The number of positive integral solution of the inequality `x+y+zle20` is

Options

A. 1008
B. 1028
C. 1108
D. 1140

Answer

To find the number of positive integral solutions for the inequality \( x + y + z \leq 20 \), we can convert this into an equation by introducing a new variable. Let's denote the new variable as \( w \), such that:

\[ x + y + z + w = 20 \]

where \( w \) is a non-negative integer. Since we are looking for positive integral solutions for \( x, y, z \), we can express them in terms of new variables \( x', y', z' \) where:

\[ x = x' + 1, \quad y = y' + 1, \quad z = z' + 1 \]

Here, \( x', y', z' \) are non-negative integers. Substituting these into our equation gives:

\[ (x' + 1) + (y' + 1) + (z' + 1) + w = 20 \]

This simplifies to:

\[ x' + y' + z' + w = 17 \]

Now, we need to find the number of non-negative integral solutions to this equation. The number of solutions to the equation \( a_1 + a_2 + a_3 + \ldots + a_k = n \) in non-negative integers is given by the "stars and bars" theorem, which states that the number of solutions is:

\[ \binom{n + k - 1}{k - 1} \]

In our case, \( n = 17 \) and \( k = 4 \) (since we have \( x', y', z', w \)). Thus, the number of solutions is:

\[ \binom{17 + 4 - 1}{4 - 1} = \binom{20}{3} \]

Now, we calculate \( \binom{20}{3} \):

\[ \binom{20}{3} = \frac{20 \times 19 \times 18}{3 \times 2 \times 1} = \frac{6840}{6} = 1140 \]

Therefore, the number of positive integral solutions of the inequality \( x + y + z \leq 20 \) is \( \boxed{1140} \).