Question

A four digit number is called a doublet if any of its digit is the same as only one neighbor. For example, 1221 is a doublet but 1222 is not a doublet. The number of such doublets are:
  $
  {\text{A}}{\text{. 2259}} \\
  {\text{B}}{\text{. 2268}} \\
  {\text{C}}{\text{. 2277}} \\
  {\text{D}}{\text{. 2349}} \\
  $

Answer 1

Hint: In order to calculate the total number of such doublets, we first write down the number of types of doublets that can be formed by 4 digits. Then we find the number of possibilities of forming each type of doublets. We pick the number of possibilities from 0 to 9, i.e. 10 digits.

Complete step-by-step answer:
Given Data,
A doublet is formed by a 4 digit number
Only a number and its neighbor should be the same.

We know that a 4 digit number forms a doublet and two adjacent numbers are repeated in it. So the total number of distinct numbers is a doublet is 3 or 2 (in case of two different number repetitions).
Let us consider a, b and c to be 3 distinct numbers. Therefore the number of ways of forming a doublet using these numbers is ‘abbc’, ‘bbac’, ‘acbb’, ‘aabb’.

Now the number of ways of forming each of these specific cases of doublets is given by –
We have a total of 10 natural numbers that exist ranging from 0 – 9.
abbc –
The number of ways of picking one digit from 9 available digits (because we cannot place 0 in the first position) for the first position is $ {}^9{{\text{C}}_1} = \dfrac{{9!}}{{\left( {9 - 1} \right)!}} = \dfrac{{{\text{9!}}}}{{8!}} = 9{\text{ ways}} $
The number of ways of picking one digit from 9 available digits (because we already picked one digit for the first position and we cannot pick it here) for the second position is $ {}^9{{\text{C}}_1} = \dfrac{{9!}}{{\left( {9 - 1} \right)!}} = \dfrac{{{\text{9!}}}}{{8!}} = 9{\text{ ways}} $
The number of ways of picking a digit for the third position is 1, as we have to pick the same digit in the second place.
The number of ways of picking one digit from 9 available digits (because we already picked one digit for the second and third positions and we cannot pick it here) for the fourth position is $ {}^9{{\text{C}}_1} = \dfrac{{9!}}{{\left( {9 - 1} \right)!}} = \dfrac{{{\text{9!}}}}{{8!}} = 9{\text{ ways}} $
Therefore the total number of ways of making a doublet of the form ‘abbc’ is $ {9^3} $ ways.

bbac –
The number of ways of picking one digit from 9 available digits (because we cannot place 0 in the first position) for the first position is $ {}^9{{\text{C}}_1} = \dfrac{{9!}}{{\left( {9 - 1} \right)!}} = \dfrac{{{\text{9!}}}}{{8!}} = 9{\text{ ways}} $
The number of ways of picking a digit for the second position is 1, as we have to pick the same digit in the first place.
The number of ways of picking one digit from 9 available digits (because we already picked one digit for the first and second position and we cannot pick it here) for the third position is $ {}^9{{\text{C}}_1} = \dfrac{{9!}}{{\left( {9 - 1} \right)!}} = \dfrac{{{\text{9!}}}}{{8!}} = 9{\text{ ways}} $
The number of ways of picking one digit from 9 available digits (because we already picked one digit for the third position and we cannot pick it here) for the fourth position is $ {}^9{{\text{C}}_1} = \dfrac{{9!}}{{\left( {9 - 1} \right)!}} = \dfrac{{{\text{9!}}}}{{8!}} = 9{\text{ ways}} $
Therefore the total number of ways of making a doublet of the form ‘bbac’ is $ {9^3} $ ways.

acbb –
The number of ways of picking one digit from 9 available digits (because we cannot place 0 in the first position) for the first position is $ {}^9{{\text{C}}_1} = \dfrac{{9!}}{{\left( {9 - 1} \right)!}} = \dfrac{{{\text{9!}}}}{{8!}} = 9{\text{ ways}} $
The number of ways of picking one digit from 9 available digits (because we already picked one digit for the first position and we cannot pick it here) for the second position is $ {}^9{{\text{C}}_1} = \dfrac{{9!}}{{\left( {9 - 1} \right)!}} = \dfrac{{{\text{9!}}}}{{8!}} = 9{\text{ ways}} $
The number of ways of picking one digit from 9 available digits (because we already picked one digit for the second position and we cannot pick it here) for the third position is $ {}^9{{\text{C}}_1} = \dfrac{{9!}}{{\left( {9 - 1} \right)!}} = \dfrac{{{\text{9!}}}}{{8!}} = 9{\text{ ways}} $
The number of ways of picking a digit for the fourth position is 1, as we have to pick the same digit in the third place.

Therefore the total number of ways of making a doublet of the form ‘acbb’ is $ {9^3} $ ways.

aabb –
The number of ways of picking one digit from 9 available digits (because we cannot place 0 in the first position) for the first position is $ {}^9{{\text{C}}_1} = \dfrac{{9!}}{{\left( {9 - 1} \right)!}} = \dfrac{{{\text{9!}}}}{{8!}} = 9{\text{ ways}} $
The number of ways of picking a digit for the second position is 1, as we have to pick the same digit in the first place.
The number of ways of picking one digit from 9 available digits (because we already picked one digit for the second position and we cannot pick it here) for the third position is $ {}^9{{\text{C}}_1} = \dfrac{{9!}}{{\left( {9 - 1} \right)!}} = \dfrac{{{\text{9!}}}}{{8!}} = 9{\text{ ways}} $
The number of ways of picking a digit for the fourth position is 1, as we have to pick the same digit in the third place.
Therefore the total number of ways of making a doublet of the form ‘aabb’ is $ {9^2} $ ways.

Any other form of a doublet we consider, falls into one of the above cases.
Therefore the total number of possibilities of forming a doublet is = $ {9^3} + {9^3} + {9^3} + {9^2} = 2268{\text{ ways}} $
Option B is the correct answer.

Note: In order to solve this type of questions the key is to carefully think and list out all the possible ways a doublet which has 4 digits can be arranged. It is an important step to calculate the number of possibilities for each position carefully. Zero cannot take the first position in order to keep the number a 4 digit one.
Combination is a way of arranging a certain number of things in the available places, where order is not important. It is given by the formula, $ {}^{\text{n}}{{\text{C}}_{\text{r}}} = \dfrac{{{\text{n}}!}}{{\left( {{\text{n}} - {\text{r}}} \right)!}} $
n! means the factorial of a number n and is given by n! = n (n-1) (n-2) ……….. (n – (n-1)).