How many 5 digit numbers can be made using the set 1 2 3 5 7 8 with no repeating?

Answer 1

There are 6 digits in the set, giving

A choice of 6 for the first digit, leaving

a choice of 5 for the second digit for each choice of the first, leaving

a choice of 4 for the third digit for each choice of the first two, leaving

a choice of 3 for the fourth digit for each choice of the first three, leaving

a choice of 2 for the fifth digit for each choice of the first four

making:

number of 5-digit numbers = 6 x 5 x 4 x 3 x 2 = 720 possible 5-digit numbers.

This is a choice where the order matters and is called a permutation. To calculate a permutation of r items from a set of n the formula:

nPr = n!/(n - r)!

will do it, where n! means "n factorial" and is calculated as the multiple of all digits less than n but greater than 0, ie:

n! = n x (n - 1) x (n - 2) x ... x 2 x 1

for example, 4! = 4 x 3 x 2 x 1 = 24. (0! - zero factorial - is defined to be 1, ie 0! = 1.)

In this question, n = 6 (set of 6 digits) and a permutation of 5 digits (for a 5-digit number) is required, thus the number of permutations (5-digit numbers) is:

6P5 = 6!/(6 - 5)!

= 6!/1!

= 6 x 5 x 4 x 3 x 2 x 1/1

= 720

If the question had been how many 4-digit numbers could be made from that set of 6 digits, then the answer would have been:

6P4 = 6!/(6 - 2)!

= 6 x 5 x 4 x 3 x 2 x 1/2 x 1

= 6 x 5 x 4 x 3

= 360