Let {n_1} < {n_2} < {n_3} < {n_4} < {n_5} be positive integers such that {n_1} + {n_2} + {n_3} + {n_4} + {n_5} = 20. Then the numbers of such distinct arrangements ({n_1}.{n_2}.{n_3}.{n_4}.{n_5}) is…………………………

We are given that \( n_1 < n_2 < n_3 < n_4 < n_5 \) are distinct positive integers such that their sum is 20, i.e.,

\[
n_1 + n_2 + n_3 + n_4 + n_5 = 20
\]

### Step 1: Redefine the variables
Since the integers are distinct, we can redefine the variables in such a way that they are all positive integers. Let:

\[
n_1 = x_1 + 1, \quad n_2 = x_2 + 2, \quad n_3 = x_3 + 3, \quad n_4 = x_4 + 4, \quad n_5 = x_5 + 5
\]

where \( x_1, x_2, x_3, x_4, x_5 \) are non-negative integers. This transformation ensures that the new variables represent positive integers.

### Step 2: Express the sum
Substitute the new variables into the sum:

\[
(x_1 + 1) + (x_2 + 2) + (x_3 + 3) + (x_4 + 4) + (x_5 + 5) = 20
\]

Simplify the equation:

\[
x_1 + x_2 + x_3 + x_4 + x_5 + 15 = 20
\]

\[
x_1 + x_2 + x_3 + x_4 + x_5 = 5
\]

### Step 3: Count the non-negative integer solutions
We now need to find the number of solutions to the equation:

\[
x_1 + x_2 + x_3 + x_4 + x_5 = 5
\]

where \( x_1, x_2, x_3, x_4, x_5 \) are non-negative integers. This is a typical "stars and bars" problem, where the number of solutions is given by the binomial coefficient:

\[
\binom{5 + 5 - 1}{5 - 1} = \binom{9}{4}
\]

### Step 4: Calculate the binomial coefficient
Now, calculate \( \binom{9}{4} \):

\[
\binom{9}{4} = \frac{9 \times 8 \times 7 \times 6}{4 \times 3 \times 2 \times 1} = \frac{3024}{24} = 126
\]

### Final Answer:
The number of distinct arrangements of \( (n_1, n_2, n_3, n_4, n_5) \) is 126.