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

Let me solve this step by step.

Given information:

• 3 committees with 40 members each
• Total positions: \(3 \times 40 = 120\)
• Total employees: 110
• Some employees must be on multiple committees

Let's define variables:

• \(\mathrm{n_1}\) = employees on exactly 1 committee
• \(\mathrm{n_2}\) = employees on exactly 2 committees
• \(\mathrm{n_3}\) = employees on exactly 3 committees

Constraints:

• \(\mathrm{n_1} + \mathrm{n_2} + \mathrm{n_3} = 110\) (total employees)
• \(\mathrm{n_1} + 2\mathrm{n_2} + 3\mathrm{n_3} = 120\) (total positions)

Subtracting the first equation from the second:
\(\mathrm{n_2} + 2\mathrm{n_3} = 10\)

We want to find the range of \(\mathrm{X} = \mathrm{n_2} + \mathrm{n_3}\) (employees on more than one committee).

For minimum X:
To minimize the number of people on multiple committees, maximize those on 3 committees.
If \(\mathrm{n_3} = 5\), then \(\mathrm{n_2} = 0\)
So \(\mathrm{X_{min}} = 0 + 5 = 5\)

For maximum X:
To maximize the number of people on multiple committees, maximize those on exactly 2 committees.
If \(\mathrm{n_2} = 10\), then \(\mathrm{n_3} = 0\)
So \(\mathrm{X_{max}} = 10 + 0 = 10\)

Therefore: Least value of \(\mathrm{X} = 5\), Greatest value of \(\mathrm{X} = 10\)