Assume that a positive integer cannot have 0 as its leading digit. (a) How many five-digit positive integers have no repeated digits at all

there are 27,216 5 digit numbers such that there are no repeated digits at all and the leading digit is not zero.

Step-by-step explanation:

We can think of a five-digit number as:

a-b-c-d-e

Where each letter is a digit.

We want to find the number of combinations such that the digits can not be repeated, and that a can not be zero.

So let's count the number of options for each digit.

a can be {1, 2, 3, 4, 5, 6, 7, 8, 9}

So we have 9 options here.

for the second digit, b, we also have 9 options, because we took one in the previous selection, but now we also can have the number 0 as an option.

For the third digit, c, we have 8 options, because we already selected two.

For the fourth digit, d, we have 7 options, because we already selected 3.

For the last digit, e, we have 6 options, because we already selected 4.

Then the total number of combinations will be equal to the product between the number of options for each selection, this is:

C = 9*9*8*7*6 = 27,216

So there are 27,216 5 digit numbers such that there are no repeated digits at all and the leading digit is not zero.

answer: haha (: i’ll go see what it is

step-by-step explanation: