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Combination with 'At Least' Constraint: Worksheet 6 - Intermediate-Advanced Practice Combination with 'At Least' Constraint INTERMEDIATE ADVANCED

Ready to master Combination with 'At Least' Constraint? This timed practice ⚡ worksheet (6/10) presents 20 intermediate-advanced-level challenges. Focus area: speed building. Learn to solve combination with 'at least' constraint tricks, handle combination with 'at least' constraint shortcut methods, and perfect combination with 'at least' constraint bank exam questions with our step-by-step solutions.

📝 Worksheet 6 of 10 • 20 questions • ⏱️ Estimated time: 20 minutes • 🎯 Intermediate Advanced level

What you'll learn in this worksheet:
Your progress through Combination with 'At Least' Constraint
Worksheet 6 of 10 (55% complete)

Question 1

A committee of 8 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** 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: 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(9,6) × C(7,2) = 84 × 21 = 1764
(7 Men, 1 Women): C(9,7) × C(7,1) = 36 × 7 = 252
(8 Men, 0 Women): C(9,8) × C(7,0) = 9 × 1 = 9

Final Calculation (Sum Rule):
Total ways = (Ways with 6 men) + (Ways with 7 men) + ...
= 1764 + 252 + 9
= 2025

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 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: 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 $k
e j$).

Question 3

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** 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: 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(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 4 men) + (Ways with 5 men) + ...
= 1400 + 840 + 168 + 8
= 2416

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 9 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: 9
- 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(9,3) × C(8,5) = 84 × 56 = 4704
(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 3 men) + (Ways with 4 men) + ...
= 4704 + 8820 + 7056 + 2352 + 288 + 9
= 23229

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 9 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: 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 $k
e j$).

Question 6

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 7

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** 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: 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(8,3) = 70 × 56 = 3920
(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 4 men) + (Ways with 5 men) + ...
= 3920 + 1568 + 224 + 8
= 5720

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 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 9

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 10

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** 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: 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(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 5 men) + (Ways with 6 men) + ...
= 1176 + 196 + 8
= 1380

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 7 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: 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 $k
e j$).

Question 12

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 13

A committee of 7 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: 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(9,4) × C(8,3) = 126 × 56 = 7056
(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 4 men) + (Ways with 5 men) + ...
= 7056 + 3528 + 672 + 36
= 11292

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 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** 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: 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(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 5 men) + (Ways with 6 men) + ...
= 7056 + 2352 + 288 + 9
= 9705

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** 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 16

A committee of 6 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** 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: 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(10,3) × C(8,3) = 120 × 56 = 6720
(4 Men, 2 Women): C(10,4) × C(8,2) = 210 × 28 = 5880
(5 Men, 1 Women): C(10,5) × C(8,1) = 252 × 8 = 2016
(6 Men, 0 Women): C(10,6) × C(8,0) = 210 × 1 = 210

Final Calculation (Sum Rule):
Total ways = (Ways with 3 men) + (Ways with 4 men) + ...
= 6720 + 5880 + 2016 + 210
= 14826

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 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: 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 $k
e j$).

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** 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: 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(9,3) × C(6,4) = 84 × 15 = 1260
(4 Men, 3 Women): C(9,4) × C(6,3) = 126 × 20 = 2520
(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 3 men) + (Ways with 4 men) + ...
= 1260 + 2520 + 1890 + 504 + 36
= 6210

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 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 20

A committee of 6 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** 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: 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 $k
e j$).
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