Partition into Unequal Groups - Absolute-Beginner Level: core concept mastery Partition into Unequal Groups ABSOLUTE BEGINNER

This skill primer 🌟 worksheet focuses on Partition into Unequal Groups - 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 partition into unequal groups problems, partition into unequal groups reasoning questions, and partition into unequal groups 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 Partition into Unequal Groups
Worksheet 1 of 10 (0% complete)

Question 1

In how many ways can 10 distinct people be divided into 3 groups of sizes 7, 2, 1 (groups are unlabeled but have different sizes)?
Step-by-Step Solution:

Concept: Partitioning into groups of specified sizes. When group sizes are different, groups are automatically distinguishable by size.

Given:
- Total people: 10
- Group sizes: 7, 2, 1
- Groups are unlabeled (no names like Team A, Team B)

Formula:
$$\frac{n!}{n_1! \cdot n_2! \cdot ... \cdot n_k!}$$

Step 1 - Choose first group:
Choose 7 people from 10: C(10, 7) = 120

Step 2 - Choose second group:
From remaining 3 people, choose 2: C(3, 2) = 3

Continue for all groups:
C(10,7) = 120
C(3,2) = 3
C(1,1) = 1 (last group)

Step 3 - Multiply:
Total ways = 120 × 3 × 1 (last group)
= 360

Simplified formula:
= 10! / (7! × 2! × 1!)
= 3628800 / (5040 × 2 × 1)
= 360

Key Insight: Since groups have different sizes, we don't divide by k! (they're naturally distinguishable by their sizes).

Contrast with equal groups:
- If groups were same size: would divide by k!
- Here, sizes differ: no division needed

Verification: Sum of group sizes = 10 = 10 ✓

Question 2

In how many ways can 11 distinct people be divided into 4 groups of sizes 5, 3, 2, 1 (groups are unlabeled but have different sizes)?
Step-by-Step Solution:

Concept: Partitioning into groups of specified sizes. When group sizes are different, groups are automatically distinguishable by size.

Given:
- Total people: 11
- Group sizes: 5, 3, 2, 1
- Groups are unlabeled (no names like Team A, Team B)

Formula:
$$\frac{n!}{n_1! \cdot n_2! \cdot ... \cdot n_k!}$$

Step 1 - Choose first group:
Choose 5 people from 11: C(11, 5) = 462

Step 2 - Choose second group:
From remaining 6 people, choose 3: C(6, 3) = 20

Continue for all groups:
C(11,5) = 462
C(6,3) = 20
C(3,2) = 3
C(1,1) = 1 (last group)

Step 3 - Multiply:
Total ways = 462 × 20 × 3 × 1 (last group)
= 27720

Simplified formula:
= 11! / (5! × 3! × 2! × 1!)
= 39916800 / (120 × 6 × 2 × 1)
= 27720

Key Insight: Since groups have different sizes, we don't divide by k! (they're naturally distinguishable by their sizes).

Contrast with equal groups:
- If groups were same size: would divide by k!
- Here, sizes differ: no division needed

Verification: Sum of group sizes = 11 = 11 ✓

Question 3

In how many ways can 10 distinct people be divided into 3 groups of sizes 5, 4, 1 (groups are unlabeled but have different sizes)?
Step-by-Step Solution:

Concept: Partitioning into groups of specified sizes. When group sizes are different, groups are automatically distinguishable by size.

Given:
- Total people: 10
- Group sizes: 5, 4, 1
- Groups are unlabeled (no names like Team A, Team B)

Formula:
$$\frac{n!}{n_1! \cdot n_2! \cdot ... \cdot n_k!}$$

Step 1 - Choose first group:
Choose 5 people from 10: C(10, 5) = 252

Step 2 - Choose second group:
From remaining 5 people, choose 4: C(5, 4) = 5

Continue for all groups:
C(10,5) = 252
C(5,4) = 5
C(1,1) = 1 (last group)

Step 3 - Multiply:
Total ways = 252 × 5 × 1 (last group)
= 1260

Simplified formula:
= 10! / (5! × 4! × 1!)
= 3628800 / (120 × 24 × 1)
= 1260

Key Insight: Since groups have different sizes, we don't divide by k! (they're naturally distinguishable by their sizes).

Contrast with equal groups:
- If groups were same size: would divide by k!
- Here, sizes differ: no division needed

Verification: Sum of group sizes = 10 = 10 ✓

Question 4

In how many ways can 12 distinct people be divided into 3 groups of sizes 6, 4, 2 (groups are unlabeled but have different sizes)?
Step-by-Step Solution:

Concept: Partitioning into groups of specified sizes. When group sizes are different, groups are automatically distinguishable by size.

