For the first digit, we have 5 options that are 4,5,6,7,8 . For the second digit, we have 4 options which are 3,4,5 or 6 and for the third digit, we have the options of all numbers except 2 or 5 that is 1,3,4,6,7,8,9,0 . SO we have 8 options for third digit . So to find the total number of options, we need to multiply all the possible options for each digit that is 5 times 4 times 8 = 160 . So the number of possible options are 160 .
The possible area codes for different plan are [tex]160[/tex].
Step-by-step explanation:
Given: In a different plan for area codes, the first digit could be any number from [tex]4[/tex] through [tex]8[/tex], the second digit was either [tex]3,\;4,\;5,\;\rm{or}\;6[/tex] and the third digit could be any number except [tex]2\;\rm{or}\;5[/tex].
As mentioned in question:
Area codes are created by digits as:
first digit could be any number from [tex]4[/tex] through [tex]8[/tex] are [tex]4,\;5,\;6,\;7,\;8[/tex].
Second digit are [tex]3,\;4,\;5,\;\rm{or}\;6[/tex]
And third digit could be any number except [tex]2\;\rm{or}\;5[/tex] are [tex]1,3,4,6,7,8,9,0[/tex].
Total number of area codes are [tex]5\times4\times8=160[/tex].
Hence, the possible area codes are [tex]160.[/tex]
Learn about possible codes here:
https://brainly.com/question/23225459?referrer=searchResults
