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Basic Combination Selection - Intermediate Level: tricky scenarios handling Basic Combination Selection INTERMEDIATE

This expert challenge 📈 worksheet focuses on Basic Combination Selection - a key topic in Permutation Combination. You'll solve 20 intermediate-level problems (Worksheet 5 of 10). The primary focus is on tricky scenarios handling. Master how to solve basic combination selection, basic combination selection tricks, and basic combination selection shortcut methods through systematic practice.

📝 Worksheet 5 of 10 • 20 questions • ⏱️ Estimated time: 20 minutes • 🎯 Intermediate level

What you'll learn in this worksheet:
Your progress through Basic Combination Selection
Worksheet 5 of 10 (44% complete)

Question 1

From 10 colors, in how many ways can we select 4 colors for a design?
Step-by-Step Solution:

Concept: This is a combination problem because the order of selection doesn't matter.

Formula: C(n,r) = n! / [r!(n-r)!]

Given:
- n = 10 (total items)
- r = 4 (items to select)

Calculation:
C(10,4) = 10! / [4! × 6!]
= 10! / [24 × 720]
= 3628800 / [24 × 720]
= 210

Alternative Method (using simplified calculation):
C(10,4) = (10 × 9 × ... × 7) / 4!

Key Distinction:
- Use COMBINATION when order doesn't matter (selecting)
- Use PERMUTATION when order matters (arranging)

Verification: The answer must be less than 10! since we're selecting, not arranging.

Question 2

From a class of 10 students, in how many ways can we select 3 students for a committee?
Step-by-Step Solution:

Concept: This is a combination problem because the order of selection doesn't matter.

Formula: C(n,r) = n! / [r!(n-r)!]

Given:
- n = 10 (total items)
- r = 3 (items to select)

Calculation:
C(10,3) = 10! / [3! × 7!]
= 10! / [6 × 5040]
= 3628800 / [6 × 5040]
= 120

Alternative Method (using simplified calculation):
C(10,3) = (10 × 9 × ... × 8) / 3!

Key Distinction:
- Use COMBINATION when order doesn't matter (selecting)
- Use PERMUTATION when order matters (arranging)

Verification: The answer must be less than 10! since we're selecting, not arranging.

Question 3

From a class of 7 students, in how many ways can we select 2 students for a committee?
Step-by-Step Solution:

Concept: This is a combination problem because the order of selection doesn't matter.

Formula: C(n,r) = n! / [r!(n-r)!]

Given:
- n = 7 (total items)
- r = 2 (items to select)

Calculation:
C(7,2) = 7! / [2! × 5!]
= 7! / [2 × 120]
= 5040 / [2 × 120]
= 21

Alternative Method (using simplified calculation):
C(7,2) = (7 × 6 × ... × 6) / 2!

Key Distinction:
- Use COMBINATION when order doesn't matter (selecting)
- Use PERMUTATION when order matters (arranging)

Verification: The answer must be less than 7! since we're selecting, not arranging.

Question 4

From 6 colors, in how many ways can we select 2 colors for a design?
Step-by-Step Solution:

Concept: This is a combination problem because the order of selection doesn't matter.

Formula: C(n,r) = n! / [r!(n-r)!]

Given:
- n = 6 (total items)
- r = 2 (items to select)

Calculation:
C(6,2) = 6! / [2! × 4!]
= 6! / [2 × 24]
= 720 / [2 × 24]
= 15

Alternative Method (using simplified calculation):
C(6,2) = (6 × 5 × ... × 5) / 2!

Key Distinction:
- Use COMBINATION when order doesn't matter (selecting)
- Use PERMUTATION when order matters (arranging)

Verification: The answer must be less than 6! since we're selecting, not arranging.

Question 5

From a class of 9 students, in how many ways can we select 4 students for a committee?
Step-by-Step Solution:

Concept: This is a combination problem because the order of selection doesn't matter.

Formula: C(n,r) = n! / [r!(n-r)!]

