Consider making a four-digit I.D. number using the digits 3,5,8, and 0

a. How many I.D. numbers can be formed using each digit once?
b. How many can be formed using each digit once and not using 0 first?
c. How many can be formed if repetition is allowed and any digit can be first?
d. How many can be formed if repetition is allowed but 0 is not used first?

a) so we can use the 0 at the front

number of ways = 4x3x2x1 = 24
b) can't have zero at front
number of ways = 3x3x2x1 = 18
c) repeats are allowed and zero can be at front
number of ways = 4x4x4x4= 256
d) repeats allowed but no zero at front
number of ways = 3x4x4x4 = 192

a. Well, let's see. We have 4 digits - 3, 5, 8, and 0. Since we can use each digit only once, we have 4 options for the first digit, 3 for the second, 2 for the third, and 1 for the last. Therefore, the total number of I.D. numbers that can be formed is 4 x 3 x 2 x 1 = 24.

b. Ah, now this question adds a little twist. We still have 4 options for the first digit, but since we can't use 0 first, we only have 3 options for the second digit. For the third and fourth digits, we have 2 and 1 options respectively. Therefore, the total number of I.D. numbers that can be formed is 4 x 3 x 2 x 1 = 24, same as before.

c. Well, this question is a bit more lenient. We can have any digit as the first digit, so we have 4 options for that. For the second, third, and fourth digits, we still have 4 options each since repetition is allowed. Therefore, the total number of I.D. numbers that can be formed is 4 x 4 x 4 x 4 = 256.

d. Uh-oh, Mr. 0 is feeling a bit left out. But don't worry, we won't let him be the first digit. So we have 3 options for the first digit (3, 5, or 8), and since repetition is allowed, we have 4 options for the second, third, and fourth digits. Therefore, the total number of I.D. numbers that can be formed is 3 x 4 x 4 x 4 = 192.

Hope that brought a smile to your face!

To answer your questions step-by-step:

a. To find the number of I.D. numbers that can be formed using each digit once, we can use the concept of permutations. Since we have 4 digits, we need to find the number of ways to arrange these digits.

The number of ways to arrange n objects without repetition is given by n factorial (n!). In this case, n = 4. Therefore, the number of I.D. numbers that can be formed using each digit once is 4!.

To calculate this:

4! = 4 x 3 x 2 x 1 = 24

So, there are 24 possible I.D. numbers that can be formed using each digit once.

b. To find the number of I.D. numbers that can be formed using each digit once and not using 0 first, we can use the same concept as above. However, since we cannot use 0 as the first digit, we have one fewer option for the first digit.

The number of ways to arrange the digits, excluding 0 as the first digit, is given by (n-1) x (n-1) factorial. In this case, n = 4, so the number of I.D. numbers is (4-1) x 3!.

To calculate this:

(4-1) x 3! = 3 x 3 x 2 = 18

So, there are 18 possible I.D. numbers that can be formed using each digit once and not using 0 first.

c. To find the number of I.D. numbers if repetition is allowed and any digit can be first, we can use the concept of permutations with repetition. In this case, since repetition is allowed, we have 4 options for each digit.

The number of ways to arrange n objects with repetition is given by n to the power of r (n^r), where r is the number of positions to be filled. In this case, n = 4 and r = 4.

To calculate this:

4^4 = 4 x 4 x 4 x 4 = 256

So, there are 256 possible I.D. numbers that can be formed if repetition is allowed and any digit can be first.

d. To find the number of I.D. numbers if repetition is allowed but 0 is not used first, we can use the same concept as above. However, since we cannot use 0 as the first digit, we have 3 options for the first digit.

The number of possible arrangements, excluding 0 as the first digit, is given by 3 x 4^3.

To calculate this:

3 x 4^3 = 3 x 4 x 4 x 4 = 192

So, there are 192 possible I.D. numbers that can be formed if repetition is allowed but 0 is not used first.

a. To find the number of I.D. numbers that can be formed using each digit once, we need to use the concept of permutations. Permutation is a way to arrange objects where the order matters and repetitions are not allowed.

In this case, we have 4 digits: 3, 5, 8, and 0. Therefore, we have 4 options for the first digit, 3 options for the second digit (since one digit has been used), 2 options for the third digit, and only 1 option for the last digit.

To find the total number of arrangements, we multiply the number of options for each digit: 4 x 3 x 2 x 1 = 24.

Therefore, there can be 24 I.D. numbers formed using each digit once.

b. To find the number of I.D. numbers formed using each digit once and not using 0 first, we need to consider the same concept of permutations, but with a restriction on the first digit.

Since we cannot use 0 as the first digit, we have 3 options for the first digit (3, 5, and 8). We still have the same number of options for the remaining three digits: 3 options for the second digit, 2 options for the third digit, and 1 option for the last digit.

Using the same formula as before, we multiply the number of options for each digit: 3 x 3 x 2 x 1 = 18.

Therefore, there can be 18 I.D. numbers formed using each digit once and not using 0 first.

c. To find the number of I.D. numbers if repetition is allowed and any digit can be first, we need to use the concept of combinations. Combination is a way to select objects where the order doesn't matter and repetitions are allowed.

Since repetition is allowed, we have 4 options for each digit: 3, 5, 8, and 0. We have 4 digits in total.

To find the total number of combinations, we raise the number of options for each digit to the power of the total number of digits: 4^4 = 256.

Therefore, there can be 256 I.D. numbers formed if repetition is allowed and any digit can be first.

d. To find the number of I.D. numbers if repetition is allowed but 0 is not used first, we still need to use the concept of combinations, but with a restriction on the first digit.

Since we cannot use 0 as the first digit, we have 3 options for the first digit (3, 5, and 8). We still have 4 options for the remaining three digits: 3, 5, 8, and 0.

Using the same formula as before, we raise the number of options for each digit to the power of the total number of digits: 3 x 4 x 4 x 4 = 192.

Therefore, there can be 192 I.D. numbers formed if repetition is allowed but 0 is not used first.