Combination with 'At Least' Constraint - Intermediate Level: tricky scenarios handling Combination with 'At Least' Constraint INTERMEDIATE

This expert challenge 📈 worksheet focuses on Combination with 'At Least' Constraint - a key topic in Permutation Combination. You'll solve 20 intermediate-level problems (Worksheet 5 of 10). The primary focus is on tricky scenarios handling. Master how to solve combination with 'at least' constraint, combination with 'at least' constraint tricks, and combination with 'at least' constraint shortcut methods through systematic practice.

📝 Worksheet 5 of 10 • 20 questions • ⏱️ Estimated time: 20 minutes • 🎯 Intermediate level

What you'll learn in this worksheet:

Master how to solve combination with 'at least' constraint through focused practice
Understand the logic behind combination with 'at least' constraint tricks
Learn step-by-step approaches to tricky scenarios handling
Improve your solving speed while maintaining accuracy
Learn to eliminate wrong options efficiently

Your progress through Combination with 'At Least' Constraint

Worksheet 5 of 10 (44% complete)

Question 1

A committee of 7 members is to be formed from 10 men and 8 women. In how many ways can this be done if the committee must have **at least** 5 men?

Step-by-Step Solution (Sum Rule):

Concept: 'At least' problems require finding the sum of ways for all valid, mutually exclusive cases.

Given:
- Men available: 10
- Women available: 8
- Committee size: 7
- Constraint: At least 5 men

Strategy: We sum the ways for all cases from exactly 5 men up to the maximum possible number of men (7).

Valid Cases (Men, Women) and Calculation:

(5 Men, 2 Women): C(10,5) × C(8,2) = 252 × 28 = 7056
(6 Men, 1 Women): C(10,6) × C(8,1) = 210 × 8 = 1680
(7 Men, 0 Women): C(10,7) × C(8,0) = 120 × 1 = 120

Final Calculation (Sum Rule):
Total ways = (Ways with 5 men) + (Ways with 6 men) + ...
= 7056 + 1680 + 120
= 8856

Key Principle: Use the Sum Rule (addition) because the cases are mutually exclusive (you cannot simultaneously select exactly $k$ men and exactly $j$ men, where $k
e j$).

Question 2

A committee of 8 members is to be formed from 9 men and 6 women. In how many ways can this be done if the committee must have **at least** 6 men?

Step-by-Step Solution (Sum Rule):

Concept: 'At least' problems require finding the sum of ways for all valid, mutually exclusive cases.

Given:
- Men available: 9
- Women available: 6
- Committee size: 8
- Constraint: At least 6 men

Strategy: We sum the ways for all cases from exactly 6 men up to the maximum possible number of men (8).

Valid Cases (Men, Women) and Calculation:

(6 Men, 2 Women): C(9,6) × C(6,2) = 84 × 15 = 1260
(7 Men, 1 Women): C(9,7) × C(6,1) = 36 × 6 = 216
(8 Men, 0 Women): C(9,8) × C(6,0) = 9 × 1 = 9

Final Calculation (Sum Rule):
Total ways = (Ways with 6 men) + (Ways with 7 men) + ...
= 1260 + 216 + 9
= 1485

Key Principle: Use the Sum Rule (addition) because the cases are mutually exclusive (you cannot simultaneously select exactly $k$ men and exactly $j$ men, where $k
e j$).

Question 3

A committee of 7 members is to be formed from 8 men and 7 women. In how many ways can this be done if the committee must have **at least** 4 men?

Step-by-Step Solution (Sum Rule):

Concept: 'At least' problems require finding the sum of ways for all valid, mutually exclusive cases.

Given:
- Men available: 8
- Women available: 7
- Committee size: 7
- Constraint: At least 4 men

Strategy: We sum the ways for all cases from exactly 4 men up to the maximum possible number of men (7).

Valid Cases (Men, Women) and Calculation:

(4 Men, 3 Women): C(8,4) × C(7,3) = 70 × 35 = 2450
(5 Men, 2 Women): C(8,5) × C(7,2) = 56 × 21 = 1176
(6 Men, 1 Women): C(8,6) × C(7,1) = 28 × 7 = 196
(7 Men, 0 Women): C(8,7) × C(7,0) = 8 × 1 = 8

Final Calculation (Sum Rule):
Total ways = (Ways with 4 men) + (Ways with 5 men) + ...
= 2450 + 1176 + 196 + 8
= 3830

Key Principle: Use the Sum Rule (addition) because the cases are mutually exclusive (you cannot simultaneously select exactly $k$ men and exactly $j$ men, where $k
e j$).