Given:
- n = 9 (total items)
- r = 4 (items to select)

Calculation:
C(9,4) = 9! / [4! × 5!]
= 9! / [24 × 120]
= 362880 / [24 × 120]
= 126

Alternative Method (using simplified calculation):
C(9,4) = (9 × 8 × ... × 6) / 4!

Key Distinction:
- Use COMBINATION when order doesn't matter (selecting)
- Use PERMUTATION when order matters (arranging)

Verification: The answer must be less than 9! since we're selecting, not arranging.

Question 6

From 7 colors, in how many ways can we select 3 colors for a design?
Step-by-Step Solution:

Concept: This is a combination problem because the order of selection doesn't matter.

Formula: C(n,r) = n! / [r!(n-r)!]

Given:
- n = 7 (total items)
- r = 3 (items to select)

Calculation:
C(7,3) = 7! / [3! × 4!]
= 7! / [6 × 24]
= 5040 / [6 × 24]
= 35

Alternative Method (using simplified calculation):
C(7,3) = (7 × 6 × ... × 5) / 3!

Key Distinction:
- Use COMBINATION when order doesn't matter (selecting)
- Use PERMUTATION when order matters (arranging)

Verification: The answer must be less than 7! since we're selecting, not arranging.

Question 7

From 7 colors, in how many ways can we select 3 colors for a design?
Step-by-Step Solution:

Concept: This is a combination problem because the order of selection doesn't matter.

Formula: C(n,r) = n! / [r!(n-r)!]

Given:
- n = 7 (total items)
- r = 3 (items to select)

Calculation:
C(7,3) = 7! / [3! × 4!]
= 7! / [6 × 24]
= 5040 / [6 × 24]
= 35

Alternative Method (using simplified calculation):
C(7,3) = (7 × 6 × ... × 5) / 3!

Key Distinction:
- Use COMBINATION when order doesn't matter (selecting)
- Use PERMUTATION when order matters (arranging)

Verification: The answer must be less than 7! since we're selecting, not arranging.

Question 8

From a class of 7 students, in how many ways can we select 2 students for a committee?
Step-by-Step Solution:

Concept: This is a combination problem because the order of selection doesn't matter.

Formula: C(n,r) = n! / [r!(n-r)!]

Given:
- n = 7 (total items)
- r = 2 (items to select)

Calculation:
C(7,2) = 7! / [2! × 5!]
= 7! / [2 × 120]
= 5040 / [2 × 120]
= 21

Alternative Method (using simplified calculation):
C(7,2) = (7 × 6 × ... × 6) / 2!

Key Distinction:
- Use COMBINATION when order doesn't matter (selecting)
- Use PERMUTATION when order matters (arranging)

Verification: The answer must be less than 7! since we're selecting, not arranging.

Question 9

From 8 colors, in how many ways can we select 2 colors for a design?
Step-by-Step Solution:

Concept: This is a combination problem because the order of selection doesn't matter.

Formula: C(n,r) = n! / [r!(n-r)!]

Given:
- n = 8 (total items)
- r = 2 (items to select)

Calculation:
C(8,2) = 8! / [2! × 6!]
= 8! / [2 × 720]
= 40320 / [2 × 720]
= 28

Alternative Method (using simplified calculation):
C(8,2) = (8 × 7 × ... × 7) / 2!

Key Distinction:
- Use COMBINATION when order doesn't matter (selecting)
- Use PERMUTATION when order matters (arranging)

Verification: The answer must be less than 8! since we're selecting, not arranging.

Question 10

From a class of 9 students, in how many ways can we select 3 students for a committee?
Step-by-Step Solution:

Concept: This is a combination problem because the order of selection doesn't matter.

Formula: C(n,r) = n! / [r!(n-r)!]

Given:
- n = 9 (total items)
- r = 3 (items to select)

Calculation:
C(9,3) = 9! / [3! × 6!]
= 9! / [6 × 720]
= 362880 / [6 × 720]
= 84

Alternative Method (using simplified calculation):
C(9,3) = (9 × 8 × ... × 7) / 3!

