If ` ([""^(n)C_(r) + 4*""^(n)C_(r+1) + 6*""^(n)C_(r+2)+ 4*""^(n)C_(r+3) + ""^(n)C_(r+4)])/([""^(n)C_(r) + 3. ""^(n)C_(r+1)+ 3*""^(n)C_(r+2) + ""^(n)C_(r +3)])=(n + lambda)/(r+lambda)`
the value of ` lambda ` is

Answer

To solve the given problem, we need to simplify the expression and find the value of \( \lambda \).

### Step-by-Step Solution:

1. **Write the Given Expression**:
We start with the expression:
\[ \frac{nC_r + 4 \cdot nC_{r+1} + 6 \cdot nC_{r+2} + 4 \cdot nC_{r+3} + nC_{r+4}}{nC_r + 3 \cdot nC_{r+1} + 3 \cdot nC_{r+2} + nC_{r+3}} = \frac{n + \lambda}{r + \lambda} \]

2. **Simplify the Numerator**:
We can use the property of binomial coefficients:
\[ nC_r + nC_{r+1} = (n+1)C_{r+1} \]
Applying this property, we can rewrite the numerator:
\[ nC_r + 4 \cdot nC_{r+1} + 6 \cdot nC_{r+2} + 4 \cdot nC_{r+3} + nC_{r+4} \]
can be rearranged as:
\[ nC_r + nC_{r+1} + 3 \cdot nC_{r+1} + 3 \cdot nC_{r+2} + nC_{r+3} + nC_{r+4} \]
This becomes:
\[ (n+1)C_{r+1} + 3 \cdot (n+1)C_{r+2} + (n+1)C_{r+3} \]
Continuing this process, we can combine terms to yield:
\[ (n+2)C_{r+2} + (n+3)C_{r+3} + (n+4)C_{r+4} \]

3. **Simplify the Denominator**:
Similarly, we simplify the denominator:
\[ nC_r + 3 \cdot nC_{r+1} + 3 \cdot nC_{r+2} + nC_{r+3} \]
This can be rewritten as:
\[ nC_r + nC_{r+1} + 2 \cdot nC_{r+1} + 2 \cdot nC_{r+2} + nC_{r+3} \]
Which simplifies to:
\[ (n+1)C_{r+1} + 2 \cdot (n+1)C_{r+2} + (n+1)C_{r+3} \]

4. **Final Simplification**:
After simplifying both the numerator and denominator, we have:
\[ \frac{(n+4)C_{r+4}}{(n+3)C_{r+3}} \]
This simplifies to:
\[ \frac{n+4}{r+4} \]

5. **Equate and Solve for \( \lambda \)**:
Now we equate:
\[ \frac{n + \lambda}{r + \lambda} = \frac{n + 4}{r + 4} \]
From this, we can see that \( \lambda = 4 \).

### Conclusion:
The value of \( \lambda \) is \( 4 \).