Question 4

A committee of 6 members is to be formed from 9 men and 7 women. In how many ways can this be done if the committee must have **at least** 3 men?

Step-by-Step Solution (Sum Rule):

Concept: 'At least' problems require finding the sum of ways for all valid, mutually exclusive cases.

Given:
- Men available: 9
- Women available: 7
- Committee size: 6
- Constraint: At least 3 men

Strategy: We sum the ways for all cases from exactly 3 men up to the maximum possible number of men (6).

Valid Cases (Men, Women) and Calculation:

(3 Men, 3 Women): C(9,3) × C(7,3) = 84 × 35 = 2940
(4 Men, 2 Women): C(9,4) × C(7,2) = 126 × 21 = 2646
(5 Men, 1 Women): C(9,5) × C(7,1) = 126 × 7 = 882
(6 Men, 0 Women): C(9,6) × C(7,0) = 84 × 1 = 84

Final Calculation (Sum Rule):
Total ways = (Ways with 3 men) + (Ways with 4 men) + ...
= 2940 + 2646 + 882 + 84
= 6552

Key Principle: Use the Sum Rule (addition) because the cases are mutually exclusive (you cannot simultaneously select exactly $k$ men and exactly $j$ men, where $k
e j$).

Question 5

A committee of 6 members is to be formed from 8 men and 8 women. In how many ways can this be done if the committee must have **at least** 3 men?

Step-by-Step Solution (Sum Rule):

Concept: 'At least' problems require finding the sum of ways for all valid, mutually exclusive cases.

Given:
- Men available: 8
- Women available: 8
- Committee size: 6
- Constraint: At least 3 men

Strategy: We sum the ways for all cases from exactly 3 men up to the maximum possible number of men (6).

Valid Cases (Men, Women) and Calculation:

(3 Men, 3 Women): C(8,3) × C(8,3) = 56 × 56 = 3136
(4 Men, 2 Women): C(8,4) × C(8,2) = 70 × 28 = 1960
(5 Men, 1 Women): C(8,5) × C(8,1) = 56 × 8 = 448
(6 Men, 0 Women): C(8,6) × C(8,0) = 28 × 1 = 28

Final Calculation (Sum Rule):
Total ways = (Ways with 3 men) + (Ways with 4 men) + ...
= 3136 + 1960 + 448 + 28
= 5572

Key Principle: Use the Sum Rule (addition) because the cases are mutually exclusive (you cannot simultaneously select exactly $k$ men and exactly $j$ men, where $k
e j$).

Question 6

A committee of 7 members is to be formed from 10 men and 6 women. In how many ways can this be done if the committee must have **at least** 5 men?

Step-by-Step Solution (Sum Rule):

Concept: 'At least' problems require finding the sum of ways for all valid, mutually exclusive cases.

Given:
- Men available: 10
- Women available: 6
- Committee size: 7
- Constraint: At least 5 men

Strategy: We sum the ways for all cases from exactly 5 men up to the maximum possible number of men (7).

Valid Cases (Men, Women) and Calculation:

(5 Men, 2 Women): C(10,5) × C(6,2) = 252 × 15 = 3780
(6 Men, 1 Women): C(10,6) × C(6,1) = 210 × 6 = 1260
(7 Men, 0 Women): C(10,7) × C(6,0) = 120 × 1 = 120

Final Calculation (Sum Rule):
Total ways = (Ways with 5 men) + (Ways with 6 men) + ...
= 3780 + 1260 + 120
= 5160

Key Principle: Use the Sum Rule (addition) because the cases are mutually exclusive (you cannot simultaneously select exactly $k$ men and exactly $j$ men, where $k
e j$).

Question 7

A committee of 6 members is to be formed from 10 men and 7 women. In how many ways can this be done if the committee must have **at least** 3 men?

Step-by-Step Solution (Sum Rule):

Concept: 'At least' problems require finding the sum of ways for all valid, mutually exclusive cases.

