Six cards are drawn one by one from a set of unlimited number of cards, each card is marked with numbers `-1, `0` or `1`. Number of different ways in which they can be drawn if the sum of the numbers shown by them vanishes is

Options

A. `111`
B. `121`
C. `141`
D. none of these

Explanation

To solve the problem of finding the number of different ways to draw six cards from a set of unlimited cards marked with numbers -1, 0, and 1 such that the sum of the numbers shown by them is zero, we can break it down into different cases based on the distribution of the cards.

### Step-by-Step Solution:

1. **Understanding the Problem**:
We need to draw 6 cards such that the sum of their values equals zero. The cards can have values of -1, 0, or 1.

2. **Case Analysis**:
We will analyze different cases based on how many cards of each type (-1, 0, and 1) we can draw.

3. **Case 1: Three -1s and Three 1s**:
- We can select 3 cards with -1 and 3 cards with 1.
- The number of arrangements is given by the formula for permutations of multiset:
\[ \text{Arrangements} = \frac{6!}{3! \times 3!} = \frac{720}{6 \times 6} = 20 \]

4. **Case 2: Two -1s, Two 1s, and Two 0s**:
- We can select 2 cards with -1, 2 cards with 1, and 2 cards with 0.
- The number of arrangements is:
\[ \text{Arrangements} = \frac{6!}{2! \times 2! \times 2!} = \frac{720}{2 \times 2 \times 2} = 90 \]

5. **Case 3: One -1, One 1, and Four 0s**:
- We can select 1 card with -1, 1 card with 1, and 4 cards with 0.
- The number of arrangements is:
\[ \text{Arrangements} = \frac{6!}{1! \times 1! \times 4!} = \frac{720}{1 \times 1 \times 24} = 30 \]

6. **Case 4: All cards are 0**:
- We can select all 6 cards as 0.
- The number of arrangements is:
\[ \text{Arrangements} = 1 \]

7. **Total Arrangements**:
Now, we sum the arrangements from all cases:
\[ \text{Total} = 20 + 90 + 30 + 1 = 141 \]

### Final Answer:
The total number of different ways to draw the cards such that their sum is zero is **141**.

---