Question

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Find the number of solutions to
x_1 + x_2 + x_3 + x_4 + x_5 \le 1
in nonnegative integers.

Answer 1

To find the number of solutions to the inequality

\[
x_1 + x_2 + x_3 + x_4 + x_5 \leq 1
\]

in nonnegative integers, we can first transform the inequality into an equation. We can introduce a new variable \( x_6 \) such that

\[
x_1 + x_2 + x_3 + x_4 + x_5 + x_6 = 1
\]

where \( x_6 \) is also a nonnegative integer, representing the "slack" in the inequality. This means we want to count the number of nonnegative integer solutions to this equation.

The number of nonnegative integer solutions to the equation

\[
x_1 + x_2 + x_3 + x_4 + x_5 + x_6 = n
\]

for nonnegative integers \( x_1, x_2, x_3, x_4, x_5, x_6 \) can be found using the "stars and bars" combinatorial method. According to the stars and bars theorem, the number of solutions is given by:

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

where \( n \) is the sum we want (which is 1 in our case) and \( k \) is the number of variables (which is 6 in our case: \( x_1, x_2, x_3, x_4, x_5, x_6 \)).

Plugging in our values \( n = 1 \) and \( k = 6 \), we have:

\[
\binom{1 + 6 - 1}{6 - 1} = \binom{1 + 5}{5} = \binom{6}{5}
\]

Calculating \( \binom{6}{5} \):

\[
\binom{6}{5} = 6
\]

Thus, the number of nonnegative integer solutions to the inequality

\[
x_1 + x_2 + x_3 + x_4 + x_5 \leq 1
\]

is

\[
\boxed{6}.
\]

To verify, the nonnegative integer solutions can be listed explicitly:

1. \( (1, 0, 0, 0, 0) \)
2. \( (0, 1, 0, 0, 0) \)
3. \( (0, 0, 1, 0, 0) \)
4. \( (0, 0, 0, 1, 0) \)
5. \( (0, 0, 0, 0, 1) \)
6. \( (0, 0, 0, 0, 0) \)

Thus, we confirm that there are indeed \( 6 \) solutions.