Problem Solving

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


oa coming when a few people respond with explanations. from difficult math doc.

800guy 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


oa coming when a few people respond with explanations. from difficult math doc.

excluding 0 the first digit can be written in 9 ways .... excluding the first digit the second digit can be written in 9 ways ... so on.... the answer is 9^5 .. C

gabriel wrote:

800guy 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


oa coming when a few people respond with explanations. from difficult math doc.

excluding 0 the first digit can be written in 9 ways .... excluding the first digit the second digit can be written in 9 ways ... so on.... the answer is 9^5 .. C

Hi Gabriel

I have a doubt. The question says that the no two consecutive digits are the same. If the second digit can be written in 9 ways, is there not a chance for repeating the first number. (Sorry if my question was stupid)

Prasanna

Prasanna wrote:
I have a doubt. The question says that the no two consecutive digits are the same. If the second digit can be written in 9 ways, is there not a chance for repeating the first number. (Sorry if my question was stupid)

Prasanna

For the first digit, we are excluding 0. For the second digit, we are
including 0 and excluding the first digit (as it reappears). So, the number
of ways remain at 9

Got it. I did not consider zero and hence went wrong. Thanks for the explanation.

oa:

C is correct.
The first place has 9 possibilities, since 0 is not to be counted. All others have 9 each, since you cannot have the digit, which is same as the preceding one.
Hence 9^5

800guy wrote:oa:

C is correct.
The first place has 9 possibilities, since 0 is not to be counted. All others have 9 each, since you cannot have the digit, which is same as the preceding one.
Hence 9^5

Hi All,

I have a small doubt in the solution. Should we consider consecutive numbers in the ascending order alone?

To explain my point further ... Say the 1st place was 7, the next cannot take 6 and 8 because 7-8 are consecutive digits and 7-6 are also consecutive but in the reverse order. In this case the solution would be "the first place can be filled in 9 ways excluding zero, after which the rest of the places can be filled in only 8 ways"

Should this option be considered? I have interpreted "consecutive" in a literal sense to refer to the immediately preceeding and following digits.

I am a newbie here ... so do excuse me if my query sounds silly.

Regards,
MJay

When you say consecutive it always means the next number. So if you put 7, 8 was your next number