle:
Total arrangements = (Ways to place Book A) × (Ways to arrange remaining books)
Total = 2 × 24 = 48

Trap Explanation: Students commonly forget to multiply by 2 (the number of end positions) and only calculate 4! = 24. They treat the constraint as fixing Book A to a single position rather than allowing it to occupy either end.

Question 11

In how many ways can a committee of 3 members be selected from a group of 8 people such that a particular person (say Person X) is always included?

Accepted Answer

Setup: This question tests combination with a mandatory inclusion constraint.

Solution:
Step 1: Identify the constraint — Person X must always be on the committee.

Step 2: Since Person X is mandatory, we need to select only 2 more members from the remaining 7 people.

Step 3: Apply the combination formula:
Number of ways = C(7, 2) = 7! / (2! × 5!) = (7 × 6) / (2 × 1) = 42 / 2 = 21

Verification by counting:
- Person X is fixed in the committee
- We need to choose 2 from the remaining 7 people
- C(7,2) = 21

Trap Explanation: Students often calculate C(8,3) = 56 by ignoring the constraint, or they incorrectly use permutation P(7,2) = 42 instead of combination C(7,2) = 21. Some may also confuse the problem and calculate C(7,3) = 35.

Question 12

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

Accepted Answer

Setup: This question tests permutation with a constraint on the last digit (even number condition).

Solution:
Step 1: Identify the constraint — A number is even if and only if its last digit is even.
Even digits available: 2, 4, 6 (three choices)

Step 2: Apply the multiplication principle by considering positions from right to left.
- Position 4 (units place): Must be even. Choices = 3 (either 2, 4, or 6)
- Position 3 (tens place): Can be any of the remaining 5 digits (since one even digit is used)
- Position 2 (hundreds place): Can be any of the remaining 4 digits
- Position 1 (thousands place): Can be any of the remaining 3 digits

Step 3: By the multiplication principle:
Total = 3 × 5 × 4 × 3 = 180

Alternative verification:
- Fix units digit as 2: Remaining 5 digits arranged in 3 positions = P(5,3) = 5 × 4 × 3 = 60
- Fix units digit as 4: P(5,3) = 60
- Fix units digit as 6: P(5,3) = 60
- Total = 60 + 60 + 60 = 180

Trap Explanation: Students often calculate P(6,4) = 360 without considering the even constraint, or they mistakenly use only 2 even digits instead of 3.

Question 13

The number of 4-digit numbers that can be formed using the digits 1, 2, 3, 4, 5, 6 (with repetition allowed) such that the digits are in strictly increasing order is:

Accepted Answer

Setup: This problem tests the relationship between combinations and permutations when order is constrained.

Solution:
Step 1: We need to form 4-digit numbers using digits from {1, 2, 3, 4, 5, 6} where digits are in strictly increasing order.

Step 2: Key insight: If we select any 4 distinct digits from the 6 available digits, there is EXACTLY ONE way to arrange them in strictly increasing order.

Step 3: For example:
- If we select {1, 3, 4, 6}, the only strictly increasing arrangement is 1346.
- If we select {2, 3, 5, 6}, the only strictly increasing arrangement is 2356.

Step 4: Therefore, the number of such 4-digit numbers equals the number of ways to choose 4 digits from 6 digits.

Step 5: This is a combination problem: C(6, 4) = 6!/(4!·2!) = (6·5)/(2·1) = 15

Step 6: Verification with small example: From {1,2,3,4,5,6}, the 4-digit strictly increasing numbers are:
1234, 1235, 1236, 1245, 1246, 1256, 1345, 1346, 1356, 1456, 2345, 2346, 2356, 2456, 3456
Counting: 15 numbers ✓

Question 14

The number of ways to arrange the letters of the word 'PERMUTATION' such that no two vowels are adjacent is:

Accepted Answer

Setup: This is a constrained permutation problem requiring us to first place consonants, then insert vowels into the gaps created.

