Question

Suppose that a "code" consists of 3 digits, none of which is repeated. (A digit is one of the 10 numbers 0, 1, 2, 3, 4, 5, 6, 7, 8, 9.) How many codes are possible?

Answer 1

There are 720 ways of codes are possible.

Explanation:

Given that the code consists of 3 digits, none of which is repeated.

We need to determine the codes that are possible.

The possible ways of codes can be determined using the permutation formula,

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

where n is the number of choices and r is the number chosen.

Hence, substituting $n=10$ and $r=3$, we have,

$^{10} P_{3}=\frac{10 !}{(10-3) !}$

Simplifying, we get,

$^{10} P_{3}=\frac{10 !}{7 !}$

       $=\frac{10 \times 9\times8\times7!}{7 !}$

Cancelling the common terms, we get,

$^{10} P_{3}=10\times9\times8$

Multiplying, we get,

$^{10} P_{3}=720$

Thus, there are 720 ways of codes are possible.