Given:
- Men available: 10
- Women available: 7
- Committee size: 6
- Constraint: At least 3 men

Strategy: We sum the ways for all cases from exactly 3 men up to the maximum possible number of men (6).

Valid Cases (Men, Women) and Calculation:

(3 Men, 3 Women): C(10,3) × C(7,3) = 120 × 35 = 4200
(4 Men, 2 Women): C(10,4) × C(7,2) = 210 × 21 = 4410
(5 Men, 1 Women): C(10,5) × C(7,1) = 252 × 7 = 1764
(6 Men, 0 Women): C(10,6) × C(7,0) = 210 × 1 = 210

Final Calculation (Sum Rule):
Total ways = (Ways with 3 men) + (Ways with 4 men) + ...
= 4200 + 4410 + 1764 + 210
= 10584

Key Principle: Use the Sum Rule (addition) because the cases are mutually exclusive (you cannot simultaneously select exactly $k$ men and exactly $j$ men, where $k
e j$).

Question 8

A committee of 8 members is to be formed from 9 men and 8 women. In how many ways can this be done if the committee must have **at least** 4 men?

Step-by-Step Solution (Sum Rule):

Concept: 'At least' problems require finding the sum of ways for all valid, mutually exclusive cases.

Given:
- Men available: 9
- Women available: 8
- Committee size: 8
- Constraint: At least 4 men

Strategy: We sum the ways for all cases from exactly 4 men up to the maximum possible number of men (8).

Valid Cases (Men, Women) and Calculation:

(4 Men, 4 Women): C(9,4) × C(8,4) = 126 × 70 = 8820
(5 Men, 3 Women): C(9,5) × C(8,3) = 126 × 56 = 7056
(6 Men, 2 Women): C(9,6) × C(8,2) = 84 × 28 = 2352
(7 Men, 1 Women): C(9,7) × C(8,1) = 36 × 8 = 288
(8 Men, 0 Women): C(9,8) × C(8,0) = 9 × 1 = 9

Final Calculation (Sum Rule):
Total ways = (Ways with 4 men) + (Ways with 5 men) + ...
= 8820 + 7056 + 2352 + 288 + 9
= 18525

Key Principle: Use the Sum Rule (addition) because the cases are mutually exclusive (you cannot simultaneously select exactly $k$ men and exactly $j$ men, where $k
e j$).

Question 9

A committee of 7 members is to be formed from 10 men and 6 women. In how many ways can this be done if the committee must have **at least** 3 men?

Step-by-Step Solution (Sum Rule):

Concept: 'At least' problems require finding the sum of ways for all valid, mutually exclusive cases.

Given:
- Men available: 10
- Women available: 6
- Committee size: 7
- Constraint: At least 3 men

Strategy: We sum the ways for all cases from exactly 3 men up to the maximum possible number of men (7).

Valid Cases (Men, Women) and Calculation:

(3 Men, 4 Women): C(10,3) × C(6,4) = 120 × 15 = 1800
(4 Men, 3 Women): C(10,4) × C(6,3) = 210 × 20 = 4200
(5 Men, 2 Women): C(10,5) × C(6,2) = 252 × 15 = 3780
(6 Men, 1 Women): C(10,6) × C(6,1) = 210 × 6 = 1260
(7 Men, 0 Women): C(10,7) × C(6,0) = 120 × 1 = 120

Final Calculation (Sum Rule):
Total ways = (Ways with 3 men) + (Ways with 4 men) + ...
= 1800 + 4200 + 3780 + 1260 + 120
= 11160

Key Principle: Use the Sum Rule (addition) because the cases are mutually exclusive (you cannot simultaneously select exactly $k$ men and exactly $j$ men, where $k
e j$).

Question 10

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

Step-by-Step Solution (Sum Rule):

Concept: 'At least' problems require finding the sum of ways for all valid, mutually exclusive cases.

Given:
- Men available: 8
- Women available: 6
- Committee size: 8
- Constraint: At least 6 men

Strategy: We sum the ways for all cases from exactly 6 men up to the maximum possible number of men (8).

