If `Delta = |(""^(5)C_0,""^(5)C_3,14),(""^(5)C_1,""^(5)C_4,1),(""^(5)C_2,""^(5)C_5,1)|` , then its value is

Options

A. 0
B. `-576`
C. 80
D. none

Answer

To find the value of the determinant

\[ \Delta = \begin{vmatrix} {^5C_0} & {^5C_3} & 14 \\ {^5C_1} & {^5C_4} & 1 \\ {^5C_2} & {^5C_5} & 1 \end{vmatrix} \]

we first need to calculate the binomial coefficients \( {^5C_r} \) for \( r = 0, 1, 2, 3, 4, 5 \).

### Step 1: Calculate the binomial coefficients

1. \( {^5C_0} = 1 \)
2. \( {^5C_1} = 5 \)
3. \( {^5C_2} = 10 \)
4. \( {^5C_3} = {^5C_2} = 10 \)
5. \( {^5C_4} = {^5C_1} = 5 \)
6. \( {^5C_5} = {^5C_0} = 1 \)

### Step 2: Substitute the values into the determinant

Now substituting these values into the determinant, we have:

\[ \Delta = \begin{vmatrix} 1 & 10 & 14 \\ 5 & 5 & 1 \\ 10 & 1 & 1 \end{vmatrix} \]

### Step 3: Expand the determinant

We can expand the determinant along the first row:

\[ \Delta = 1 \cdot \begin{vmatrix} 5 & 1 \\ 1 & 1 \end{vmatrix} - 10 \cdot \begin{vmatrix} 5 & 1 \\ 10 & 1 \end{vmatrix} + 14 \cdot \begin{vmatrix} 5 & 5 \\ 10 & 1 \end{vmatrix} \]

### Step 4: Calculate the 2x2 determinants

1. For the first determinant:

\[ \begin{vmatrix} 5 & 1 \\ 1 & 1 \end{vmatrix} = (5 \cdot 1) - (1 \cdot 1) = 5 - 1 = 4 \]

2. For the second determinant:

\[ \begin{vmatrix} 5 & 1 \\ 10 & 1 \end{vmatrix} = (5 \cdot 1) - (1 \cdot 10) = 5 - 10 = -5 \]

3. For the third determinant:

\[ \begin{vmatrix} 5 & 5 \\ 10 & 1 \end{vmatrix} = (5 \cdot 1) - (5 \cdot 10) = 5 - 50 = -45 \]

### Step 5: Substitute back into the determinant expansion

Now substituting these values back into the expansion:

\[ \Delta = 1 \cdot 4 - 10 \cdot (-5) + 14 \cdot (-45) \]

Calculating this gives:

\[ \Delta = 4 + 50 - 630 \]

\[ \Delta = 54 - 630 = -576 \]

### Final Answer

Thus, the value of the determinant \( \Delta \) is

\[ \Delta = -576 \]