A certain company has the following policy regarding membership in committees A, B, C: each of the committees have exactly...

Let's visualize this problem to make it crystal clear...

Phase 1: Owning the Dataset

Visualization Selection

This is a set membership problem, so I'll use a Venn diagram approach with careful accounting.

Total Committee Memberships = \(40 + 40 + 40 = 120\)
Total Employees = \(110\)

Key insight: Since there are 120 total memberships but only 110 employees, some employees must be counted multiple times (in multiple committees).

Setting Up the Problem

Let me denote:

• Employees in exactly 1 committee = \(110 - \mathrm{X}\)
• Employees in more than 1 committee = \(\mathrm{X}\)

Since each person in multiple committees creates extra "membership counts":

• Someone in 2 committees counts as 2 memberships
• Someone in 3 committees counts as 3 memberships

Phase 2: Understanding the Question

Using Inclusion-Exclusion Principle

For sets A, B, C with \(|\mathrm{A}| = |\mathrm{B}| = |\mathrm{C}| = 40\):
\(|\mathrm{A} \cup \mathrm{B} \cup \mathrm{C}| = 110\) (total employees)

By inclusion-exclusion:
\(110 = 120 - |\mathrm{A} \cap \mathrm{B}| - |\mathrm{A} \cap \mathrm{C}| - |\mathrm{B} \cap \mathrm{C}| + |\mathrm{A} \cap \mathrm{B} \cap \mathrm{C}|\)

Rearranging:
\(|\mathrm{A} \cap \mathrm{B}| + |\mathrm{A} \cap \mathrm{C}| + |\mathrm{B} \cap \mathrm{C}| - |\mathrm{A} \cap \mathrm{B} \cap \mathrm{C}| = 10\)

Key Relationship

The number of employees in more than one committee is:
\(\mathrm{X} = |\mathrm{A} \cap \mathrm{B}| + |\mathrm{A} \cap \mathrm{C}| + |\mathrm{B} \cap \mathrm{C}| - 2|\mathrm{A} \cap \mathrm{B} \cap \mathrm{C}|\)

Let me denote \(|\mathrm{A} \cap \mathrm{B} \cap \mathrm{C}| = \mathrm{a}\)

Since \(|\mathrm{A} \cap \mathrm{B}| + |\mathrm{A} \cap \mathrm{C}| + |\mathrm{B} \cap \mathrm{C}| = 10 + \mathrm{a}\):
\(\mathrm{X} = (10 + \mathrm{a}) - 2\mathrm{a} = 10 - \mathrm{a}\)

Phase 3: Finding the Answer

Finding the Range of X

Since \(\mathrm{X} = 10 - \mathrm{a}\), we need the range of a (employees in all 3 committees).

Maximum value of a:
Each pairwise intersection must be at least a, so:
\(|\mathrm{A} \cap \mathrm{B}| + |\mathrm{A} \cap \mathrm{C}| + |\mathrm{B} \cap \mathrm{C}| \geq 3\mathrm{a}\)

But we know this sum equals \(10 + \mathrm{a}\), so:
\(10 + \mathrm{a} \geq 3\mathrm{a}\)
\(10 \geq 2\mathrm{a}\)
\(\mathrm{a} \leq 5\)

Minimum value of a:
Clearly \(\mathrm{a} \geq 0\) (can't be negative)

Verifying Feasibility

For minimum X (when a = 5):

• \(\mathrm{X} = 10 - 5 = 5\)
• All 5 are in all three committees
• The remaining 105 employees are each in exactly one committee
• Each committee needs 35 more members (\(40 - 5 = 35\))
• Total single-committee members needed: \(3 \times 35 = 105\) ✓

For maximum X (when a = 0):

• \(\mathrm{X} = 10 - 0 = 10\)
• No one is in all three committees
• 10 employees are in exactly 2 committees
• 100 employees are in exactly 1 committee

This can be distributed (for example):

• 5 in A∩B only
• 3 in A∩C only
• 2 in B∩C only
• Plus 100 in single committees

Phase 4: Solution

The least value of \(\mathrm{X} = 5\)
The greatest value of \(\mathrm{X} = 10\)

Both values appear in our answer choices, confirming our solution.