Given:
- Total people: 12
- Group sizes: 6, 4, 2
- Groups are unlabeled (no names like Team A, Team B)

Formula:
$$\frac{n!}{n_1! \cdot n_2! \cdot ... \cdot n_k!}$$

Step 1 - Choose first group:
Choose 6 people from 12: C(12, 6) = 924

Step 2 - Choose second group:
From remaining 6 people, choose 4: C(6, 4) = 15

Continue for all groups:
C(12,6) = 924
C(6,4) = 15
C(2,2) = 1 (last group)

Step 3 - Multiply:
Total ways = 924 × 15 × 1 (last group)
= 13860

Simplified formula:
= 12! / (6! × 4! × 2!)
= 479001600 / (720 × 24 × 2)
= 13860

Key Insight: Since groups have different sizes, we don't divide by k! (they're naturally distinguishable by their sizes).

Contrast with equal groups:
- If groups were same size: would divide by k!
- Here, sizes differ: no division needed

Verification: Sum of group sizes = 12 = 12 ✓

Question 5

In how many ways can 10 distinct people be divided into 4 groups of sizes 5, 4, 1 (groups are unlabeled but have different sizes)?
Step-by-Step Solution:

Concept: Partitioning into groups of specified sizes. When group sizes are different, groups are automatically distinguishable by size.

Given:
- Total people: 10
- Group sizes: 5, 4, 1
- Groups are unlabeled (no names like Team A, Team B)

Formula:
$$\frac{n!}{n_1! \cdot n_2! \cdot ... \cdot n_k!}$$

Step 1 - Choose first group:
Choose 5 people from 10: C(10, 5) = 252

Step 2 - Choose second group:
From remaining 5 people, choose 4: C(5, 4) = 5

Continue for all groups:
C(10,5) = 252
C(5,4) = 5
C(1,1) = 1 (last group)

Step 3 - Multiply:
Total ways = 252 × 5 × 1 (last group)
= 1260

Simplified formula:
= 10! / (5! × 4! × 1!)
= 3628800 / (120 × 24 × 1)
= 1260

Key Insight: Since groups have different sizes, we don't divide by k! (they're naturally distinguishable by their sizes).

Contrast with equal groups:
- If groups were same size: would divide by k!
- Here, sizes differ: no division needed

Verification: Sum of group sizes = 10 = 10 ✓

Question 6

In how many ways can 11 distinct people be divided into 3 groups of sizes 5, 4, 2 (groups are unlabeled but have different sizes)?
Step-by-Step Solution:

Concept: Partitioning into groups of specified sizes. When group sizes are different, groups are automatically distinguishable by size.

Given:
- Total people: 11
- Group sizes: 5, 4, 2
- Groups are unlabeled (no names like Team A, Team B)

Formula:
$$\frac{n!}{n_1! \cdot n_2! \cdot ... \cdot n_k!}$$

Step 1 - Choose first group:
Choose 5 people from 11: C(11, 5) = 462

Step 2 - Choose second group:
From remaining 6 people, choose 4: C(6, 4) = 15

Continue for all groups:
C(11,5) = 462
C(6,4) = 15
C(2,2) = 1 (last group)

Step 3 - Multiply:
Total ways = 462 × 15 × 1 (last group)
= 6930

Simplified formula:
= 11! / (5! × 4! × 2!)
= 39916800 / (120 × 24 × 2)
= 6930

Key Insight: Since groups have different sizes, we don't divide by k! (they're naturally distinguishable by their sizes).

Contrast with equal groups:
- If groups were same size: would divide by k!
- Here, sizes differ: no division needed

Verification: Sum of group sizes = 11 = 11 ✓

Question 7

In how many ways can 11 distinct people be divided into 4 groups of sizes 8, 2, 1 (groups are unlabeled but have different sizes)?
Step-by-Step Solution:

Concept: Partitioning into groups of specified sizes. When group sizes are different, groups are automatically distinguishable by size.

