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 is 30, we can follow these steps:

### Step 1: Define the problem
We need to find three numbers \( a, b, c \) such that:
\[ a + b + c = 30 \]
where \( 0 \leq a, b, c \leq 29 \).

### Step 2: Set constraints
Since \( a, b, c \) are tokens numbered from 0 to 29, we can assume \( a \leq b \leq c \) to avoid counting the same combination multiple times. This means we can rearrange our equation as:
\[ a + b + c = 30 \]
with the condition \( a \leq b \leq c \).

### Step 3: Express \( c \) in terms of \( a \) and \( b \)
From the equation, we can express \( c \) as:
\[ c = 30 - a - b \]
Now, we need to ensure that \( c \) is still within the range of the tokens:
\[ 30 - a - b \geq b \]
This simplifies to:
\[ 30 - a \geq 2b \]
or
\[ b \leq \frac{30 - a}{2} \]

### Step 4: Determine the range for \( a \)
Since \( a \) must also be a non-negative integer, we can find the maximum value for \( a \):
- If \( a = 0 \), then \( b \) can go up to 15 (as \( b \leq 15 \)).
- If \( a = 1 \), then \( b \) can go up to 14.
- Continuing this way, we see that as \( a \) increases, the maximum value of \( b \) decreases.

### Step 5: Count the valid combinations
Now, we can count the valid combinations for each value of \( a \):
- For \( a = 0 \): \( b \) can be \( 0 \) to \( 15 \) (16 options).
- For \( a = 1 \): \( b \) can be \( 1 \) to \( 14 \) (14 options).
- For \( a = 2 \): \( b \) can be \( 2 \) to \( 13 \) (12 options).
- For \( a = 3 \): \( b \) can be \( 3 \) to \( 12 \) (10 options).
- For \( a = 4 \): \( b \) can be \( 4 \) to \( 11 \) (8 options).
- For \( a = 5 \): \( b \) can be \( 5 \) to \( 10 \) (6 options).
- For \( a = 6 \): \( b \) can be \( 6 \) to \( 9 \) (4 options).
- For \( a = 7 \): \( b \) can be \( 7 \) to \( 8 \) (2 options).
- For \( a = 8 \): \( b \) can only be \( 8 \) (1 option).

### Step 6: Sum the combinations
Now we add all the valid combinations:
\[ 16 + 14 + 12 + 10 + 8 + 6 + 4 + 2 + 1 = 73 \]

### Conclusion
Thus, the total number of ways to choose 3 tokens such that their sum is 30 is **75**.