Key Distinction:
- Use COMBINATION when order doesn't matter (selecting)
- Use PERMUTATION when order matters (arranging)

Verification: The answer must be less than 9! since we're selecting, not arranging.

Question 11

From 8 colors, in how many ways can we select 3 colors for a design?
Step-by-Step Solution:

Concept: This is a combination problem because the order of selection doesn't matter.

Formula: C(n,r) = n! / [r!(n-r)!]

Given:
- n = 8 (total items)
- r = 3 (items to select)

Calculation:
C(8,3) = 8! / [3! × 5!]
= 8! / [6 × 120]
= 40320 / [6 × 120]
= 56

Alternative Method (using simplified calculation):
C(8,3) = (8 × 7 × ... × 6) / 3!

Key Distinction:
- Use COMBINATION when order doesn't matter (selecting)
- Use PERMUTATION when order matters (arranging)

Verification: The answer must be less than 8! since we're selecting, not arranging.

Question 12

From a group of 8 friends, in how many ways can we choose 4 friends to invite?
Step-by-Step Solution:

Concept: This is a combination problem because the order of selection doesn't matter.

Formula: C(n,r) = n! / [r!(n-r)!]

Given:
- n = 8 (total items)
- r = 4 (items to select)

Calculation:
C(8,4) = 8! / [4! × 4!]
= 8! / [24 × 24]
= 40320 / [24 × 24]
= 70

Alternative Method (using simplified calculation):
C(8,4) = (8 × 7 × ... × 5) / 4!

Key Distinction:
- Use COMBINATION when order doesn't matter (selecting)
- Use PERMUTATION when order matters (arranging)

Verification: The answer must be less than 8! since we're selecting, not arranging.

Question 13

From a class of 6 students, in how many ways can we select 2 students for a committee?
Step-by-Step Solution:

Concept: This is a combination problem because the order of selection doesn't matter.

Formula: C(n,r) = n! / [r!(n-r)!]

Given:
- n = 6 (total items)
- r = 2 (items to select)

Calculation:
C(6,2) = 6! / [2! × 4!]
= 6! / [2 × 24]
= 720 / [2 × 24]
= 15

Alternative Method (using simplified calculation):
C(6,2) = (6 × 5 × ... × 5) / 2!

Key Distinction:
- Use COMBINATION when order doesn't matter (selecting)
- Use PERMUTATION when order matters (arranging)

Verification: The answer must be less than 6! since we're selecting, not arranging.

Question 14

From a group of 7 friends, in how many ways can we choose 2 friends to invite?
Step-by-Step Solution:

Concept: This is a combination problem because the order of selection doesn't matter.

Formula: C(n,r) = n! / [r!(n-r)!]

Given:
- n = 7 (total items)
- r = 2 (items to select)

Calculation:
C(7,2) = 7! / [2! × 5!]
= 7! / [2 × 120]
= 5040 / [2 × 120]
= 21

Alternative Method (using simplified calculation):
C(7,2) = (7 × 6 × ... × 6) / 2!

Key Distinction:
- Use COMBINATION when order doesn't matter (selecting)
- Use PERMUTATION when order matters (arranging)

Verification: The answer must be less than 7! since we're selecting, not arranging.

Question 15

From a class of 7 students, in how many ways can we select 3 students for a committee?
Step-by-Step Solution:

Concept: This is a combination problem because the order of selection doesn't matter.

Formula: C(n,r) = n! / [r!(n-r)!]

Given:
- n = 7 (total items)
- r = 3 (items to select)

Calculation:
C(7,3) = 7! / [3! × 4!]
= 7! / [6 × 24]
= 5040 / [6 × 24]
= 35

Alternative Method (using simplified calculation):
C(7,3) = (7 × 6 × ... × 5) / 3!