Solution:
Step 1: Identify vowels and consonants in 'PERMUTATION'.
Vowels: E, U, A, I (4 vowels, all distinct)
Consonants: P, R, M, T, T, N (6 consonants, with T repeated twice)

Step 2: Arrange consonants first.
Number of ways to arrange 6 consonants where T appears twice = 6!/2! = 720/2 = 360

Step 3: When 6 consonants are arranged, they create 7 possible positions for vowels (before first, between each pair, after last):
_C_C_C_C_C_C_

Step 4: We need to place 4 distinct vowels in these 7 available positions such that no two vowels are adjacent. This is equivalent to choosing 4 positions from 7 and arranging 4 distinct vowels in them.
Number of ways = P(7,4) = 7!/(7-4)! = 7!/3! = 7 × 6 × 5 × 4 = 840

Step 5: Total arrangements = 360 × 840 = 302,400

Question 15

A committee of 5 people is to be formed from 6 men and 4 women. The number of ways to form the committee such that it contains at least 2 women is:

Accepted Answer

Setup: Form a 5-person committee from 6 men and 4 women with the constraint that at least 2 women must be included.

Solution:
Step 1: 'At least 2 women' means the committee can have 2, 3, or 4 women (not 0 or 1).

Step 2: Since we're choosing 5 people total and need at least 2 women:
- If 2 women: 3 men needed. Ways = C(4,2) × C(6,3)
- If 3 women: 2 men needed. Ways = C(4,3) × C(6,2)
- If 4 women: 1 man needed. Ways = C(4,4) × C(6,1)

Step 3: Calculate each case.
Case 1 (2W, 3M): C(4,2) × C(6,3) = 6 × 20 = 120
Case 2 (3W, 2M): C(4,3) × C(6,2) = 4 × 15 = 60
Case 3 (4W, 1M): C(4,4) × C(6,1) = 1 × 6 = 6

Step 4: Total = 120 + 60 + 6 = 186

Alternative approach (using complement):
Total ways to form committee of 5 from 10 people = C(10,5) = 252
Ways with fewer than 2 women (0 or 1 women):
- 0 women, 5 men: C(4,0) × C(6,5) = 1 × 6 = 6
- 1 woman, 4 men: C(4,1) × C(6,4) = 4 × 15 = 60
Ways with fewer than 2 women = 6 + 60 = 66
Ways with at least 2 women = 252 - 66 = 186

Question 16

The number of ways to distribute 10 identical balls into 3 distinct boxes such that each box contains at least 2 balls is:

Accepted Answer

Setup: Distribute 10 identical balls into 3 distinct boxes with each box containing at least 2 balls.

Solution:
Step 1: Since each box must contain at least 2 balls, place 2 balls in each box first.
Balls used: 3 × 2 = 6 balls
Remaining balls: 10 - 6 = 4 balls

Step 2: Now distribute the remaining 4 identical balls into 3 distinct boxes with no restrictions (boxes can be empty).

Step 3: Use the stars and bars formula. The number of ways to distribute n identical objects into k distinct boxes is C(n+k-1, k-1).
Here, n = 4 (remaining balls), k = 3 (boxes)
Number of ways = C(4+3-1, 3-1) = C(6, 2) = 6!/(2!×4!) = (6×5)/(2×1) = 15

Step 4: Verification by enumeration. Let (x₁, x₂, x₃) denote the number of additional balls in boxes 1, 2, 3 respectively, where x₁ + x₂ + x₃ = 4 and xᵢ ≥ 0.
Valid distributions:
(4,0,0), (0,4,0), (0,0,4),
(3,1,0), (3,0,1), (1,3,0), (0,3,1), (1,0,3), (0,1,3),
(2,2,0), (2,0,2), (0,2,2),
(2,1,1), (1,2,1), (1,1,2)
Total: 3 + 6 + 3 + 3 = 15 ✓

Question 17

In how many ways can 5 distinct books be arranged on a shelf such that a specific book (say Book A) is never at either end?

