The problem asks for the smallest non-negative value of the expression \(5 * 4 * 3 * 2 * 1\), where each asterisk '*' can be replaced by an operator from \(\{+, -, \times\}\). The constraint is that each operator (\(+\), \(-\), \(\times\)) can be used at most two times.

There are four asterisks in the expression \(5 * 4 * 3 * 2 * 1\), so we need to use exactly four operators in total. Let \(n_+\), \(n_-\), and \(n_\times\) be the number of times the operators \(+\), \(-\), and \(\times\) are used, respectively. The constraints are:

• \(n_+ + n_- + n_\times = 4\)
• \(0 \le n_+ \le 2\)
• \(0 \le n_- \le 2\)
• \(0 \le n_\times \le 2\)

We need to find a combination of operators that satisfies these conditions and, when placed in the expression, results in the smallest possible non-negative value following the standard order of operations (multiplication before addition and subtraction).

Exploring Operator Combinations

Let's list the possible combinations of \((n_+, n_-, n_\times)\) that satisfy the sum and individual limits:

• \((2, 2, 0)\): Two '+', two '-', zero '×'.
• \((2, 1, 1)\): Two '+', one '-', one '×'.
• \((2, 0, 2)\): Two '+', zero '-', two '×'.
• \((1, 2, 1)\): One '+', two '-', one '×'.
• \((1, 1, 2)\): One '+', one '-', two '×'.
• \((0, 2, 2)\): Zero '+', two '-', two '×'.

Any other combination summing to 4 would require at least one operator to be used more than twice (e.g., (3, 1, 0), (0, 3, 1), (1, 0, 3), etc.), which is not allowed.

Finding the Smallest Non-Negative Value

We need to test arrangements of operators for these combinations to find the smallest non-negative result. The goal is a value close to zero, ideally zero itself if possible.

Let's systematically explore the combinations and placements of operators in the expression \(5 \_ 4 \_ 3 \_ 2 \_ 1\), where the underscores represent the four operator slots.

Case 1: Two '+', Two '-', Zero '×' (\(n_+ = 2, n_- = 2, n_\times = 0\))

We need to place two '+' and two '-' in the four slots. Let's try some arrangements:

• \(5 + 4 + 3 - 2 - 1 = 9 + 3 - 2 - 1 = 12 - 3 = 9\)
• \(5 + 4 - 3 + 2 - 1 = 9 - 3 + 2 - 1 = 6 + 2 - 1 = 7\)
• \(5 + 4 - 3 - 2 + 1 = 9 - 3 - 2 + 1 = 6 - 2 + 1 = 5\)
• \(5 - 4 + 3 + 2 - 1 = 1 + 3 + 2 - 1 = 4 + 2 - 1 = 5\)
• \(5 - 4 + 3 - 2 + 1 = 1 + 3 - 2 + 1 = 4 - 2 + 1 = 3\)
• \(5 - 4 - 3 + 2 + 1 = 1 - 3 + 2 + 1 = -2 + 3 = 1\)

The smallest non-negative value found in this case is 1.

Case 2: One '+', Two '-', One '×' (\(n_+ = 1, n_- = 2, n_\times = 1\))

We need to place one '+', two '-', and one '×'. Remember multiplication takes precedence.

Let's consider where the '×' can be placed (between adjacent numbers) and then arrange the remaining operators.