Key Distinction:
- Use COMBINATION when order doesn't matter (selecting)
- Use PERMUTATION when order matters (arranging)

Verification: The answer must be less than 7! since we're selecting, not arranging.

Question 16

From a class of 10 students, in how many ways can we select 2 students for a committee?
Step-by-Step Solution:

Concept: This is a combination problem because the order of selection doesn't matter.

Formula: C(n,r) = n! / [r!(n-r)!]

Given:
- n = 10 (total items)
- r = 2 (items to select)

Calculation:
C(10,2) = 10! / [2! × 8!]
= 10! / [2 × 40320]
= 3628800 / [2 × 40320]
= 45

Alternative Method (using simplified calculation):
C(10,2) = (10 × 9 × ... × 9) / 2!

Key Distinction:
- Use COMBINATION when order doesn't matter (selecting)
- Use PERMUTATION when order matters (arranging)

Verification: The answer must be less than 10! since we're selecting, not arranging.

Question 17

From 6 colors, in how many ways can we select 4 colors for a design?
Step-by-Step Solution:

Concept: This is a combination problem because the order of selection doesn't matter.

Formula: C(n,r) = n! / [r!(n-r)!]

Given:
- n = 6 (total items)
- r = 4 (items to select)

Calculation:
C(6,4) = 6! / [4! × 2!]
= 6! / [24 × 2]
= 720 / [24 × 2]
= 15

Alternative Method (using simplified calculation):
C(6,4) = (6 × 5 × ... × 3) / 4!

Key Distinction:
- Use COMBINATION when order doesn't matter (selecting)
- Use PERMUTATION when order matters (arranging)

Verification: The answer must be less than 6! since we're selecting, not arranging.

Question 18

From 8 colors, in how many ways can we select 4 colors for a design?
Step-by-Step Solution:

Concept: This is a combination problem because the order of selection doesn't matter.

Formula: C(n,r) = n! / [r!(n-r)!]

Given:
- n = 8 (total items)
- r = 4 (items to select)

Calculation:
C(8,4) = 8! / [4! × 4!]
= 8! / [24 × 24]
= 40320 / [24 × 24]
= 70

Alternative Method (using simplified calculation):
C(8,4) = (8 × 7 × ... × 5) / 4!

Key Distinction:
- Use COMBINATION when order doesn't matter (selecting)
- Use PERMUTATION when order matters (arranging)

Verification: The answer must be less than 8! since we're selecting, not arranging.

Question 19

From a group of 7 friends, in how many ways can we choose 4 friends to invite?
Step-by-Step Solution:

Concept: This is a combination problem because the order of selection doesn't matter.

Formula: C(n,r) = n! / [r!(n-r)!]

Given:
- n = 7 (total items)
- r = 4 (items to select)

Calculation:
C(7,4) = 7! / [4! × 3!]
= 7! / [24 × 6]
= 5040 / [24 × 6]
= 35

Alternative Method (using simplified calculation):
C(7,4) = (7 × 6 × ... × 4) / 4!

Key Distinction:
- Use COMBINATION when order doesn't matter (selecting)
- Use PERMUTATION when order matters (arranging)

Verification: The answer must be less than 7! since we're selecting, not arranging.

Question 20

From a class of 9 students, in how many ways can we select 2 students for a committee?
Step-by-Step Solution:

Concept: This is a combination problem because the order of selection doesn't matter.

Formula: C(n,r) = n! / [r!(n-r)!]

Given:
- n = 9 (total items)
- r = 2 (items to select)

Calculation:
C(9,2) = 9! / [2! × 7!]
= 9! / [2 × 5040]
= 362880 / [2 × 5040]
= 36

Alternative Method (using simplified calculation):
C(9,2) = (9 × 8 × ... × 8) / 2!

Key Distinction:
- Use COMBINATION when order doesn't matter (selecting)
- Use PERMUTATION when order matters (arranging)

Verification: The answer must be less than 9! since we're selecting, not arranging.
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