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Combination with 'At Least' Constraint Advanced Worksheet: Focus on exam-oriented approach Combination with 'At Least' Constraint ADVANCED

Level up your Combination with 'At Least' Constraint skills! You're at Worksheet 8 of 10 (77% through this series). This exam hall simulation worksheet features 20 advanced-level problems with a focus on exam-oriented approach. Topics covered: combination with 'at least' constraint bank exam questions, combination with 'at least' constraint ssc cgl, combination with 'at least' constraint reasoning tricks.

📝 Worksheet 8 of 10 • 20 questions • ⏱️ Estimated time: 20 minutes • 🎯 Advanced level

What you'll learn in this worksheet:
Your progress through Combination with 'At Least' Constraint
Worksheet 8 of 10 (77% 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 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** 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: 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(8,4) = 56 × 70 = 3920
(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 3 men) + (Ways with 4 men) + ...
= 3920 + 3920 + 1568 + 224 + 8
= 9640

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 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 4

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 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 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 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** 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: 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(7,4) = 84 × 35 = 2940
(4 Men, 3 Women): C(9,4) × C(7,3) = 126 × 35 = 4410
(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 3 men) + (Ways with 4 men) + ...
= 2940 + 4410 + 2646 + 588 + 36
= 10620

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 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: 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(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 4 men) + (Ways with 5 men) + ...
= 1050 + 336 + 28
= 1414

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 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: 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(8,4) = 56 × 70 = 3920
(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 3 men) + (Ways with 4 men) + ...
= 3920 + 3920 + 1568 + 224 + 8
= 9640

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 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** 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: 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(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 4 men) + (Ways with 5 men) + ...
= 5880 + 2016 + 210
= 8106

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 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** 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: 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(8,4) = 84 × 70 = 5880
(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 3 men) + (Ways with 4 men) + ...
= 5880 + 7056 + 3528 + 672 + 36
= 17172

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 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 13

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 14

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 15

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 16

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 17

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 18

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 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** 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 20

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** 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: 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(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 6 men) + (Ways with 7 men) + ...
= 2352 + 288 + 9
= 2649

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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