Given:
- Total people: 11
- Group sizes: 8, 2, 1
- Groups are unlabeled (no names like Team A, Team B)

Formula:
$$\frac{n!}{n_1! \cdot n_2! \cdot ... \cdot n_k!}$$

Step 1 - Choose first group:
Choose 8 people from 11: C(11, 8) = 165

Step 2 - Choose second group:
From remaining 3 people, choose 2: C(3, 2) = 3

Continue for all groups:
C(11,8) = 165
C(3,2) = 3
C(1,1) = 1 (last group)

Step 3 - Multiply:
Total ways = 165 × 3 × 1 (last group)
= 495

Simplified formula:
= 11! / (8! × 2! × 1!)
= 39916800 / (40320 × 2 × 1)
= 495

Key Insight: Since groups have different sizes, we don't divide by k! (they're naturally distinguishable by their sizes).

Contrast with equal groups:
- If groups were same size: would divide by k!
- Here, sizes differ: no division needed

Verification: Sum of group sizes = 11 = 11 ✓

Question 8

In how many ways can 9 distinct people be divided into 4 groups of sizes 6, 2, 1 (groups are unlabeled but have different sizes)?
Step-by-Step Solution:

Concept: Partitioning into groups of specified sizes. When group sizes are different, groups are automatically distinguishable by size.

Given:
- Total people: 9
- Group sizes: 6, 2, 1
- Groups are unlabeled (no names like Team A, Team B)

Formula:
$$\frac{n!}{n_1! \cdot n_2! \cdot ... \cdot n_k!}$$

Step 1 - Choose first group:
Choose 6 people from 9: C(9, 6) = 84

Step 2 - Choose second group:
From remaining 3 people, choose 2: C(3, 2) = 3

Continue for all groups:
C(9,6) = 84
C(3,2) = 3
C(1,1) = 1 (last group)

Step 3 - Multiply:
Total ways = 84 × 3 × 1 (last group)
= 252

Simplified formula:
= 9! / (6! × 2! × 1!)
= 362880 / (720 × 2 × 1)
= 252

Key Insight: Since groups have different sizes, we don't divide by k! (they're naturally distinguishable by their sizes).

Contrast with equal groups:
- If groups were same size: would divide by k!
- Here, sizes differ: no division needed

Verification: Sum of group sizes = 9 = 9 ✓

Question 9

In how many ways can 8 distinct people be divided into 4 groups of sizes 5, 2, 1 (groups are unlabeled but have different sizes)?
Step-by-Step Solution:

Concept: Partitioning into groups of specified sizes. When group sizes are different, groups are automatically distinguishable by size.

Given:
- Total people: 8
- Group sizes: 5, 2, 1
- Groups are unlabeled (no names like Team A, Team B)

Formula:
$$\frac{n!}{n_1! \cdot n_2! \cdot ... \cdot n_k!}$$

Step 1 - Choose first group:
Choose 5 people from 8: C(8, 5) = 56

Step 2 - Choose second group:
From remaining 3 people, choose 2: C(3, 2) = 3

Continue for all groups:
C(8,5) = 56
C(3,2) = 3
C(1,1) = 1 (last group)

Step 3 - Multiply:
Total ways = 56 × 3 × 1 (last group)
= 168

Simplified formula:
= 8! / (5! × 2! × 1!)
= 40320 / (120 × 2 × 1)
= 168

Key Insight: Since groups have different sizes, we don't divide by k! (they're naturally distinguishable by their sizes).

Contrast with equal groups:
- If groups were same size: would divide by k!
- Here, sizes differ: no division needed

Verification: Sum of group sizes = 8 = 8 ✓

Question 10

In how many ways can 12 distinct people be divided into 3 groups of sizes 6, 4, 2 (groups are unlabeled but have different sizes)?
Step-by-Step Solution:

Concept: Partitioning into groups of specified sizes. When group sizes are different, groups are automatically distinguishable by size.

Given:
- Total people: 12
- Group sizes: 6, 4, 2
- Groups are unlabeled (no names like Team A, Team B)

Formula:
$$\frac{n!}{n_1! \cdot n_2! \cdot ... \cdot n_k!}$$

Step 1 - Choose first group:
Choose 6 people from 12: C(12, 6) = 924

Step 2 - Choose second group:
From remaining 6 people, choose 4: C(6, 4) = 15

Continue for all groups:
C(12,6) = 924
C(6,4) = 15
C(2,2) = 1 (last group)

Step 3 - Multiply:
Total ways = 924 × 15 × 1 (last group)
= 13860

Simplified formula:
= 12! / (6! × 4! × 2!)
= 479001600 / (720 × 24 × 2)
= 13860

Key Insight: Since groups have different sizes, we don't divide by k! (they're naturally distinguishable by their sizes).

