How many 4-digit positive integers are there, where each

Question

How many 4-digit positive integers are there, where each digit is positive, and no two adjacent digits are same?

A. 1236
B. 3024
C. 4096
D. 4608
E. 6561

dave785 · Answer

I'm amazed I actually got this one right..

First digit.. 9 posibilities
Second digit, 8 possibilities
third digit, 8 possibilities
Fourth digit, 8 possibilities.

9*8*8*8 = 4608.

But that can't be right... for instance, let's say that the number is 1212 ... if the first and third digits are the same, there could be 9 possibilities for the second digit.

avinashrao9 · Answer

That cannot be true.
The 2nd digit has 8 possibilities because it excludes the 1st digit out of the total 9 digits possible.
Similarly, 3rd digit has 8 possibilities because it exclues the 2nd digit out of the total 9 digits possible.

Hope it's clear !!!

MacFauz · Answer

Every digit has to be positive.. So "0" would not be valid... In 1212, the second digit can only be 2,3,4,5,6,7,8 or 9...

pradhan · Answer

First digit takes numbers from 1 to 9 --- 9 nos
second digit onwards 0 to 10 --- 10 nos......but to avoid repetition of the previous one ----- 9 nos

hence total nos = 9*9*9*9 =  6561

mattce · Answer

Just to elaborate a little since there was a bit of confusion up there:

We need 4-digit numbers, where each digit is positive. There are 9 positive digits.

Starting from right to left, we have:

9 possibilities * 
(9 - the digit just used = 8 possibilities) * 
(Again, 9 - the digit just used) = 8 possibilities) *
(Again, 9 - the digit just used) = 8 possibilities)
= 9 * 8 * 8 * 8 = 4608.

If anyone is thrown off by the fact that if the first and third numbers are different, the second digit only has 7 possibilities, this should clarify:

Suppose we didn't choose numbers from left to right, and consider the problem in the following way:

-(1)- -(2)- -(3)- -(4)-

(1) has 9 possibilities; nothing has been chosen
(3) has 9 possibilities; nothing has been chosen
(4) has 8 possibilities; it cannot be the same as (3).
(2) now has 7 possibilities right? since (1) and (3) could be different?   Nope! .. Well, sort of.

Here, we have to split the problem into two scenarios:
Case A: 1 and 3 have the same digit --> there are 8 options for the second digit.
Case B: 1 and 3 have different digits --> there are 7 options for the second digit.

So, adding up the two cases, we get: (Case A: 9 * 8 * 1 * 8) + (Case B: 9 * 7 * 8 * 8) = 9 * 8 * 8 * 8 = again, 4608.

Same logic applies to any way that you want to consider the problem.

Apologies if that was obvious to everyone else; I was thrown off for a second before realizing how it worked, so thought I'd post just in case anyone else had the same confusion.

dave785 · Answer

Ah that makes sense. thank you.

What is the source of this question?

Posted from my mobile device

onindokamboz · Answer

why we are not including 0 in second and third place

KarishmaB · Answer

Because you have been asked to use only positive digits. 0 is neither negative nor positive.

deya · Answer

I got some different figure. My approach is as follows: There can be 4 scenarios.
First, 1212----> 9*8*1*1 = 72
Second, 1534-----> 9*8*7*6 = 3024
Third, 1231-------> 9*8*7*1 = 504
Fourth, 2131------> 8*9*7*1 = 504
Total= 72+3024+504+504 = 4104. I know this is wrong. Kindly shed some light on it.

Thank You in advance. :-D

KarishmaB · Answer

You forgot a case: 
Fifth, 2325 (1st and 3rd digits are the same, other two distinct) ---------> 9*8*1*7 = 504

Now the total is 4104 + 504 = 4608

But note that this is a very convoluted approach. Try to understand the simpler approach mentioned above.

How many 4-digit positive integers are there, where each

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