The number of ways three different natural numbers cab be drawn from the set `{1, 2, 3, 4,……….., 10}`, if minimum of the chosen numbers is smaller than 4, is

Answer

To solve the problem of finding the number of ways to choose three different natural numbers from the set {1, 2, 3, ..., 10} such that the minimum of the chosen numbers is smaller than 4, we can break it down into cases based on the minimum number.

### Step-by-Step Solution:

1. **Identify the Cases Based on Minimum Number:**
- The minimum number can be 1, 2, or 3 since we want it to be smaller than 4.

2. **Case 1: Minimum Number is 1**
- If the minimum number is 1, the other two numbers can be chosen from the remaining numbers {2, 3, 4, 5, 6, 7, 8, 9, 10}.
- This gives us 9 options (2 to 10).
- We need to choose 2 numbers from these 9 options.
- The number of ways to choose 2 numbers from 9 is given by the combination formula:
\[ \binom{9}{2} = \frac{9 \times 8}{2} = 36 \]

3. **Case 2: Minimum Number is 2**
- If the minimum number is 2, the other two numbers can be chosen from the remaining numbers {3, 4, 5, 6, 7, 8, 9, 10}.
- This gives us 8 options (3 to 10).
- We need to choose 2 numbers from these 8 options.
- The number of ways to choose 2 numbers from 8 is:
\[ \binom{8}{2} = \frac{8 \times 7}{2} = 28 \]

4. **Case 3: Minimum Number is 3**
- If the minimum number is 3, the other two numbers can be chosen from the remaining numbers {4, 5, 6, 7, 8, 9, 10}.
- This gives us 7 options (4 to 10).
- We need to choose 2 numbers from these 7 options.
- The number of ways to choose 2 numbers from 7 is:
\[ \binom{7}{2} = \frac{7 \times 6}{2} = 21 \]

5. **Total Number of Ways:**
- Now, we sum the number of ways from all three cases:
\[ \text{Total} = \binom{9}{2} + \binom{8}{2} + \binom{7}{2} = 36 + 28 + 21 = 85 \]

### Final Answer:
The total number of ways to choose three different natural numbers from the set {1, 2, 3, ..., 10} such that the minimum number is smaller than 4 is **85**.