Question

A lock has a three number code made up of 17 numbers if none of the numbers are allowed to repeat how many different ways can you choose three different numbers in order for unit code?

Answer 1

Answer:

680

Step-by-step explanation:

Use the binomial coefficient where you choose $k=3$ numbers out of $n=17$ possible numbers and find the total amount of combinations since order does not matter:

$\displaystyle \binom{n}{k}=\frac{n!}{k!(n-k)!}\\ \\\binom{17}{3}=\frac{17!}{3!(17-3)!}\\\\\binom{17}{3}=\frac{17!}{3!(14)!}\\\\\binom{17}{3}=\frac{17*16*15}{3*2*1}\\\\\binom{17}{3}=680$

Thus, you can make 680 three-non-repeating-number codes