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andrezito [222]
3 years ago
15

In a different plan for area​ codes, the first digit could be any number from 4 through 8​, the second digit was either 3, 4, 5,

or 6​, and the third digit could be any number exceptnbsp 2 or 5. with this​ plan, how many different area codes are​ possible?
Mathematics
1 answer:
Snezhnost [94]3 years ago
7 0

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 .

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