Question

Find the number of integers from 0 to 999999 that have no two equal neighbouring digits in their decimal representation.
(a) $10+{{9}^{2}}+{{9}^{3}}+{{9}^{4}}+{{9}^{5}}+{{9}^{6}}$
(b) $9+{{9}^{2}}+{{9}^{3}}{{9}^{4}}{{9}^{5}}{{9}^{6}}$
(c) ${{9}^{2}}+{{9}^{3}}+{{9}^{4}}+{{9}^{5}}+{{9}^{6}}$
(d) None of these

Answer 1

Hint: We will use permutation here. By the term of permutation we mean taking out numbers of choices by placing the numbers into the form like so, for two digits we will form – and place the choices for each space created. Similarly, for other digits also we will do the same. With the help of this we will be able to solve the question.

Complete step-by-step answer:
According to the question we need to find the integers from 0 to 999999. But there is a restriction here. The restriction is that no two equal neighbouring digits should be in their decimal representation. We cannot find the numbers in a bunch. So, we will divide the numbers from 0 to 999999 into sets one digit number, two digit numbers, three digit numbers and so on.
So, the numbers that we will consider first are from 0 to 9 which are all single digit numbers. Now we can see that there are no numbers which have common neighbouring digits in them. This is also because we are considering here the single digits. Thus, the count of the numbers without any repetitive neighbouring digits in their decimal representation is 10.
Now, we will consider two digit numbers. These start from 10 to 99. So, clearly the form of these types of numbers will be -. We will use the concept of permutation and combination here. For the first place we will consider numbers from 1 to 9 only. So, we have choices for the first place is 9. Now the second place will be filled by choices from 0 to 9 but without repetition. So, the choices are 9. Hence, $9\times 9={{9}^{2}}$.
Now, we will consider three digit numbers. So, now we will consider 100 numbers here. The numbers that we will consider here are from 100 to 999. We will use the concept of permutation and combination here. As there are 3 digit numbers so we have -- form. Now the first number can be any number without 0. Therefore there are 9 digits only that can be placed here. In the second space we will have the numbers from 0 to 9 but without repetition that is 9 only. Now finding the choices for the last place is a bit tricky here. As we need to form those numbers in which no two equal neighbouring digits in their decimal representation. So because of this the third place will be filled by 9 choices instead of 8. . So, the choices are 9. Hence, $9\times 9\times 9={{9}^{3}}$.

Similarly, for four digit numbers starting from 1000 to 9999 we have the required count as ${{9}^{4}}$. And for the numbers 10000 to 99999 the choices are ${{9}^{5}}$. Moreover, the count of numbers starting from 100000 to 999999 is ${{9}^{6}}$. Therefore, by adding all these counts we will get that the number of integers from 0 to 999999 that have no two equal neighbouring digits in their decimal representation is $10+{{9}^{2}}+{{9}^{3}}+{{9}^{4}}+{{9}^{5}}+{{9}^{6}}$.
Hence, the correct option is (a).

Note: The concept of permutation here is due to the fact that there are so many numbers to deal with. And after dividing the numbers from 0 to 999999 into single digit numbers, two digit numbers, three digits numbers and so on will give us the usage of permutations even more. Also, the counting of the numbers is difficult. For example even after dividing the numbers into sections and we consider two digits numbers then the numbers from 10 to 99 cannot be found. Here, it should be noted that the numbers can be found easily from 10 to 99 by just doing 99 – 10 – 1 which are 88. But we should not forget that we are restricted about choosing the numbers. Thus, this calculation is wrong here.