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Combination with 'At Least' Constraint - Absolute-Beginner Level: core concept mastery Combination with 'At Least' Constraint ABSOLUTE BEGINNER

This skill primer 🌟 worksheet focuses on Combination with 'At Least' Constraint - a key topic in Permutation Combination. You'll solve 20 absolute-beginner-level problems (Worksheet 1 of 10). The primary focus is on core concept mastery. Master combination with 'at least' constraint problems, combination with 'at least' constraint reasoning questions, and combination with 'at least' constraint practice through systematic practice.

📝 Worksheet 1 of 10 • 20 questions • ⏱️ Estimated time: 20 minutes • 🎯 Absolute Beginner level

What you'll learn in this worksheet:
Your progress through Combination with 'At Least' Constraint
Worksheet 1 of 10 (0% complete)

Question 1

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** 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: 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(8,3) × C(7,3) = 56 × 35 = 1960
(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 3 men) + (Ways with 4 men) + ...
= 1960 + 1470 + 392 + 28
= 3850

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 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 3

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 4

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 5

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** 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: 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(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 4 men) + (Ways with 5 men) + ...
= 2520 + 1890 + 504 + 36
= 4950

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 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 7

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 8

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** 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: 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(7,4) = 126 × 35 = 4410
(5 Men, 3 Women): C(9,5) × C(7,3) = 126 × 35 = 4410
(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 4 men) + (Ways with 5 men) + ...
= 4410 + 4410 + 1764 + 252 + 9
= 10845

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 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: 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(8,4) = 120 × 70 = 8400
(4 Men, 3 Women): C(10,4) × C(8,3) = 210 × 56 = 11760
(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 3 men) + (Ways with 4 men) + ...
= 8400 + 11760 + 7056 + 1680 + 120
= 29016

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 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 11

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 12

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 13

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 $k
e j$).

Question 14

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** 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: 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 $k
e j$).

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 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: 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(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 4 men) + (Ways with 5 men) + ...
= 3150 + 5040 + 3150 + 720 + 45
= 12105

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 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 18

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 $k
e j$).

Question 19

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 20

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 $k
e j$).
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