The value of `(""^(47)C_(4))/(""^(57)C_(4))+sum_(j=0)^(3)(""^(50-j)C_(3))/(""^(57)C_(53))+sum_(j=0)^(5)""^((56-k)C_(53-k))/(""^(57)C_(4))` is :

Options

A. 0
B. 2
C. 1
D. None of these

Answer

To solve the expression

\[ \frac{{^{47}C_4}}{{^{57}C_4}} + \sum_{j=0}^{3} \frac{{^{50-j}C_3}}{{^{57}C_{53}}} + \sum_{k=0}^{5} \frac{{^{56-k}C_{53-k}}}{{^{57}C_4}}, \]

let's break it down step by step.

### Step 1: Simplifying the first term

The first term is

\[ \frac{{^{47}C_4}}{{^{57}C_4}}. \]

Using the property of combinations, we can express this as:

\[ \frac{{^{47}C_4}}{{^{57}C_4}} = \frac{{^{47}C_4}}{{^{57}C_4}} = \frac{{^{47}C_4}}{{^{57}C_4}} = \frac{{^{47}C_4}}{{^{57}C_4}} = \frac{47!}{4!(47-4)!} \cdot \frac{(57-4)!}{57!}. \]

This simplifies to:

\[ \frac{47! \cdot (53)!}{4! \cdot 43! \cdot 57!} = \frac{1}{^{57}C_4} \cdot ^{50}C_3 + ^{49}C_3 + ^{48}C_3 + ^{47}C_3. \]

### Step 2: Simplifying the second term

Now we look at the second term:

\[ \sum_{j=0}^{3} \frac{{^{50-j}C_3}}{{^{57}C_{53}}}. \]

This can be rewritten as:

\[ \frac{1}{{^{57}C_{53}}} \left(^{50}C_3 + ^{49}C_3 + ^{48}C_3 + ^{47}C_3\right). \]

### Step 3: Simplifying the third term

The third term is:

\[ \sum_{k=0}^{5} \frac{{^{56-k}C_{53-k}}}{{^{57}C_4}}. \]

This can be rewritten as:

\[ \frac{1}{{^{57}C_4}} \left(^{56}C_{53} + ^{55}C_{52} + ^{54}C_{51} + ^{53}C_{50} + ^{52}C_{49} + ^{51}C_{48}\right). \]

### Step 4: Combining all terms

Now we combine all the terms:

\[ \frac{1}{{^{57}C_4}} \left(^{47}C_4 + ^{50}C_3 + ^{49}C_3 + ^{48}C_3 + ^{47}C_3 + ^{56}C_{53} + ^{55}C_{52} + ^{54}C_{51} + ^{53}C_{50} + ^{52}C_{49} + ^{51}C_{48}\right). \]

### Step 5: Using Hockey Stick Identity

Using the hockey stick identity in combinatorics, we can simplify the sum of combinations:

\[ ^{n}C_r + ^{n-1}C_r + ... + ^{r}C_r = ^{n+1}C_{r+1}. \]

Applying this identity, we find that:

\[ ^{50}C_3 + ^{49}C_3 + ^{48}C_3 + ^{47}C_3 = ^{51}C_4. \]

And similarly for the other sums.

### Step 6: Final simplification

Putting everything together, we find that:

\[ \frac{1}{{^{57}C_4}} \cdot ^{56}C_4 = 1. \]

Thus, the final value of the entire expression is:

\[ \boxed{1}. \]