Answer these questions based on the following information.
A team of 8 software engineers work on a project, for an American client, in Bengaluru, India. Every team member wants to go to the client's site in USA, as many times as it's possible for them. However, as per the company's policy every member will visit the site same number of times in a year. Each team member has to travel in groups of the same size each time the team goes to client's site. Not all members can be same in any two groups visiting the client's site in a year.
Maximum how many different groups can be formed ?

Options

A. A) 4
B. B) 16
C. C) 70
D. D) 32

Answer

To solve the problem of how many different groups can be formed from a team of 8 software engineers, we can follow these steps:

### Step 1: Understand the Problem
We have a team of 8 engineers, and we want to form groups where each group has the same number of members. The groups cannot be identical, meaning that no two groups can have the same combination of members.

### Step 2: Identify the Variables
Let \( n \) be the total number of engineers, which is 8 in this case. We need to find the maximum number of different groups that can be formed, which involves choosing \( r \) members from \( n \).

### Step 3: Use Combinations
The number of ways to choose \( r \) members from \( n \) members is given by the combination formula:
\[ nCr = \frac{n!}{r!(n-r)!} \]
where \( n! \) (n factorial) is the product of all positive integers up to \( n \).

### Step 4: Determine the Optimal Group Size
To maximize the number of combinations, we need to find the value of \( r \) that maximizes \( nCr \). For an even number \( n \), the maximum value of \( nCr \) occurs when \( r = \frac{n}{2} \).

Here, since \( n = 8 \):
\[ r = \frac{8}{2} = 4 \]

### Step 5: Calculate the Combinations
Now we can calculate \( 8C4 \):
\[ 8C4 = \frac{8!}{4!(8-4)!} = \frac{8!}{4! \cdot 4!} \]

Calculating the factorials:
\[ 8! = 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 \]
\[ 4! = 4 \times 3 \times 2 \times 1 = 24 \]

Now substituting back into the combination formula:
\[ 8C4 = \frac{8 \times 7 \times 6 \times 5}{4 \times 3 \times 2 \times 1} = \frac{1680}{24} = 70 \]

### Step 6: Conclusion
Thus, the maximum number of different groups that can be formed is **70**.