The symbols ` + , + , xx , xx , *** ,cdot` are placed in the squares of the adjoining figure. The number of ways of placing symbols so that no row remains empty is

Options

A. (a)16580
B. (b)14580
C. (c)5400
D. (d)4860

Answer

To solve the problem of placing the symbols ` + , + , xx , xx , *** , cdot` in the squares of the figure while ensuring that no row remains empty, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Symbols and Rows**:
We have 6 symbols: ` + , + , xx , xx , *** , cdot`. We need to place these symbols in a way that no row is empty.

2. **Determine the Arrangement**:
We can think of this as distributing the symbols into rows. Let's assume there are 3 rows (since we have 3 types of symbols: `+`, `xx`, `*`, and `cdot`).

3. **Choose Symbols for Each Row**:
We need to choose one symbol for each row. Since we have 3 symbols of the same type (`+`), we can place one `+` in one of the rows, and we can do the same for `xx` and `*`.

4. **Calculate Combinations**:
The number of ways to choose one symbol from each type for each row can be calculated using combinations. We can use the formula for combinations \( nCr = \frac{n!}{r!(n-r)!} \).

- For the `+` symbols: We can choose 1 from 3, which is \( 3C1 = 3 \).
- For the `xx` symbols: We can choose 1 from 3, which is \( 3C1 = 3 \).
- For the `*` symbols: We can choose 1 from 3, which is \( 3C1 = 3 \).

5. **Calculate Total Arrangements**:
Now, we multiply the combinations for each symbol type:
\[ \text{Total Ways} = 3C1 \times 3C1 \times 3C1 = 3 \times 3 \times 3 = 27 \]

6. **Arrange Remaining Symbols**:
After placing one symbol from each type, we have 3 symbols left. We need to arrange these remaining symbols (which are 2 `+`, 2 `xx`, and 1 `*`).

The total arrangements of these symbols can be calculated using the formula for permutations of multiset:
\[ \text{Arrangements} = \frac{n!}{n1! \times n2! \times ...} \]
Here, \( n = 6 \) (total symbols), \( n1 = 2 \) (for `+`), \( n2 = 2 \) (for `xx`), and \( n3 = 1 \) (for `*`):
\[ \text{Arrangements} = \frac{6!}{2! \times 2! \times 1!} = \frac{720}{2 \times 2 \times 1} = \frac{720}{4} = 180 \]

7. **Combine Both Parts**:
Finally, we combine the number of ways to choose the symbols and the arrangements of the remaining symbols:
\[ \text{Total Ways} = 27 \times 180 = 4860 \]

### Final Answer:
The total number of ways to place the symbols so that no row remains empty is **4860**.