In a certain an algebraical exercise book there and 4 examples on arithmetical progression, 5 examples on permutation and combination, and 6 examples on binomial theorem. Find the number of ways a teacher can select or his pupils at least one but not more than 2 examples from each of these sets.

Options

A. 4500
B. 2550
C. 2850
D. 3150

Answer

To solve the problem, we need to find the number of ways a teacher can select at least one but not more than two examples from three different sets of examples: arithmetic progression (AP), permutation and combination (P&C), and binomial theorem (BT).

### Step-by-step Solution:

1. **Identify the number of examples in each category:**
- Arithmetic Progression (AP): 4 examples
- Permutation and Combination (P&C): 5 examples
- Binomial Theorem (BT): 6 examples

2. **Calculate the number of ways to select examples from AP:**
- The teacher can select either 1 or 2 examples from the 4 examples in AP.
- The number of ways to select 1 example from 4 is given by \( \binom{4}{1} \).
- The number of ways to select 2 examples from 4 is given by \( \binom{4}{2} \).

\[ \binom{4}{1} = 4 \]
\[ \binom{4}{2} = \frac{4!}{2!(4-2)!} = \frac{4 \times 3}{2 \times 1} = 6 \]

- Total ways to select from AP:
\[ 4 + 6 = 10 \]

3. **Calculate the number of ways to select examples from P&C:**
- The teacher can select either 1 or 2 examples from the 5 examples in P&C.
- The number of ways to select 1 example from 5 is given by \( \binom{5}{1} \).
- The number of ways to select 2 examples from 5 is given by \( \binom{5}{2} \).

\[ \binom{5}{1} = 5 \]
\[ \binom{5}{2} = \frac{5!}{2!(5-2)!} = \frac{5 \times 4}{2 \times 1} = 10 \]

- Total ways to select from P&C:
\[ 5 + 10 = 15 \]

4. **Calculate the number of ways to select examples from BT:**
- The teacher can select either 1 or 2 examples from the 6 examples in BT.
- The number of ways to select 1 example from 6 is given by \( \binom{6}{1} \).
- The number of ways to select 2 examples from 6 is given by \( \binom{6}{2} \).

\[ \binom{6}{1} = 6 \]
\[ \binom{6}{2} = \frac{6!}{2!(6-2)!} = \frac{6 \times 5}{2 \times 1} = 15 \]

- Total ways to select from BT:
\[ 6 + 15 = 21 \]

5. **Calculate the total number of ways to select examples from all categories:**
- The total number of ways is the product of the number of ways from each category.

\[ \text{Total ways} = (\text{Ways from AP}) \times (\text{Ways from P&C}) \times (\text{Ways from BT}) \]
\[ \text{Total ways} = 10 \times 15 \times 21 \]

- Calculate:
\[ 10 \times 15 = 150 \]
\[ 150 \times 21 = 3150 \]

### Final Answer:
The total number of ways the teacher can select examples is **3150**.