In a conference 10 speakers are present . If `S_1` wants to speak before `S_2` and `S_2` wants to speak after `S_3` , then the number of ways all the 10 speakers can give their speeches with the above restriction if the remaining seven speakers have no objection to speak at any number is

Options

A. A) `""^(10) C_(3)`
B. B) ` ""^(10 )P_(8)`
C. C) `""^(10)P_(3)`
D. D) `10! //3`

Answer

To solve the problem step by step, we need to consider the restrictions placed on the speaking order of the speakers.

### Step 1: Understand the restrictions
We have 10 speakers: S1, S2, S3, and 7 others. The restrictions are:
- S1 must speak before S2.
- S2 must speak after S3.

This means the order must be S3, S1, S2 in terms of their speaking positions.

### Step 2: Determine the arrangement of S1, S2, and S3
Since S1 must be before S2 and S2 must be after S3, we can think of S3 being in a position before both S1 and S2.

The possible arrangements of S1, S2, and S3 can be visualized as:
- S3 can occupy any of the first 8 positions (1 to 8), allowing S1 and S2 to occupy the remaining positions after S3.

### Step 3: Choose positions for S1, S2, and S3
We need to choose 3 positions out of the 10 available for S1, S2, and S3. The number of ways to choose 3 positions from 10 is given by the combination formula \( \binom{n}{r} \):
\[ \text{Number of ways to choose positions} = \binom{10}{3} \]

### Step 4: Arrange S1, S2, and S3 in the chosen positions
Once we have chosen 3 positions, there is only one valid arrangement of S1, S2, and S3 that satisfies the restrictions, which is S3 in the first chosen position, S1 in the second, and S2 in the third.

### Step 5: Arrange the remaining 7 speakers
The remaining 7 speakers can occupy the remaining 7 positions without any restrictions. The number of ways to arrange these 7 speakers is given by \( 7! \).

### Step 6: Combine the results
Now we can combine the number of ways to choose the positions for S1, S2, and S3 with the arrangements of the remaining speakers:
\[ \text{Total arrangements} = \binom{10}{3} \times 7! \]

### Step 7: Calculate the final answer
Substituting the values:
\[ \binom{10}{3} = \frac{10!}{3!(10-3)!} = \frac{10!}{3! \cdot 7!} \]
Thus, the total number of arrangements becomes:
\[ \text{Total arrangements} = \frac{10!}{3! \cdot 7!} \times 7! = \frac{10!}{3!} \]

### Final Answer
The total number of ways all 10 speakers can give their speeches with the given restrictions is:
\[ \frac{10!}{3!} \]