You are setting up a system of programming codes for a universal TV remote. Each code will have five digits and cannot include a 0 or repeat any digits. How many codes:

A) do not contain a 1?
My answer: 8x7x6x5x4 = 6720 codes

B)have 2 as the 2nd digit and 4 as the fourth digit?
- I don't know how to do this one

C)begin with an even number
My answer: 4x8x7x6x5 = 6720

D) begin and end with an even number
My answer: 4x3x8x7x6 = 4032

Are my answers correct... And how do I do B)

A - correct

B - I assume we can now use the 1?2?4? , so we can use only 7 other digitsnumber of ways= 7x1x6x1x5 = 210
c - correct
D - evens at front and end4x?x?x?x3 , leaves 7 for the other places= 4x7x6x5x3 = 2520
also the other person stole my answer

Your answer for part A) is correct.

B) To find the number of codes that have 2 as the second digit and 4 as the fourth digit, we can break down the problem into two steps.

Step 1: Fix the positions of 2 and 4.
In this step, we fix the positions for 2 and 4, so we have _ 2 _ 4 _.
Since there are no restrictions for the first, third, and fifth digits, we have 8 choices for each of those positions. So, there are 8x1x8 = 64 possibilities.

Step 2: Choose the remaining digits.
In this step, we need to choose three digits for the remaining three positions, excluding 2 and 4. Since we cannot have a 0 or repeat any digits, we have 7 choices for the first remaining digit, 6 choices for the second, and finally 5 choices for the last digit.

Multiplying the possibilities from both steps together, we get: 64x7x6x5 = 11,200 codes.

Therefore, the number of codes that have 2 as the second digit and 4 as the fourth digit is 11,200.

C) Your answer for part C) is incorrect. The correct calculation should be:

To find the number of codes that begin with an even number, we can consider two cases:

Case 1: The first digit is 2.
In this case, we have only one choice for the first digit, 2. For the remaining digits, we have 7 choices for the second digit, 6 choices for the third, 5 choices for the fourth, and 4 choices for the last digit. So, there are 1x7x6x5x4 = 840 codes.

Case 2: The first digit is 4, 6, or 8 (three even digits left).
In this case, we have 3 choices for the first digit. For the remaining digits, we have 8 choices for the second digit, 7 choices for the third, 6 choices for the fourth, and 5 choices for the last digit. So, there are 3x8x7x6x5 = 5040 codes.

Adding the number of codes from both cases together, we get 840 + 5040 = 5880 codes.

Therefore, the number of codes that begin with an even number is 5880.

D) Your answer for part D) is incorrect. The correct calculation should be:

To find the number of codes that begin and end with an even number, we can consider two cases:

Case 1: The first digit is 2.
In this case, we have only one choice for the first digit, 2. For the remaining digits (excluding the last digit), we have 7 choices for the second digit, 6 choices for the third, 5 choices for the fourth, and 3 choices for the fifth digit (as we cannot repeat the last even number). So, there are 1x7x6x5x3 = 630 codes.

Case 2: The first digit is 4, 6, or 8 (three even digits left).
In this case, we have 3 choices for the first digit. For the remaining digits (excluding the last digit), we have 8 choices for the second digit, 7 choices for the third, 6 choices for the fourth, and 3 choices for the fifth digit (as we cannot repeat the last even number). So, there are 3x8x7x6x3 = 3,024 codes.

Adding the number of codes from both cases together, we get 630 + 3,024 = 3,654 codes.

Therefore, the number of codes that begin and end with an even number is 3,654.

Your answers for parts A, C, and D are correct. Now let's move on to part B:

To find the number of codes that have 2 as the second digit and 4 as the fourth digit, we can break it down into steps:

Step 1: Place the fixed digits
Since 2 is the second digit and 4 is the fourth digit, we start with _ 2 _ 4 _. We have fixed two digits.

Step 2: Place the remaining digits
We have three remaining digits to place. We cannot use 0 or repeat any digits, so we have 8 choices for the first digit (since 1 is not allowed), 7 choices for the third digit, and 6 choices for the fifth digit.

Step 3: Multiply the choices
To find the total number of codes, we multiply the number of choices for each step.

Total number of codes = 8 * 1 * 7 * 1 * 6 = 336

Therefore, the correct answer for part B is 336 codes.

A - correct

B - I assume we can now use the 1
?2?4? , so we can use only 7 other digits
number of ways
= 7x1x6x1x5 = 210

c - correct

D - evens at front and end
4x?x?x?x3 , leaves 7 for the other places= 4x7x6x5x3 = 2520