Valid Cases (Men, Women) and Calculation:

(6 Men, 2 Women): C(8,6) × C(6,2) = 28 × 15 = 420
(7 Men, 1 Women): C(8,7) × C(6,1) = 8 × 6 = 48
(8 Men, 0 Women): C(8,8) × C(6,0) = 1 × 1 = 1

Final Calculation (Sum Rule):
Total ways = (Ways with 6 men) + (Ways with 7 men) + ...
= 420 + 48 + 1
= 469

Key Principle: Use the Sum Rule (addition) because the cases are mutually exclusive (you cannot simultaneously select exactly $k$ men and exactly $j$ men, where $k
e j$).

Question 11

A committee of 6 members is to be formed from 9 men and 7 women. In how many ways can this be done if the committee must have **at least** 4 men?

Step-by-Step Solution (Sum Rule):

Concept: 'At least' problems require finding the sum of ways for all valid, mutually exclusive cases.

Given:
- Men available: 9
- Women available: 7
- Committee size: 6
- Constraint: At least 4 men

Strategy: We sum the ways for all cases from exactly 4 men up to the maximum possible number of men (6).

Valid Cases (Men, Women) and Calculation:

(4 Men, 2 Women): C(9,4) × C(7,2) = 126 × 21 = 2646
(5 Men, 1 Women): C(9,5) × C(7,1) = 126 × 7 = 882
(6 Men, 0 Women): C(9,6) × C(7,0) = 84 × 1 = 84

Final Calculation (Sum Rule):
Total ways = (Ways with 4 men) + (Ways with 5 men) + ...
= 2646 + 882 + 84
= 3612

Key Principle: Use the Sum Rule (addition) because the cases are mutually exclusive (you cannot simultaneously select exactly $k$ men and exactly $j$ men, where $k
e j$).

Question 12

A committee of 6 members is to be formed from 10 men and 6 women. In how many ways can this be done if the committee must have **at least** 3 men?

Step-by-Step Solution (Sum Rule):

Concept: 'At least' problems require finding the sum of ways for all valid, mutually exclusive cases.

Given:
- Men available: 10
- Women available: 6
- Committee size: 6
- Constraint: At least 3 men

Strategy: We sum the ways for all cases from exactly 3 men up to the maximum possible number of men (6).

Valid Cases (Men, Women) and Calculation:

(3 Men, 3 Women): C(10,3) × C(6,3) = 120 × 20 = 2400
(4 Men, 2 Women): C(10,4) × C(6,2) = 210 × 15 = 3150
(5 Men, 1 Women): C(10,5) × C(6,1) = 252 × 6 = 1512
(6 Men, 0 Women): C(10,6) × C(6,0) = 210 × 1 = 210

Final Calculation (Sum Rule):
Total ways = (Ways with 3 men) + (Ways with 4 men) + ...
= 2400 + 3150 + 1512 + 210
= 7272

Key Principle: Use the Sum Rule (addition) because the cases are mutually exclusive (you cannot simultaneously select exactly $k$ men and exactly $j$ men, where $k
e j$).

Question 13

A committee of 8 members is to be formed from 10 men and 8 women. In how many ways can this be done if the committee must have **at least** 6 men?

Step-by-Step Solution (Sum Rule):

Concept: 'At least' problems require finding the sum of ways for all valid, mutually exclusive cases.

Given:
- Men available: 10
- Women available: 8
- Committee size: 8
- Constraint: At least 6 men

Strategy: We sum the ways for all cases from exactly 6 men up to the maximum possible number of men (8).

Valid Cases (Men, Women) and Calculation:

(6 Men, 2 Women): C(10,6) × C(8,2) = 210 × 28 = 5880
(7 Men, 1 Women): C(10,7) × C(8,1) = 120 × 8 = 960
(8 Men, 0 Women): C(10,8) × C(8,0) = 45 × 1 = 45

Final Calculation (Sum Rule):
Total ways = (Ways with 6 men) + (Ways with 7 men) + ...
= 5880 + 960 + 45
= 6885

Key Principle: Use the Sum Rule (addition) because the cases are mutually exclusive (you cannot simultaneously select exactly $k$ men and exactly $j$ men, where $k
e j$).

Question 14

A committee of 6 members is to be formed from 8 men and 6 women. In how many ways can this be done if the committee must have **at least** 3 men?

Step-by-Step Solution (Sum Rule):

Concept: 'At least' problems require finding the sum of ways for all valid, mutually exclusive cases.

