There are 16 executives including two brothers, Lehman and Mckinsey. In how many ways can they be arranged around the circular table if the two brothers can not be seated together?

Options

A. a. `(14! ) .13`
B. b. `""^(14) P_3`
C. c. `(141)/(3)`
D. d. none of these

Answer

To solve the problem of arranging 16 executives around a circular table such that two brothers, Lehman and McKinsey, do not sit together, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Total Arrangement**:
In a circular arrangement, the number of ways to arrange \( n \) distinct objects is given by \( (n-1)! \). Here, we have 16 executives, so the total arrangements without any restrictions would be \( (16-1)! = 15! \).

2. **Calculate the Arrangements with Brothers Together**:
To find the arrangements where Lehman and McKinsey are sitting together, we can treat them as a single unit or block. This means we now have 15 units to arrange (the block of brothers + 14 other executives).

The number of ways to arrange these 15 units in a circle is \( (15-1)! = 14! \).

3. **Arrange the Brothers Within Their Block**:
Within their block, Lehman and McKinsey can be arranged in 2 ways (Lehman first or McKinsey first). Thus, the total arrangements with the brothers together is:
\[ 14! \times 2 \]

4. **Calculate the Arrangements Where Brothers Are Not Together**:
To find the arrangements where the brothers are not sitting together, we subtract the arrangements where they are together from the total arrangements:
\[ \text{Arrangements where brothers are not together} = 15! - (14! \times 2) \]

5. **Simplify the Expression**:
We can factor out \( 14! \) from the expression:
\[ 15! = 15 \times 14! \]
Therefore, we have:
\[ 15! - (14! \times 2) = 15 \times 14! - 2 \times 14! = (15 - 2) \times 14! = 13 \times 14! \]

6. **Final Result**:
The number of ways to arrange the 16 executives around the table such that the two brothers do not sit together is:
\[ 13 \times 14! \]