A person invites a party of 10 friends at dinner and place them so that 4 are on one round table and 6 on the other round table. The number of ways in which he can arrange the guests is

Options

A. A. 45!
B. B. 35!/3!
C. C. 10!/(24)
D. D. can't be determined

Answer

To solve the problem of arranging 10 friends at two round tables, we can break it down into a series of steps:

### Step 1: Choose 4 Friends for the First Table
We need to select 4 friends out of the 10 to sit at the first round table. This can be done using combinations.

\[ \text{Number of ways to choose 4 from 10} = \binom{10}{4} = \frac{10!}{4!(10-4)!} = \frac{10!}{4!6!} \]

### Step 2: Arrange the 4 Friends at the First Round Table
Once we have chosen 4 friends, we need to arrange them around the round table. The number of arrangements for \( n \) people around a round table is given by \( (n-1)! \).

For our 4 friends, the arrangements will be:

\[ \text{Arrangements of 4 friends} = (4-1)! = 3! = 6 \]

### Step 3: Arrange the Remaining 6 Friends at the Second Round Table
The remaining 6 friends will sit at the second round table. Similarly, the arrangements for these 6 friends will be:

\[ \text{Arrangements of 6 friends} = (6-1)! = 5! = 120 \]

### Step 4: Calculate the Total Arrangements
Now, we can combine all these calculations to find the total number of ways to arrange the guests:

\[ \text{Total arrangements} = \binom{10}{4} \times (4-1)! \times (6-1)! \]

Substituting the values we calculated:

\[ \text{Total arrangements} = \binom{10}{4} \times 3! \times 5! \]

Calculating \( \binom{10}{4} \):

\[ \binom{10}{4} = \frac{10!}{4!6!} = \frac{10 \times 9 \times 8 \times 7}{4 \times 3 \times 2 \times 1} = 210 \]

Now substituting back into the total arrangements:

\[ \text{Total arrangements} = 210 \times 6 \times 120 \]

Calculating this gives:

\[ = 210 \times 720 = 151200 \]

### Final Answer
The total number of ways in which the person can arrange the guests is **151200**.

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