The value of the determinant `|{:(""^5C_0,""^5C_3,14),(""^5C_1,""^5C_4,1),(""^5C_2,""^5C_5,1):}|` is

Options

A. 0
B. `-(6!)`
C. 80
D. -576

Answer

To solve the determinant

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

we will first calculate the binomial coefficients and then evaluate the determinant.

### Step 1: Calculate the Binomial Coefficients

- \( {^5C_0} = 1 \)
- \( {^5C_1} = 5 \)
- \( {^5C_2} = 10 \)
- \( {^5C_3} = 10 \)
- \( {^5C_4} = 5 \)
- \( {^5C_5} = 1 \)

Now substituting these values into the determinant, we have:

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

### Step 2: Evaluate the Determinant

We can use the method of cofactor expansion along the first row:

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

### Step 3: 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 4: Substitute Back into the Determinant

Now substituting these values back into the expression for \( D \):

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

Calculating each term:

\[ D = 4 + 50 - 630 \]

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

### Final Answer

Thus, the value of the determinant is

\[ \boxed{-576} \]

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