Accepted Answer

Setup: We need to count arrangements of 5 distinct objects with a positional restriction on one object.

Solution:
Total arrangements of 5 books without restriction = 5! = 120

Arrangements where Book A is at an end:
- Book A at position 1: remaining 4 books in 4! = 24 ways
- Book A at position 5: remaining 4 books in 4! = 24 ways
- Total with Book A at an end = 24 + 24 = 48

Arrangements where Book A is NOT at either end:
= Total arrangements − Arrangements with Book A at an end
= 120 − 48 = 72

Alternative direct method:
- Book A can be placed in positions 2, 3, or 4 (3 choices)
- Remaining 4 books arranged in remaining 4 positions: 4! = 24 ways
- Total = 3 × 24 = 72 ✓

Question 18

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 5?

Accepted Answer

Setup: A number is divisible by 5 if and only if its last digit is 0 or 5. Since we only have digits 1–6, the last digit must be 5.

Solution:
For a 4-digit number to be divisible by 5:
- Last digit (units place) must be 5 (only option from our set)
- Remaining 3 positions must be filled with 3 digits chosen from {1, 2, 3, 4, 6}

Step 1: Fix the last digit = 5 (1 way)

Step 2: Fill the first 3 positions with digits from {1, 2, 3, 4, 6}:
- First position: 5 choices
- Second position: 4 choices (one digit already used)
- Third position: 3 choices (two digits already used)

Total = 5 × 4 × 3 = 60

Alternatively: P(5, 3) = 5!/(5−3)! = 5!/2! = 120/2 = 60

Question 19

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

Setup: We need to count combinations with a constraint on the composition of the committee. 'At least 2 women' means 2, 3, or 4 women.

Solution:
Total people = 6 men + 5 women = 11
Committee size = 4
Constraint: At least 2 women

This means:
- Case 1: Exactly 2 women and 2 men
- Case 2: Exactly 3 women and 1 man
- Case 3: Exactly 4 women and 0 men

Case 1: C(5,2) × C(6,2) = 10 × 15 = 150

Case 2: C(5,3) × C(6,1) = 10 × 6 = 60

Case 3: C(5,4) × C(6,0) = 5 × 1 = 5

Total = 150 + 60 + 5 = 215

Verification using complement:
Total committees without restriction = C(11,4) = 330
Committees with fewer than 2 women (0 or 1 woman):
- 0 women, 4 men: C(5,0) × C(6,4) = 1 × 15 = 15
- 1 woman, 3 men: C(5,1) × C(6,3) = 5 × 20 = 100
- Total with <2 women = 15 + 100 = 115

Committees with ≥2 women = 330 − 115 = 215 ✓

Question 20

The number of ways to arrange the letters of the word 'MISSISSIPPI' is:

Accepted Answer

Setup: We count permutations of a word with repeated letters using the formula: n! / (n₁! × n₂! × ... × nₖ!), where nᵢ is the frequency of the i-th distinct letter.

Solution:
Word: MISSISSIPPI
Total letters: 11

Frequency of each letter:
- M: 1
- I: 4
- S: 4
- P: 2

Number of arrangements = 11! / (1! × 4! × 4! × 2!)

Calculation:
11! = 39,916,800
1! = 1
4! = 24
4! = 24
2! = 2

Denominator = 1 × 24 × 24 × 2 = 1,152

Number of arrangements = 39,916,800 / 1,152 = 34,650

Step-by-step division:
39,916,800 ÷ 24 = 1,663,200
1,663,200 ÷ 24 = 69,300
69,300 ÷ 2 = 34,650

NDA Permutation & Combination

Permutation & Combination— Rules & Concept

Key Points to Remember

Exam-Specific Tips

Permutation & Combination — Practice Questions

60-Second Revision — Permutation & Combination