A term of randomly chosen from the expansion of `(root(6)(4) + 1/(root(4)(5)))^(20)`. If the probability that it is a rational term is P, then 420P is euqal to

Answer

To solve the problem, we need to find the probability \( P \) that a randomly chosen term from the expansion of \( \left( \sqrt[6]{4} + \frac{1}{\sqrt[4]{5}} \right)^{20} \) is a rational term. We will then calculate \( 420P \).

### Step-by-Step Solution:

1. **Rewrite the expression**:
\[ \sqrt[6]{4} = 4^{1/6} = (2^2)^{1/6} = 2^{1/3} \]
\[ \frac{1}{\sqrt[4]{5}} = 5^{-1/4} \]
Thus, we can rewrite the expression as:
\[ \left( 2^{1/3} + 5^{-1/4} \right)^{20} \]

2. **Use the Binomial Theorem**:
The expansion of \( (a + b)^n \) can be expressed as:
\[ \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k \]
In our case, \( a = 2^{1/3} \) and \( b = 5^{-1/4} \) with \( n = 20 \).

3. **Identify a general term**:
The general term in the expansion is given by:
\[ T_k = \binom{20}{k} (2^{1/3})^{20-k} (5^{-1/4})^k = \binom{20}{k} 2^{(20-k)/3} 5^{-k/4} \]

4. **Determine when the term is rational**:
For \( T_k \) to be rational, both exponents \( \frac{20-k}{3} \) and \( -\frac{k}{4} \) must be integers. This leads us to the following conditions:
- \( 20 - k \) must be divisible by 3.
- \( k \) must be divisible by 4.

5. **Set up the equations**:
Let \( k = 4m \) for some integer \( m \). Then substituting this into the first condition:
\[ 20 - 4m \equiv 0 \, (\text{mod } 3) \]
Simplifying gives:
\[ 20 \equiv 2 \, (\text{mod } 3) \implies -4m \equiv 0 \, (\text{mod } 3) \implies m \equiv 0 \, (\text{mod } 3) \]
Thus, \( m \) can take values \( 0, 3, 6 \) (since \( 4m \) must be less than or equal to 20).

6. **Calculate possible values of \( k \)**:
- If \( m = 0 \), then \( k = 0 \).
- If \( m = 3 \), then \( k = 12 \).
- If \( m = 6 \), then \( k = 24 \) (not valid since \( k \) must be \( \leq 20 \)).
Thus, valid \( k \) values are \( 0 \) and \( 12 \).

7. **Count the rational terms**:
The valid \( k \) values that yield rational terms are \( k = 0 \) and \( k = 12 \). Therefore, there are 2 favorable outcomes.

8. **Total outcomes**:
The total number of terms in the expansion is \( n + 1 = 20 + 1 = 21 \).

9. **Calculate the probability \( P \)**:
\[ P = \frac{\text{Number of favorable outcomes}}{\text{Total outcomes}} = \frac{2}{21} \]

10. **Calculate \( 420P \)**:
\[ 420P = 420 \times \frac{2}{21} = 40 \]

### Final Answer:
\[ \boxed{40} \]