How many 6 digit positive integers exist such that no two consecutive

Question

How many 6 digit positive integers exist such that no two consecutive digits are the same?

A. 10*9*8*7*6*5
B. 9*8*7*6*5*4
C. 9*9*8*7*6*5
D. 9*9*8^4
E. 9^6

AdlaT · Answer

no two consecutive integers are same, so the digit is in the form of  XYXYXY .

so leftmost digit can be any digit but 0. we can choose in 9 ways. next digit can be chosen in 9 ways, remaining each digit can be chosen in 9 ways.

total digits are  9*9*9*9*9*9 = 9^6

Ans. E

PolarCatastrophe · Answer

Just to play Devil's Advocate for the learning process. Let's say you build your formula using the ones place first.

You'd have:  10*9*9*9*9*8  =  9^4*8*10  which is one less than  9^6 . Where'd the missing case go!? =)

KarishmaB · Answer

If you start from the ones place, when you reach the fifth digit and give it 9 options, you include 0 as one of the possibilities. Now, how many possibilities does the sixth place (leftmost digit) have depends on what the fifth place is. If it is something other than 0, there are 8 options for the sixth place. If it is 0, there are 9 options for the sixth place.

Hence, you must start from the leftmost digit to keep matters simple.

sudhirmadaan · Answer

please let me know whether i missed anything.
if you start from left to right 
left most will have 9 case to fill as 0 cannnot be at right most , next to left most is again 9 , leaving the one which is already at left most place and including 0 now.. similarly till right most place you will to chance to fill again 9 leaving by leaving the one which used on just next to it.. 9^6

but if i start with right most digit , right most i can fill with 10 digit. next to that will be 9 cases and next to that will again be 9 cases ..and go on ...leads to 9^5 *10 .....

please correct me what missed in this

KarishmaB · Answer

Look at the post right above yours.

sudhirmadaan · Answer

so is it a practice to start always from left most part in such questions or we have some rule of this one.

PolarCatastrophe · Answer

I don't think it's a good idea to slate this as a "rule". The way this problem is structured it is definitely better to start from the left due to the confusion that arises from the zero as the second digit coming from the right. However, I'd like to say it is possible that a problem could be constructed where starting from the left would result in the same problem. Say for example it is stated that the ones digit "can not be zero" in this problem. Now, which way are you going to go? The idea is to understand WHY this happens and the best way to do that is to throw away any idea of "rules".

KarishmaB · Answer

It is not a rule but a matter of convenience based on the type of question. Questions requiring you to create number with fixed number of digits say 5 digit number will have the constraint that the leftmost digit cannot be 0. So you take care of that and then move forward. Even if it doesn't come to mind and you start from right, by the time you hit the leftmost digit, you will realise the complications involved. So you will switch to starting with the leftmost digit. A question requiring you to create a number of at most 5 digits will not have this constraint. So you needn't start from the left. Your technique will vary according to the needs of the question.
The more you practice with different questions, the more techniques you will come across and that is how you will develop an intuition for what works with what.

arvind910619 · Answer

It is little tricky question 
No two consecutive integers can be same but the next integer can take any value 
First digit can not have 0 so we 9 choices next will also have 9 choices leaving the first digit as 0 can now be used so therefore again we have 9 choices
9*9*9*9*9*9=9^6

energetics · Answer

Breaking it down into steps:

First digit can be anything but 0, so 9 choices
Second digit can be anything but the first digit, so 9 choices (since we can use 0 now)
Third digit can be anything but the second digit, so 9 again (we can use the first digit again)
Fourth digit can be anything but the 3rd digit, so 9 again, etc...
So it's 9*9*9*9*9*9

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How many 6 digit positive integers exist such that no two consecutive

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How many 6 digit positive integers exist such that no two consecutive

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