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.
Quick Answer
The number of different tests required is 126.
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