Question

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

Answer 1

Answer:

The no of tests required shall be 495.

Explanation:

The possible pairing shall be the number of different ways in which 4 wires can be selected among a group of 12 different wires which is given by combinatorics as

$\binom{n}{r}=\frac{n!}{r!(n-r)!}$

Applying values we get

No of required ways are = $\binom{12}{4}=\frac{12!}{4!(12-4)!}=495$