Given:
- Men available: 8
- Women available: 6
- Committee size: 6
- Constraint: At least 3 men

Strategy: We sum the ways for all cases from exactly 3 men up to the maximum possible number of men (6).

Valid Cases (Men, Women) and Calculation:

(3 Men, 3 Women): C(8,3) × C(6,3) = 56 × 20 = 1120
(4 Men, 2 Women): C(8,4) × C(6,2) = 70 × 15 = 1050
(5 Men, 1 Women): C(8,5) × C(6,1) = 56 × 6 = 336
(6 Men, 0 Women): C(8,6) × C(6,0) = 28 × 1 = 28

Final Calculation (Sum Rule):
Total ways = (Ways with 3 men) + (Ways with 4 men) + ...
= 1120 + 1050 + 336 + 28
= 2534

Key Principle: Use the Sum Rule (addition) because the cases are mutually exclusive (you cannot simultaneously select exactly $k$ men and exactly $j$ men, where $k
e j$).

Question 15

A committee of 6 members is to be formed from 9 men and 7 women. In how many ways can this be done if the committee must have **at least** 3 men?

Step-by-Step Solution (Sum Rule):

Concept: 'At least' problems require finding the sum of ways for all valid, mutually exclusive cases.

Given:
- Men available: 9
- Women available: 7
- Committee size: 6
- Constraint: At least 3 men

Strategy: We sum the ways for all cases from exactly 3 men up to the maximum possible number of men (6).

Valid Cases (Men, Women) and Calculation:

(3 Men, 3 Women): C(9,3) × C(7,3) = 84 × 35 = 2940
(4 Men, 2 Women): C(9,4) × C(7,2) = 126 × 21 = 2646
(5 Men, 1 Women): C(9,5) × C(7,1) = 126 × 7 = 882
(6 Men, 0 Women): C(9,6) × C(7,0) = 84 × 1 = 84

Final Calculation (Sum Rule):
Total ways = (Ways with 3 men) + (Ways with 4 men) + ...
= 2940 + 2646 + 882 + 84
= 6552

Key Principle: Use the Sum Rule (addition) because the cases are mutually exclusive (you cannot simultaneously select exactly $k$ men and exactly $j$ men, where $k
e j$).

Question 16

A committee of 8 members is to be formed from 10 men and 6 women. In how many ways can this be done if the committee must have **at least** 5 men?

Step-by-Step Solution (Sum Rule):

Concept: 'At least' problems require finding the sum of ways for all valid, mutually exclusive cases.

Given:
- Men available: 10
- Women available: 6
- Committee size: 8
- Constraint: At least 5 men

Strategy: We sum the ways for all cases from exactly 5 men up to the maximum possible number of men (8).

Valid Cases (Men, Women) and Calculation:

(5 Men, 3 Women): C(10,5) × C(6,3) = 252 × 20 = 5040
(6 Men, 2 Women): C(10,6) × C(6,2) = 210 × 15 = 3150
(7 Men, 1 Women): C(10,7) × C(6,1) = 120 × 6 = 720
(8 Men, 0 Women): C(10,8) × C(6,0) = 45 × 1 = 45

Final Calculation (Sum Rule):
Total ways = (Ways with 5 men) + (Ways with 6 men) + ...
= 5040 + 3150 + 720 + 45
= 8955

Key Principle: Use the Sum Rule (addition) because the cases are mutually exclusive (you cannot simultaneously select exactly $k$ men and exactly $j$ men, where $k
e j$).

Question 17

A committee of 6 members is to be formed from 9 men and 8 women. In how many ways can this be done if the committee must have **at least** 4 men?

Step-by-Step Solution (Sum Rule):

Concept: 'At least' problems require finding the sum of ways for all valid, mutually exclusive cases.

Given:
- Men available: 9
- Women available: 8
- Committee size: 6
- Constraint: At least 4 men

Strategy: We sum the ways for all cases from exactly 4 men up to the maximum possible number of men (6).

Valid Cases (Men, Women) and Calculation:

(4 Men, 2 Women): C(9,4) × C(8,2) = 126 × 28 = 3528
(5 Men, 1 Women): C(9,5) × C(8,1) = 126 × 8 = 1008
(6 Men, 0 Women): C(9,6) × C(8,0) = 84 × 1 = 84

Final Calculation (Sum Rule):
Total ways = (Ways with 4 men) + (Ways with 5 men) + ...
= 3528 + 1008 + 84
= 4620

Key Principle: Use the Sum Rule (addition) because the cases are mutually exclusive (you cannot simultaneously select exactly $k$ men and exactly $j$ men, where $k
e j$).