Contrast with equal groups:
- If groups were same size: would divide by k!
- Here, sizes differ: no division needed

Verification: Sum of group sizes = 12 = 12 ✓

Question 11

In how many ways can 10 distinct people be divided into 4 groups of sizes 4, 3, 2, 1 (groups are unlabeled but have different sizes)?
Step-by-Step Solution:

Concept: Partitioning into groups of specified sizes. When group sizes are different, groups are automatically distinguishable by size.

Given:
- Total people: 10
- Group sizes: 4, 3, 2, 1
- Groups are unlabeled (no names like Team A, Team B)

Formula:
$$\frac{n!}{n_1! \cdot n_2! \cdot ... \cdot n_k!}$$

Step 1 - Choose first group:
Choose 4 people from 10: C(10, 4) = 210

Step 2 - Choose second group:
From remaining 6 people, choose 3: C(6, 3) = 20

Continue for all groups:
C(10,4) = 210
C(6,3) = 20
C(3,2) = 3
C(1,1) = 1 (last group)

Step 3 - Multiply:
Total ways = 210 × 20 × 3 × 1 (last group)
= 12600

Simplified formula:
= 10! / (4! × 3! × 2! × 1!)
= 3628800 / (24 × 6 × 2 × 1)
= 12600

Key Insight: Since groups have different sizes, we don't divide by k! (they're naturally distinguishable by their sizes).

Contrast with equal groups:
- If groups were same size: would divide by k!
- Here, sizes differ: no division needed

Verification: Sum of group sizes = 10 = 10 ✓

Question 12

In how many ways can 10 distinct people be divided into 4 groups of sizes 5, 4, 1 (groups are unlabeled but have different sizes)?
Step-by-Step Solution:

Concept: Partitioning into groups of specified sizes. When group sizes are different, groups are automatically distinguishable by size.

Given:
- Total people: 10
- Group sizes: 5, 4, 1
- Groups are unlabeled (no names like Team A, Team B)

Formula:
$$\frac{n!}{n_1! \cdot n_2! \cdot ... \cdot n_k!}$$

Step 1 - Choose first group:
Choose 5 people from 10: C(10, 5) = 252

Step 2 - Choose second group:
From remaining 5 people, choose 4: C(5, 4) = 5

Continue for all groups:
C(10,5) = 252
C(5,4) = 5
C(1,1) = 1 (last group)

Step 3 - Multiply:
Total ways = 252 × 5 × 1 (last group)
= 1260

Simplified formula:
= 10! / (5! × 4! × 1!)
= 3628800 / (120 × 24 × 1)
= 1260

Key Insight: Since groups have different sizes, we don't divide by k! (they're naturally distinguishable by their sizes).

Contrast with equal groups:
- If groups were same size: would divide by k!
- Here, sizes differ: no division needed

Verification: Sum of group sizes = 10 = 10 ✓

Question 13

In how many ways can 11 distinct people be divided into 3 groups of sizes 5, 4, 2 (groups are unlabeled but have different sizes)?
Step-by-Step Solution:

Concept: Partitioning into groups of specified sizes. When group sizes are different, groups are automatically distinguishable by size.

Given:
- Total people: 11
- Group sizes: 5, 4, 2
- Groups are unlabeled (no names like Team A, Team B)

Formula:
$$\frac{n!}{n_1! \cdot n_2! \cdot ... \cdot n_k!}$$

Step 1 - Choose first group:
Choose 5 people from 11: C(11, 5) = 462

Step 2 - Choose second group:
From remaining 6 people, choose 4: C(6, 4) = 15

Continue for all groups:
C(11,5) = 462
C(6,4) = 15
C(2,2) = 1 (last group)

Step 3 - Multiply:
Total ways = 462 × 15 × 1 (last group)
= 6930

Simplified formula:
= 11! / (5! × 4! × 2!)
= 39916800 / (120 × 24 × 2)
= 6930

Key Insight: Since groups have different sizes, we don't divide by k! (they're naturally distinguishable by their sizes).

Contrast with equal groups:
- If groups were same size: would divide by k!
- Here, sizes differ: no division needed

Verification: Sum of group sizes = 11 = 11 ✓

Question 14

In how many ways can 7 distinct people be divided into 3 groups of sizes 4, 2, 1 (groups are unlabeled but have different sizes)?
Step-by-Step Solution:

Concept: Partitioning into groups of specified sizes. When group sizes are different, groups are automatically distinguishable by size.

