Combination with 'At Least' Constraint: Worksheet 2 - Beginner Practice Combination with 'At Least' Constraint BEGINNER

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: 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(6,2) = 126 × 15 = 1890
(5 Men, 1 Women): C(9,5) × C(6,1) = 126 × 6 = 756
(6 Men, 0 Women): C(9,6) × C(6,0) = 84 × 1 = 84

Final Calculation (Sum Rule):
Total ways = (Ways with 4 men) + (Ways with 5 men) + ...
= 1890 + 756 + 84
= 2730

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 $k
e j$).

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 $k
e j$).

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 $k
e j$).

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: 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(7,3) = 210 × 35 = 7350
(5 Men, 2 Women): C(10,5) × C(7,2) = 252 × 21 = 5292
(6 Men, 1 Women): C(10,6) × C(7,1) = 210 × 7 = 1470
(7 Men, 0 Women): C(10,7) × C(7,0) = 120 × 1 = 120

Final Calculation (Sum Rule):
Total ways = (Ways with 4 men) + (Ways with 5 men) + ...
= 7350 + 5292 + 1470 + 120
= 14232

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 $k
e j$).

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: 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(6,3) = 84 × 20 = 1680
(4 Men, 2 Women): C(9,4) × C(6,2) = 126 × 15 = 1890
(5 Men, 1 Women): C(9,5) × C(6,1) = 126 × 6 = 756
(6 Men, 0 Women): C(9,6) × C(6,0) = 84 × 1 = 84

Final Calculation (Sum Rule):
Total ways = (Ways with 3 men) + (Ways with 4 men) + ...
= 1680 + 1890 + 756 + 84
= 4410

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 $k
e j$).

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 $k
e j$).

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: 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(8,2) = 126 × 28 = 3528
(6 Men, 1 Women): C(9,6) × C(8,1) = 84 × 8 = 672
(7 Men, 0 Women): C(9,7) × C(8,0) = 36 × 1 = 36

Final Calculation (Sum Rule):
Total ways = (Ways with 5 men) + (Ways with 6 men) + ...
= 3528 + 672 + 36
= 4236

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 $k
e j$).

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 $k
e j$).

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 $k
e j$).

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 $k
e j$).

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 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(10,3) × C(6,5) = 120 × 6 = 720
(4 Men, 4 Women): C(10,4) × C(6,4) = 210 × 15 = 3150
(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 3 men) + (Ways with 4 men) + ...
= 720 + 3150 + 5040 + 3150 + 720 + 45
= 12825

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 $k
e j$).

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: 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(7,3) = 252 × 35 = 8820
(6 Men, 2 Women): C(10,6) × C(7,2) = 210 × 21 = 4410
(7 Men, 1 Women): C(10,7) × C(7,1) = 120 × 7 = 840
(8 Men, 0 Women): C(10,8) × C(7,0) = 45 × 1 = 45

Final Calculation (Sum Rule):
Total ways = (Ways with 5 men) + (Ways with 6 men) + ...
= 8820 + 4410 + 840 + 45
= 14115

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 $k
e j$).

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 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(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 5 men) + (Ways with 6 men) + ...
= 3136 + 784 + 64 + 1
= 3985

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 $k
e j$).

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: 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(10,3) × C(7,5) = 120 × 21 = 2520
(4 Men, 4 Women): C(10,4) × C(7,4) = 210 × 35 = 7350
(5 Men, 3 Women): C(10,5) × C(7,3) = 252 × 35 = 8820
(6 Men, 2 Women): C(10,6) × C(7,2) = 210 × 21 = 4410
(7 Men, 1 Women): C(10,7) × C(7,1) = 120 × 7 = 840
(8 Men, 0 Women): C(10,8) × C(7,0) = 45 × 1 = 45

Final Calculation (Sum Rule):
Total ways = (Ways with 3 men) + (Ways with 4 men) + ...
= 2520 + 7350 + 8820 + 4410 + 840 + 45
= 23985

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 $k
e j$).

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: 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(7,2) = 210 × 21 = 4410
(7 Men, 1 Women): C(10,7) × C(7,1) = 120 × 7 = 840
(8 Men, 0 Women): C(10,8) × C(7,0) = 45 × 1 = 45

Final Calculation (Sum Rule):
Total ways = (Ways with 6 men) + (Ways with 7 men) + ...
= 4410 + 840 + 45
= 5295

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 $k
e j$).

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 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(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 5 men) + (Ways with 6 men) + ...
= 3136 + 784 + 64 + 1
= 3985

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 $k
e j$).

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: 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(6,3) = 84 × 20 = 1680
(4 Men, 2 Women): C(9,4) × C(6,2) = 126 × 15 = 1890
(5 Men, 1 Women): C(9,5) × C(6,1) = 126 × 6 = 756
(6 Men, 0 Women): C(9,6) × C(6,0) = 84 × 1 = 84

Final Calculation (Sum Rule):
Total ways = (Ways with 3 men) + (Ways with 4 men) + ...
= 1680 + 1890 + 756 + 84
= 4410

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 $k
e j$).

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 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(8,3) = 252 × 56 = 14112
(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 5 men) + (Ways with 6 men) + ...
= 14112 + 5880 + 960 + 45
= 20997

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 $k
e j$).

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 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(8,3) = 84 × 56 = 4704
(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 3 men) + (Ways with 4 men) + ...
= 4704 + 3528 + 1008 + 84
= 9324

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 $k
e j$).

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: 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(7,2) = 252 × 21 = 5292
(6 Men, 1 Women): C(10,6) × C(7,1) = 210 × 7 = 1470
(7 Men, 0 Women): C(10,7) × C(7,0) = 120 × 1 = 120

Final Calculation (Sum Rule):
Total ways = (Ways with 5 men) + (Ways with 6 men) + ...
= 5292 + 1470 + 120
= 6882

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 $k
e j$).