A bag contains 30 tokens numbered serially from 0 to 29. The number of ways of choosing 3 tokens from the bag, such that the sum on them is 30, is

Options

A. 56
B. 44
C. 75
D. 81

Answer

To solve the problem of finding the number of ways to choose 3 tokens from a bag containing tokens numbered from 0 to 29 such that their sum equals 30, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Problem**:
We need to find three distinct numbers \( a, b, c \) such that \( a + b + c = 30 \) and \( 0 \leq a, b, c \leq 29 \).

2. **Fixing the First Token**:
Let's denote the smallest of the three chosen tokens as \( a \). We can fix \( a \) and then find the possible pairs \( (b, c) \) such that \( b + c = 30 - a \).

3. **Finding Valid Pairs**:
For each fixed value of \( a \), \( b \) and \( c \) must be greater than \( a \) to ensure they are distinct. Thus, we have:
\[ b + c = 30 - a \quad \text{and} \quad b > a, c > a \]
This leads to:
\[ b + c = 30 - a \implies b + c > 2a \implies 30 - a > 2a \implies 30 > 3a \implies a < 10 \]
Therefore, \( a \) can take values from 0 to 9.

4. **Counting the Pairs for Each Fixed \( a \)**:
For each value of \( a \), we can calculate the number of valid pairs \( (b, c) \):
- When \( a = 0 \): \( b + c = 30 \) → Pairs: (1, 29), (2, 28), ..., (14, 16) → 14 pairs.
- When \( a = 1 \): \( b + c = 29 \) → Pairs: (2, 27), (3, 26), ..., (14, 15) → 13 pairs.
- When \( a = 2 \): \( b + c = 28 \) → Pairs: (3, 25), (4, 24), ..., (14, 14) → 11 pairs.
- When \( a = 3 \): \( b + c = 27 \) → Pairs: (4, 23), (5, 22), ..., (14, 13) → 10 pairs.
- When \( a = 4 \): \( b + c = 26 \) → Pairs: (5, 21), (6, 20), ..., (14, 12) → 8 pairs.
- When \( a = 5 \): \( b + c = 25 \) → Pairs: (6, 19), (7, 18), ..., (14, 11) → 7 pairs.
- When \( a = 6 \): \( b + c = 24 \) → Pairs: (7, 17), (8, 16), ..., (14, 10) → 5 pairs.
- When \( a = 7 \): \( b + c = 23 \) → Pairs: (8, 15), (9, 14), ..., (14, 9) → 4 pairs.
- When \( a = 8 \): \( b + c = 22 \) → Pairs: (9, 13), (10, 12) → 2 pairs.
- When \( a = 9 \): \( b + c = 21 \) → Pairs: (10, 11) → 1 pair.

5. **Summing the Total Pairs**:
Now, we sum all the pairs:
\[ 14 + 13 + 11 + 10 + 8 + 7 + 5 + 4 + 2 + 1 = 75 \]

### Final Answer:
The total number of ways to choose 3 tokens such that their sum is 30 is **75**.