Given:
- Total people: 7
- Group sizes: 4, 2, 1
- Groups are unlabeled (no names like Team A, Team B)

Formula:
$$\frac{n!}{n_1! \cdot n_2! \cdot ... \cdot n_k!}$$

Step 1 - Choose first group:
Choose 4 people from 7: C(7, 4) = 35

Step 2 - Choose second group:
From remaining 3 people, choose 2: C(3, 2) = 3

Continue for all groups:
C(7,4) = 35
C(3,2) = 3
C(1,1) = 1 (last group)

Step 3 - Multiply:
Total ways = 35 × 3 × 1 (last group)
= 105

Simplified formula:
= 7! / (4! × 2! × 1!)
= 5040 / (24 × 2 × 1)
= 105

Key Insight: Since groups have different sizes, we don't divide by k! (they're naturally distinguishable by their sizes).

Contrast with equal groups:
- If groups were same size: would divide by k!
- Here, sizes differ: no division needed

Verification: Sum of group sizes = 7 = 7 ✓

Question 15

In how many ways can 12 distinct people be divided into 3 groups of sizes 7, 3, 2 (groups are unlabeled but have different sizes)?
Step-by-Step Solution:

Concept: Partitioning into groups of specified sizes. When group sizes are different, groups are automatically distinguishable by size.

Given:
- Total people: 12
- Group sizes: 7, 3, 2
- Groups are unlabeled (no names like Team A, Team B)

Formula:
$$\frac{n!}{n_1! \cdot n_2! \cdot ... \cdot n_k!}$$

Step 1 - Choose first group:
Choose 7 people from 12: C(12, 7) = 792

Step 2 - Choose second group:
From remaining 5 people, choose 3: C(5, 3) = 10

Continue for all groups:
C(12,7) = 792
C(5,3) = 10
C(2,2) = 1 (last group)

Step 3 - Multiply:
Total ways = 792 × 10 × 1 (last group)
= 7920

Simplified formula:
= 12! / (7! × 3! × 2!)
= 479001600 / (5040 × 6 × 2)
= 7920

Key Insight: Since groups have different sizes, we don't divide by k! (they're naturally distinguishable by their sizes).

Contrast with equal groups:
- If groups were same size: would divide by k!
- Here, sizes differ: no division needed

Verification: Sum of group sizes = 12 = 12 ✓

Question 16

In how many ways can 10 distinct people be divided into 3 groups of sizes 5, 4, 1 (groups are unlabeled but have different sizes)?
Step-by-Step Solution:

Concept: Partitioning into groups of specified sizes. When group sizes are different, groups are automatically distinguishable by size.

Given:
- Total people: 10
- Group sizes: 5, 4, 1
- Groups are unlabeled (no names like Team A, Team B)

Formula:
$$\frac{n!}{n_1! \cdot n_2! \cdot ... \cdot n_k!}$$

Step 1 - Choose first group:
Choose 5 people from 10: C(10, 5) = 252

Step 2 - Choose second group:
From remaining 5 people, choose 4: C(5, 4) = 5

Continue for all groups:
C(10,5) = 252
C(5,4) = 5
C(1,1) = 1 (last group)

Step 3 - Multiply:
Total ways = 252 × 5 × 1 (last group)
= 1260

Simplified formula:
= 10! / (5! × 4! × 1!)
= 3628800 / (120 × 24 × 1)
= 1260

Key Insight: Since groups have different sizes, we don't divide by k! (they're naturally distinguishable by their sizes).

Contrast with equal groups:
- If groups were same size: would divide by k!
- Here, sizes differ: no division needed

Verification: Sum of group sizes = 10 = 10 ✓

Question 17

In how many ways can 11 distinct people be divided into 4 groups of sizes 7, 3, 1 (groups are unlabeled but have different sizes)?
Step-by-Step Solution:

Concept: Partitioning into groups of specified sizes. When group sizes are different, groups are automatically distinguishable by size.

Given:
- Total people: 11
- Group sizes: 7, 3, 1
- Groups are unlabeled (no names like Team A, Team B)

Formula:
$$\frac{n!}{n_1! \cdot n_2! \cdot ... \cdot n_k!}$$

Step 1 - Choose first group:
Choose 7 people from 11: C(11, 7) = 330

Step 2 - Choose second group:
From remaining 4 people, choose 3: C(4, 3) = 4

Continue for all groups:
C(11,7) = 330
C(4,3) = 4
C(1,1) = 1 (last group)

Step 3 - Multiply:
Total ways = 330 × 4 × 1 (last group)
= 1320

Simplified formula:
= 11! / (7! × 3! × 1!)
= 39916800 / (5040 × 6 × 1)
= 1320

Key Insight: Since groups have different sizes, we don't divide by k! (they're naturally distinguishable by their sizes).

