mrsmarthi wrote:

I guess the answer would be D - 9 * 8^4 And here is the explanation

Let _ _ _ _ _ be five digits positive integers (Means they can range from 1 to 9. I call them as first digit, second digit etc starting from Left side.

So the first digit can be picked from 9 positive integers = 9

Since no two consecutive digits should be same, second digit can be picked from 8 different numbers(eliminating the number picked from first digit = 8

Similary third, fourth and fifth digits can be picked from 8 different numbers each(eliminating the number picked from previous digit ).

So the answer is 9 * 8 * 8 * 8 * 8 = 9 * 8^4.

First digit from 1 to 9 = 9 ways

Second digit from 0 to 9 but exclude one that is already in first digit = 9 ways

Third digit from 0 to 9 but exclude one that is already in second digit = 9 ways

Fourth digit is also from 0 to 9 but exclude one that is already in third digit = 9 ways

Fifth (final) digit is also from 0 to 9 but exclude one that is already in fourth digit = 9 ways

So the possibilities are 9x9x9x9x9 = 9^5 ways..

vivektripathi wrote:

How many 5-digit positive integers exist where no two consecutive digits are the same?

A.) 9*9*8*7*6

B.) 9*9*8*8*8

C.) 9^5

D.) 9*8^4

E.) 10*9^4

_________________

Verbal: http://gmatclub.com/forum/new-to-the-verbal-forum-please-read-this-first-77546.html

Math: http://gmatclub.com/forum/new-to-the-math-forum-please-read-this-first-77764.html

Gmat: http://gmatclub.com/forum/everything-you-need-to-prepare-for-the-gmat-revised-77983.html

GT