Question 18

A committee of 7 members is to be formed from 10 men and 6 women. In how many ways can this be done if the committee must have **at least** 4 men?

Step-by-Step Solution (Sum Rule):

Concept: 'At least' problems require finding the sum of ways for all valid, mutually exclusive cases.

Given:
- Men available: 10
- Women available: 6
- Committee size: 7
- Constraint: At least 4 men

Strategy: We sum the ways for all cases from exactly 4 men up to the maximum possible number of men (7).

Valid Cases (Men, Women) and Calculation:

(4 Men, 3 Women): C(10,4) × C(6,3) = 210 × 20 = 4200
(5 Men, 2 Women): C(10,5) × C(6,2) = 252 × 15 = 3780
(6 Men, 1 Women): C(10,6) × C(6,1) = 210 × 6 = 1260
(7 Men, 0 Women): C(10,7) × C(6,0) = 120 × 1 = 120

Final Calculation (Sum Rule):
Total ways = (Ways with 4 men) + (Ways with 5 men) + ...
= 4200 + 3780 + 1260 + 120
= 9360

Key Principle: Use the Sum Rule (addition) because the cases are mutually exclusive (you cannot simultaneously select exactly $k$ men and exactly $j$ men, where $k
e j$).

Question 19

A committee of 6 members is to be formed from 8 men and 7 women. In how many ways can this be done if the committee must have **at least** 4 men?

Step-by-Step Solution (Sum Rule):

Concept: 'At least' problems require finding the sum of ways for all valid, mutually exclusive cases.

Given:
- Men available: 8
- Women available: 7
- Committee size: 6
- Constraint: At least 4 men

Strategy: We sum the ways for all cases from exactly 4 men up to the maximum possible number of men (6).

Valid Cases (Men, Women) and Calculation:

(4 Men, 2 Women): C(8,4) × C(7,2) = 70 × 21 = 1470
(5 Men, 1 Women): C(8,5) × C(7,1) = 56 × 7 = 392
(6 Men, 0 Women): C(8,6) × C(7,0) = 28 × 1 = 28

Final Calculation (Sum Rule):
Total ways = (Ways with 4 men) + (Ways with 5 men) + ...
= 1470 + 392 + 28
= 1890

Key Principle: Use the Sum Rule (addition) because the cases are mutually exclusive (you cannot simultaneously select exactly $k$ men and exactly $j$ men, where $k
e j$).

Question 20

A committee of 8 members is to be formed from 8 men and 6 women. In how many ways can this be done if the committee must have **at least** 3 men?

Step-by-Step Solution (Sum Rule):

Concept: 'At least' problems require finding the sum of ways for all valid, mutually exclusive cases.

Given:
- Men available: 8
- Women available: 6
- Committee size: 8
- Constraint: At least 3 men

Strategy: We sum the ways for all cases from exactly 3 men up to the maximum possible number of men (8).

Valid Cases (Men, Women) and Calculation:

(3 Men, 5 Women): C(8,3) × C(6,5) = 56 × 6 = 336
(4 Men, 4 Women): C(8,4) × C(6,4) = 70 × 15 = 1050
(5 Men, 3 Women): C(8,5) × C(6,3) = 56 × 20 = 1120
(6 Men, 2 Women): C(8,6) × C(6,2) = 28 × 15 = 420
(7 Men, 1 Women): C(8,7) × C(6,1) = 8 × 6 = 48
(8 Men, 0 Women): C(8,8) × C(6,0) = 1 × 1 = 1

Final Calculation (Sum Rule):
Total ways = (Ways with 3 men) + (Ways with 4 men) + ...
= 336 + 1050 + 1120 + 420 + 48 + 1
= 2975

Key Principle: Use the Sum Rule (addition) because the cases are mutually exclusive (you cannot simultaneously select exactly $k$ men and exactly $j$ men, where $k
e j$).

📝 Continue your Combination with 'At Least' Constraint practice. Worksheet 5 focuses on tricky scenarios handling.

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