Contrast with equal groups:
- If groups were same size: would divide by k!
- Here, sizes differ: no division needed

Verification: Sum of group sizes = 11 = 11 ✓

Question 18

In how many ways can 11 distinct people be divided into 3 groups of sizes 6, 3, 2 (groups are unlabeled but have different sizes)?
Step-by-Step Solution:

Concept: Partitioning into groups of specified sizes. When group sizes are different, groups are automatically distinguishable by size.

Given:
- Total people: 11
- Group sizes: 6, 3, 2
- Groups are unlabeled (no names like Team A, Team B)

Formula:
$$\frac{n!}{n_1! \cdot n_2! \cdot ... \cdot n_k!}$$

Step 1 - Choose first group:
Choose 6 people from 11: C(11, 6) = 462

Step 2 - Choose second group:
From remaining 5 people, choose 3: C(5, 3) = 10

Continue for all groups:
C(11,6) = 462
C(5,3) = 10
C(2,2) = 1 (last group)

Step 3 - Multiply:
Total ways = 462 × 10 × 1 (last group)
= 4620

Simplified formula:
= 11! / (6! × 3! × 2!)
= 39916800 / (720 × 6 × 2)
= 4620

Key Insight: Since groups have different sizes, we don't divide by k! (they're naturally distinguishable by their sizes).

Contrast with equal groups:
- If groups were same size: would divide by k!
- Here, sizes differ: no division needed

Verification: Sum of group sizes = 11 = 11 ✓

Question 19

In how many ways can 10 distinct people be divided into 4 groups of sizes 4, 3, 2, 1 (groups are unlabeled but have different sizes)?
Step-by-Step Solution:

Concept: Partitioning into groups of specified sizes. When group sizes are different, groups are automatically distinguishable by size.

Given:
- Total people: 10
- Group sizes: 4, 3, 2, 1
- Groups are unlabeled (no names like Team A, Team B)

Formula:
$$\frac{n!}{n_1! \cdot n_2! \cdot ... \cdot n_k!}$$

Step 1 - Choose first group:
Choose 4 people from 10: C(10, 4) = 210

Step 2 - Choose second group:
From remaining 6 people, choose 3: C(6, 3) = 20

Continue for all groups:
C(10,4) = 210
C(6,3) = 20
C(3,2) = 3
C(1,1) = 1 (last group)

Step 3 - Multiply:
Total ways = 210 × 20 × 3 × 1 (last group)
= 12600

Simplified formula:
= 10! / (4! × 3! × 2! × 1!)
= 3628800 / (24 × 6 × 2 × 1)
= 12600

Key Insight: Since groups have different sizes, we don't divide by k! (they're naturally distinguishable by their sizes).

Contrast with equal groups:
- If groups were same size: would divide by k!
- Here, sizes differ: no division needed

Verification: Sum of group sizes = 10 = 10 ✓

Question 20

In how many ways can 11 distinct people be divided into 3 groups of sizes 6, 3, 2 (groups are unlabeled but have different sizes)?
Step-by-Step Solution:

Concept: Partitioning into groups of specified sizes. When group sizes are different, groups are automatically distinguishable by size.

Given:
- Total people: 11
- Group sizes: 6, 3, 2
- Groups are unlabeled (no names like Team A, Team B)

Formula:
$$\frac{n!}{n_1! \cdot n_2! \cdot ... \cdot n_k!}$$

Step 1 - Choose first group:
Choose 6 people from 11: C(11, 6) = 462

Step 2 - Choose second group:
From remaining 5 people, choose 3: C(5, 3) = 10

Continue for all groups:
C(11,6) = 462
C(5,3) = 10
C(2,2) = 1 (last group)

Step 3 - Multiply:
Total ways = 462 × 10 × 1 (last group)
= 4620

Simplified formula:
= 11! / (6! × 3! × 2!)
= 39916800 / (720 × 6 × 2)
= 4620

Key Insight: Since groups have different sizes, we don't divide by k! (they're naturally distinguishable by their sizes).

Contrast with equal groups:
- If groups were same size: would divide by k!
- Here, sizes differ: no division needed

Verification: Sum of group sizes = 11 = 11 ✓
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