You are setting a system programing codes for a universal TV remote control. Each code will have five digits and cannot include a 0 or repeat any digits. How many codes
a)do not contain a 7?
b)have 3 as the third digit, and 5 as the fifth digit
c)begin with an even digit?
d)begin and end with an even digit?
Please show all work.
Thank you for all the help that is provided:)
To find the number of codes that meet each condition, we can break down the problem into smaller steps.
a) How many codes do not contain a 7?
Since we cannot include 0 or repeat any digits, there are only 8 possible digits for each position (1-9 except for 7). Since each code has five digits, the number of codes that do not contain a 7 would be 8 * 8 * 8 * 8 * 8 = 8^5 = 32,768.
b) How many codes have 3 as the third digit and 5 as the fifth digit?
We have fixed positions for the third and fifth digits, so the other three positions can be any of the remaining digits. Again, there are 8 possible digits (1-9 except for 7), so the number of codes that meet the condition would be 1 * 1 * 1 * 8 * 1 = 8.
c) How many codes begin with an even digit?
Since the first digit must be even, we have 4 possible choices (2, 4, 6, 8). The remaining four digits can be any of the remaining 7 digits (1, 3, 4, 5, 6, 8, 9). Therefore, the number of codes that meet the condition would be 4 * 7 * 7 * 7 * 7 = 4 * 7^4 = 4 * 2401 = 9604.
d) How many codes begin and end with an even digit?
Similar to the previous condition, we have 4 choices for the first and last digit (2, 4, 6, 8). For the three middle digits, we have 7 choices each (1, 3, 4, 5, 6, 8, 9). Therefore, the number of codes that meet the condition would be 4 * 7 * 7 * 7 * 4 = 4^2 * 7^3 = 16 * 343 = 5488.
In summary:
a) Codes that do not contain a 7: 32,768
b) Codes with 3 as the third digit and 5 as the fifth digit: 8
c) Codes that begin with an even digit: 9,604
d) Codes that begin and end with an even digit: 5,488