Question

A lock has a code of 5 numbers from 1 to 20. If no numbers in the code are allowed to repeat, how many different codes could be made?

Answer 1

HI,

Explanation:

We must choice 5 numbers in a set of 20

number of possibilities=
`\left(\begin{array}{c}20\\ 5\end{array}\right)=(20*19*18*17*16)/(1*2*3*4*5)`

Each possibility gives 5!=5*4*3*2*1=120 possibilities.

Number of different codes=20*19*18*17*16=826880.

Answer 2

1,860,480

Explanation:

There are 20 options for the first number.

Since there's no repetition, there are now 19 options for the second number.

Similarly, there are 18 options for the third number, 17 options for the fourth number, and 16 options for the final number.

So the number of permutations is:

20 × 19 × 18 × 17 × 16

= 1,860,480