Roorkee University has to send 10 professors to 5 centers for its entrance examination, 2 to each center. Two of the enters are in Roorkee and the others are outside. Two of the professors prefer to work in Roorkee while three prefer to work outside. In how many ways can this be made if the preferences are to be satisfied?

Options

A. 5400
B. 5200
C. 4800
D. 4000

Answer

To solve the problem of sending 10 professors to 5 centers while satisfying their preferences, we can break down the solution into a series of logical steps.

### Step-by-Step Solution:

1. **Identify the Centers and Professors:**
- There are 5 centers: 2 in Roorkee (let's call them Center A and Center B) and 3 outside Roorkee (let's call them Center C, Center D, and Center E).
- There are 10 professors in total.
- Out of these, 2 professors (let's call them P1 and P2) prefer to work in Roorkee, and 3 professors (let's call them P3, P4, and P5) prefer to work outside.

2. **Assign Professors to Roorkee Centers:**
- Since there are 2 professors who prefer Roorkee, they must be assigned to the Roorkee centers.
- We can assign P1 and P2 to Center A and Center B. There are 2 ways to assign them (P1 to A and P2 to B or P1 to B and P2 to A).

3. **Assign Professors to Outside Centers:**
- There are 3 professors who prefer to work outside (P3, P4, and P5). We need to assign 6 professors to the 3 outside centers (2 professors per center).
- First, we assign the 3 professors who prefer outside to the outside centers. We can choose 3 out of the 6 remaining professors (after assigning P1 and P2) to work outside.
- The number of ways to choose 3 professors from the remaining 8 professors (10 total - 2 in Roorkee) is given by the combination formula \( \binom{n}{r} \), where \( n \) is the total number of items to choose from, and \( r \) is the number of items to choose.

\[ \text{Ways to choose 3 professors from 8} = \binom{8}{3} = \frac{8!}{3!(8-3)!} = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = 56 \]

4. **Distributing the Chosen Professors to the Centers:**
- After choosing 3 professors to go outside, we need to assign them to the 3 centers (C, D, E).
- The number of ways to assign 3 professors to 3 centers (2 professors per center) is given by the multinomial coefficient. Since we have 3 professors and need to assign them to 3 centers, we can do this in \( 3! \) ways.

\[ \text{Ways to assign 3 professors to 3 centers} = 3! = 6 \]

5. **Assigning Remaining Professors:**
- After assigning the 3 professors who prefer outside, we have 5 professors left (10 total - 2 in Roorkee - 3 outside).
- We need to assign these 5 professors to the 3 centers (C, D, E) with 2 professors per center.
- The number of ways to assign these remaining professors is given by the multinomial coefficient as well.

\[ \text{Ways to assign 5 professors to 3 centers} = \frac{5!}{2! \times 2! \times 1!} = \frac{120}{2 \times 2 \times 1} = 30 \]

6. **Total Ways:**
- Now, we can calculate the total number of ways to assign the professors while satisfying their preferences.

\[ \text{Total ways} = (\text{Ways to assign Roorkee}) \times (\text{Ways to choose outside}) \times (\text{Ways to assign outside}) \times (\text{Ways to assign remaining}) \]

\[ \text{Total ways} = 2 \times 56 \times 6 \times 30 \]

\[ = 2 \times 56 = 112 \]
\[ = 112 \times 6 = 672 \]
\[ = 672 \times 30 = 20160 \]

### Final Answer:
The total number of ways to assign the professors while satisfying their preferences is **20160**.