• If \(\times\) is between 5 and 4: \(5 \times 4 \_ 3 \_ 2 \_ 1 \Rightarrow 20 \_ 3 \_ 2 \_ 1\). Remaining operators: +, -, -.
  • \(20 + 3 - 2 - 1 = 23 - 3 = 20\)
  • \(20 - 3 + 2 - 1 = 17 + 2 - 1 = 18\)
  • \(20 - 3 - 2 + 1 = 17 - 2 + 1 = 16\)
• If \(\times\) is between 4 and 3: \(5 \_ 4 \times 3 \_ 2 \_ 1 \Rightarrow 5 \_ 12 \_ 2 \_ 1\). Remaining operators: +, -, -.
  • \(5 + 12 - 2 - 1 = 17 - 3 = 14\)
  • \(5 - 12 + 2 - 1 = -7 + 2 - 1 = -6\) (Negative)
  • \(5 - 12 - 2 + 1 = -7 - 2 + 1 = -8\) (Negative)
• If \(\times\) is between 3 and 2: \(5 \_ 4 \_ 3 \times 2 \_ 1 \Rightarrow 5 \_ 4 \_ 6 \_ 1\). Remaining operators: +, -, -.
  • \(5 + 4 - 6 - 1 = 9 - 7 = 2\)
  • \(5 - 4 + 6 - 1 = 1 + 6 - 1 = 6\)
  • \(5 - 4 - 6 + 1 = 1 - 6 + 1 = -4\) (Negative)
• If \(\times\) is between 2 and 1: \(5 \_ 4 \_ 3 \_ 2 \times 1 \Rightarrow 5 \_ 4 \_ 3 \_ 2\). Remaining operators: +, -, -.
  • \(5 + 4 - 3 - 2 = 9 - 5 = 4\)
  • \(5 - 4 + 3 - 2 = 1 + 3 - 2 = 2\)
  • \(5 - 4 - 3 + 2 = 1 - 3 + 2 = -2 + 2 = 0\)

In this case, we found a value of 0: \(5 - 4 - 3 + (2 \times 1) = 5 - 4 - 3 + 2 = 1 - 3 + 2 = 0\). This uses one '×', two '-', and one '+', which fits the \((1, 2, 1)\) criteria.

Checking Other Combinations (Briefly)

Since we found a way to achieve 0, and 0 is the smallest non-negative number, it is highly likely that 0 is the smallest possible non-negative value. We don't need to exhaustively check all permutations for other combinations if we are confident 0 is achievable and non-negative.

• Two '+', Zero '-', Two '×' (\(n_+ = 2, n_- = 0, n_\times = 2\)): e.g., \(5 + 4 + 3 \times 2 \times 1 = 9 + 6 = 15\). Values are likely to be larger due to more multiplications and additions.
• One '+', One '-', Two '×' (\(n_+ = 1, n_- = 1, n_\times = 2\)): e.g., \(5 + 4 - 3 \times 2 \times 1 = 9 - 6 = 3\). e.g., \(5 - 4 + 3 \times 2 \times 1 = 1 + 6 = 7\). Smallest non-negative found here is 3.
• Zero '+', Two '-', Two '×' (\(n_+ = 0, n_- = 2, n_\times = 2\)): e.g., \(5 \times 4 - 3 - 2 \times 1 = 20 - 3 - 2 = 15\). Values are likely positive or negative. \(5 - 4 - 3 \times 2 \times 1 = 1 - 6 = -5\). Smallest non-negative found here is 15.

Comparing the smallest non-negative values found across different allowed operator combinations (1, 0, 15+, 3, 15+), the smallest value achieved is 0.

The expression \(5 - 4 - 3 + (2 \times 1)\) evaluates to 0, is non-negative, and uses the allowed operators within the given constraints (one \(\times\), two \(-\), one \(+\)).

Thus, the smallest non-negative value of the expression is 0.

Comparing the smallest non-negative values found (1, 4, 15, 0, 3, 15), the overall smallest non-negative value is 0.

Revision Table: Key Concepts

Additional Information: Operator Permutations

When solving problems like this, systematically exploring the possibilities is key. The number of ways to arrange operators increases with the number of slots and the number of distinct operators available. For \(n\) slots and \(k\) types of operators with counts \(c_1, c_2, ..., c_k\) such that \(\sum c_i = n\), the number of distinct arrangements is given by the multinomial coefficient \(\frac{n!}{c_1! c_2! ... c_k!}\). In this problem, for 4 slots and operator counts like (2, 2, 0), the number of arrangements is \(\frac{4!}{2!2!0!} = \frac{24}{2 \times 2 \times 1} = 6\). For (1, 2, 1), it's \(\frac{4!}{1!2!1!} = \frac{24}{1 \times 2 \times 1} = 12\). We need to evaluate each valid arrangement. However, due to the order of operations, placements of multiplication are often the most impactful and were explored first in the relevant cases.