How many 7-digit telephone numbers are possible if the first digit cannot be zero and

a. every other digit is odd

b. the telephone number must start with 226 and end with 5

c. no repetitions are allowed

Guest Feb 14, 2020

#1**+1 **

How many 7-digit telephone numbers are possible if the first digit cannot be zero and

a. every other digit is odd

If every other digit is odd either :

positions 1 - 3 - 5 -7 are odd or positions 2 - 4 - 6 are odd

Assuming that the other positions must be even we have

(5 odd)(5 even)(5 odd) (5 even) (5 odd) (5 even) (5odd) = 5^7 or

(4 even- we can't have a zero) (5 odd) (5even) ( 5 odd) (5 even) (5 odd) (5 even) = 4 * 5^6

So....the total number = 5^7 + 4*5^6 = 140,625

b. the telephone number must start with 226 and end with 5

We have (10)^3 = 1000 possible numbers here [ positions 4, 5, 6 can be even or odd ]

c. no repetitions are allowed

(9 ways to choose the first number - zero is not allowed) ( 9 ways to choose the second number, we can choose 0 or any of the other 8 digits we didn't choose in the first position) * ( 8) * (7) *(6) * (5) * (4) =

(9) (9) (8)(7)(6)(5)(4) = 544,320

CPhill Feb 14, 2020