Page Watkins
02/14/2023 · Junior High School

When testing for current in a cable with nine color-coded wires, the author used a meter to test five wires at a time. How many different tests are required for every possible pairing of five wires? The number of tests required is

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To find the number of different tests required for every possible pairing of five wires out of nine, we can use the combination formula. The combination formula is given by: \[ C(n, k) = \frac{n!}{k!(n-k)!} \] Where: - \( n \) is the total number of items (in this case, 9 wires) - \( k \) is the number of items to choose at a time (in this case, 5 wires) - \( n! \) represents the factorial of \( n \), which is the product of all positive integers up to \( n \) In this case, we want to find the number of ways to choose 5 wires out of 9, so we will calculate \( C(9, 5) \). Let's calculate this using the combination formula. Simplify the expression by following steps: - step0: Solution: \(\frac{9!}{\left(5!\times \left(9-5\right)!\right)}\) - step1: Remove the parentheses: \(\frac{9!}{5!\times \left(9-5\right)!}\) - step2: Subtract the numbers: \(\frac{9!}{5!\times 4!}\) - step3: Expand the expression: \(\frac{9\times 8\times 7\times 6\times 5!}{5!\times 4!}\) - step4: Simplify: \(\frac{9\times 8\times 7\times 6}{4!}\) - step5: Multiply the terms: \(\frac{3024}{4!}\) - step6: Calculate the value: \(\frac{3024}{24}\) - step7: Reduce the fraction: \(126\) The number of different tests required for every possible pairing of five wires out of nine is 126.

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The number of different tests required is 126.
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