ReasoningAbility.com

Combination with 'At Least' Constraint - Expert Level: conceptual clarity Combination with 'At Least' Constraint EXPERT

This skill evaluation ⚡ worksheet focuses on Combination with 'At Least' Constraint - a key topic in Permutation Combination. You'll solve 20 expert-level problems (Worksheet 9 of 10). The primary focus is on conceptual clarity. Master combination with 'at least' constraint ssc cgl, combination with 'at least' constraint reasoning tricks, and fast combination with 'at least' constraint solving through systematic practice.

📝 Worksheet 9 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 9 of 10 (88% complete)

Question 1

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** 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: 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(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 4 men) + (Ways with 5 men) + ...
= 4410 + 2646 + 588 + 36
= 7680

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** 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: 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(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 4 men) + (Ways with 5 men) + ...
= 4410 + 2646 + 588 + 36
= 7680

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 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 4

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 5

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 6

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 7

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 8

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 9

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** 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: 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(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 5 men) + (Ways with 6 men) + ...
= 840 + 168 + 8
= 1016

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 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: 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(7,4) = 120 × 35 = 4200
(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 3 men) + (Ways with 4 men) + ...
= 4200 + 7350 + 5292 + 1470 + 120
= 18432

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 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: 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(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 4 men) + (Ways with 5 men) + ...
= 11760 + 7056 + 1680 + 120
= 20616

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 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 13

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 14

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 15

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** 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: 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(7,4) = 120 × 35 = 4200
(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 3 men) + (Ways with 4 men) + ...
= 4200 + 7350 + 5292 + 1470 + 120
= 18432

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 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 17

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 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** 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 19

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 20

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 $k
e j$).
Previous Worksheet Next Worksheet