`""^(15)C_(0)*""^(5)C_(5)+""^(15)C_(1)*""^(5)C_(4)+""^(15)C_(2)*""^(5)C_(3)+""^(15)C_(3)*""^(5)C_(2)+""^(15)C_(4)*""^(5)C_(1)` is equal to :

Options

A. `2^(20)-2^(5)`
B. `(20!)/(5!15!)`
C. `(20!)/(5!15!)-1`
D. `(20!)/(5!15!)-(15!)/(5!10!)`

Answer

To solve the expression:

\[ ^{15}C_{0} \cdot ^{5}C_{5} + ^{15}C_{1} \cdot ^{5}C_{4} + ^{15}C_{2} \cdot ^{5}C_{3} + ^{15}C_{3} \cdot ^{5}C_{2} + ^{15}C_{4} \cdot ^{5}C_{1} \]

we can use the Binomial Theorem. The Binomial Theorem states that:

\[ (1 + x)^n = \sum_{k=0}^{n} {n \choose k} x^k \]

### Step 1: Write the Binomial expansions

We can express the two binomials as:

\[ (1 + x)^{15} = \sum_{k=0}^{15} {15 \choose k} x^k \]
\[ (1 + x)^{5} = \sum_{j=0}^{5} {5 \choose j} x^j \]

### Step 2: Multiply the two expansions

When we multiply these two expansions, we get:

\[ (1 + x)^{15} \cdot (1 + x)^{5} = (1 + x)^{20} \]

### Step 3: Identify the coefficients

We want to find the coefficient of \(x^5\) in the expansion of \((1 + x)^{20}\). This coefficient is given by:

\[ {20 \choose 5} \]

### Step 4: Calculate the coefficient of \(x^5\)

Using the binomial coefficient formula:

\[ {n \choose k} = \frac{n!}{k!(n-k)!} \]

we can calculate:

\[ {20 \choose 5} = \frac{20!}{5! \cdot 15!} \]

### Step 5: Subtract the unwanted term

However, our original expression also includes the term \(^{15}C_{5} \cdot ^{5}C_{0}\), which corresponds to \(x^5\) in the expansion. We need to subtract this term from our previous result:

\[ {15 \choose 5} \cdot {5 \choose 0} = {15 \choose 5} \cdot 1 = {15 \choose 5} \]

### Step 6: Final calculation

Thus, the final result is:

\[ {20 \choose 5} - {15 \choose 5} \]

Calculating these values:

\[ {20 \choose 5} = \frac{20!}{5! \cdot 15!} \]
\[ {15 \choose 5} = \frac{15!}{5! \cdot 10!} \]

### Step 7: Simplify the expression

Now, we can simplify:

\[ {20 \choose 5} - {15 \choose 5} = \frac{20!}{5! \cdot 15!} - \frac{15!}{5! \cdot 10!} \]

### Step 8: Final result

The final answer is:

\[ {20 \choose 5} - {15 \choose 5} \]