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Combination with 'At Least' Constraint: Worksheet 10 - Expert Practice Combination with 'At Least' Constraint EXPERT

Ready to master Combination with 'At Least' Constraint? This accuracy focus 👑 worksheet (10/10) presents 20 expert-level challenges. Focus area: application-based learning. Learn to solve combination with 'at least' constraint reasoning tricks, handle fast combination with 'at least' constraint solving, and perfect combination with 'at least' constraint mastery with our step-by-step solutions.

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

What you'll learn in this worksheet:
Your progress through Combination with 'At Least' Constraint
Worksheet 10 of 10 (100% complete)

Question 1

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 $k
e j$).

Question 2

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 3

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 4

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 5

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** 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: 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(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 4 men) + (Ways with 5 men) + ...
= 4410 + 1764 + 210
= 6384

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 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 7

A committee of 7 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** 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: 9
- Women available: 7
- 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(9,5) × C(7,2) = 126 × 21 = 2646
(6 Men, 1 Women): C(9,6) × C(7,1) = 84 × 7 = 588
(7 Men, 0 Women): C(9,7) × C(7,0) = 36 × 1 = 36

Final Calculation (Sum Rule):
Total ways = (Ways with 5 men) + (Ways with 6 men) + ...
= 2646 + 588 + 36
= 3270

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 7 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: 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(8,3) × C(6,4) = 56 × 15 = 840
(4 Men, 3 Women): C(8,4) × C(6,3) = 70 × 20 = 1400
(5 Men, 2 Women): C(8,5) × C(6,2) = 56 × 15 = 840
(6 Men, 1 Women): C(8,6) × C(6,1) = 28 × 6 = 168
(7 Men, 0 Women): C(8,7) × C(6,0) = 8 × 1 = 8

Final Calculation (Sum Rule):
Total ways = (Ways with 3 men) + (Ways with 4 men) + ...
= 840 + 1400 + 840 + 168 + 8
= 3256

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 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** 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: 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(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 4 men) + (Ways with 5 men) + ...
= 4410 + 1764 + 210
= 6384

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 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: 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(8,5) = 56 × 56 = 3136
(4 Men, 4 Women): C(8,4) × C(8,4) = 70 × 70 = 4900
(5 Men, 3 Women): C(8,5) × C(8,3) = 56 × 56 = 3136
(6 Men, 2 Women): C(8,6) × C(8,2) = 28 × 28 = 784
(7 Men, 1 Women): C(8,7) × C(8,1) = 8 × 8 = 64
(8 Men, 0 Women): C(8,8) × C(8,0) = 1 × 1 = 1

Final Calculation (Sum Rule):
Total ways = (Ways with 3 men) + (Ways with 4 men) + ...
= 3136 + 4900 + 3136 + 784 + 64 + 1
= 12021

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 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** 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: 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(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 6 men) + (Ways with 7 men) + ...
= 3150 + 720 + 45
= 3915

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 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 14

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 15

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** 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: 9
- 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(9,5) × C(6,3) = 126 × 20 = 2520
(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 5 men) + (Ways with 6 men) + ...
= 2520 + 1260 + 216 + 9
= 4005

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 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 17

A committee of 7 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** 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: 8
- 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(8,5) × C(8,2) = 56 × 28 = 1568
(6 Men, 1 Women): C(8,6) × C(8,1) = 28 × 8 = 224
(7 Men, 0 Women): C(8,7) × C(8,0) = 8 × 1 = 8

Final Calculation (Sum Rule):
Total ways = (Ways with 5 men) + (Ways with 6 men) + ...
= 1568 + 224 + 8
= 1800

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 9 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: 9
- 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(9,5) × C(6,2) = 126 × 15 = 1890
(6 Men, 1 Women): C(9,6) × C(6,1) = 84 × 6 = 504
(7 Men, 0 Women): C(9,7) × C(6,0) = 36 × 1 = 36

Final Calculation (Sum Rule):
Total ways = (Ways with 5 men) + (Ways with 6 men) + ...
= 1890 + 504 + 36
= 2430

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 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 20

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** 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: 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(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 4 men) + (Ways with 5 men) + ...
= 1960 + 448 + 28
= 2436

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 $k
e j$).
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