SOLUTION: Assume a telephone number consists of 8 digits, and the first and last digits cannot be 0. (a) If repetitions are allowed, how many telephone numbers are possible if (i) there is

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Question 406031: Assume a telephone number consists of 8 digits, and the first and last digits cannot be 0.
(a) If repetitions are allowed, how many telephone numbers are possible if
(i) there is no restriction;
(ii) they are odd and start with 2?
(b) If repetitions are not allowed, how many telephone numbers are possible if
(i) there is no restriction;
(ii) they are divisible by 3 and the first digit is less than 5.

Answer by sudhanshu_kmr(1152)   (Show Source): You can put this solution on YOUR website!

A) (i) for first and last position any digit except 0, for other any of 10 digits
total no. of possible numbers = 9*10*10*10*10*10*10*9 = 81000000 = 81*10^6
(ii)
for odd numbers odd number must in at unit place....
total no. of odd numbers = 5
we can put digit 2 at first place by 1 way.
total no. of possible numbers = 1*10*10*10*10*10*10*5 = 5000000 = 5 * 10^6



B) if repetitions are not allowed...

(i) as first place and last can't be zero,
no. of ways to put digit at first and last position = 9*8 = 72

( now we have 8 digits and 6 positions to fill )
no. of ways to put digit at remaining 6 positions = 8P6 = 20160
total no. of ways = 72*20160 = 1451520




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