Using the digits 0 through 9, find out how many 4-digit numbers can be configured based on the stated conditions: a. [1 pt] the number cannot start with zero and no digits can be repeated. b. [1 pt] the number must begin and end with an odd digit. (repeated digits are okay.) c. [1 pt] the number must be at least 5000 and be divisible by 10. (repeated digits are okay.) d. [2 pt] the number must be less than 3000 and must be even. no digits may be repeated in the last 3 digits. (that is, 2234 would be okay, but 2334 would not be okay.)

a. 4536, b. 2500, c. 500, d. 1080

Step-by-step explanation:

for every question we can use the numbers 0 - 9, this is 10 numbers in total (0, 1, 2, ,3 , 4, 5, 6, 7, 8, and 9).

Fore every case we have to check how many numbers we are allow to use in each digit.

"The number cannot start with zero" this left us with 9 options of numers for the first digit.

"no digits can be repeated." So if we already use one numer for the first, we can't repeat this in the second, and so on. So for the second numer we have 9 numbers as option (now we can use 0), for the third, since we already use 2 digits, we have 8 options, and for the last, we have 7

that results in: 9*9*8*7 = 4536

"The number must begin and end with an odd digit. " for odd digits we have 1, 3, 5, 7, 9 --> 5 options for the fist and last digit, and for the two middle digits, we have 10 options for each since it is allowed to repeat numbers

that results in: 5*10*10*5=2500

"The number must be at least 5000" So the first digit can use 5, 6, 7, 8, or 9 -->5 options for the fist digit

"and be divisible by 10." so it has to end in a 0, --> 1 option fot the last numer

and for the middle digits there are no restrictions, so we use the 10 options for each.

that results in: 5*10*10*1=500

"The number must be less than 3000 and must be even" for it to be less than 3000 it has to start with 0, 1 or 2 -> 3 options for the first digit

"and must be even" so the number has to end in 0, 2, 4, 6 or 8 --> 5 options for the last number

"No digits may be repeated in the last 3 digits." So we can't repeat in the middle number the one we put in the last digit, this gives us 9 options for the fist middle number and 8 for the second middle number

that results in: 3*8*9*5 = 1080

1

step-by-step explanation:

it's the only multiple of both and the least

10 you would take 50/ .2

step-by-step explanation: