SOLUTION: Comparing Arithmetic and Geometric Sequences - Studypool - Free Printable

Problem:

A company has 10 employees, including a manager and his assistant. Now, 4 people are to be selected from these 10 employees to participate in an activity. The conditions are as follows:

1. Question (1): How many different ways can the selection be made if the manager must participate?
2. Question (2): How many different ways can the selection be made if both the manager and his assistant must participate?
3. Question (3): How many different ways can the selection be made if at least one of the manager or his assistant must participate?

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Solution:

#### Step 1: Total number of ways to select 4 people from 10
The total number of ways to choose 4 people from 10 is given by the combination formula:
\[
\binom{n}{r} = \frac{n!}{r!(n-r)!}
\]
Here, \( n = 10 \) and \( r = 4 \):
\[
\binom{10}{4} = \frac{10!}{4! \cdot 6!} = \frac{10 \times 9 \times 8 \times 7}{4 \times 3 \times 2 \times 1} = 210
\]

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#### Question (1): Manager must participate

If the manager must participate, we need to select the remaining 3 people from the other 9 employees (since the manager is already included).

The number of ways to choose 3 people from 9 is:
\[
\binom{9}{3} = \frac{9!}{3! \cdot 6!} = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} = 84
\]

Thus, the number of ways to select 4 people with the manager included is:
\[
\boxed{84}
\]

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#### Question (2): Both the manager and his assistant must participate

If both the manager and his assistant must participate, we need to select the remaining 2 people from the other 8 employees (since both the manager and his assistant are already included).

The number of ways to choose 2 people from 8 is:
\[
\binom{8}{2} = \frac{8!}{2! \cdot 6!} = \frac{8 \times 7}{2 \times 1} = 28
\]

Thus, the number of ways to select 4 people with both the manager and his assistant included is:
\[
\boxed{28}
\]

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#### Question (3): At least one of the manager or his assistant must participate

To solve this, we use the principle of complementary counting. First, we calculate the number of ways to select 4 people such that neither the manager nor his assistant participates. Then, we subtract this from the total number of ways to select 4 people from 10.

1. Number of ways to select 4 people without the manager or his assistant:
If neither the manager nor his assistant participates, we are left with 8 employees to choose from. The number of ways to choose 4 people from these 8 is:
\[
\binom{8}{4} = \frac{8!}{4! \cdot 4!} = \frac{8 \times 7 \times 6 \times 5}{4 \times 3 \times 2 \times 1} = 70
\]

2. Number of ways to select 4 people with at least one of the manager or his assistant:
The total number of ways to select 4 people from 10 is 210. Subtracting the number of ways to select 4 people without the manager or his assistant gives:
\[
210 - 70 = 140
\]

Thus, the number of ways to select 4 people such that at least one of the manager or his assistant participates is:
\[
\boxed{140}
\]

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Final Answers:

1. \(\boxed{84}\)
2. \(\boxed{28}\)
3. \(\boxed{140}\)