+0  
 
0
65
1
avatar

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

 Feb 14, 2020
 #1
avatar+109740 
+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

 

 

 

cool cool cool 

 Feb 14, 2020

65 Online Users

avatar
avatar
avatar