Question

There are $10$ different books on the shelf. Find the number of ways in which $3$ books can be selected so that exactly two of them are consecutive.

Answer 1

If we take two consecutive books, at the end the third book can be chosen from $7$ different books. Now if we are taking two books from the middle and it must not be consecutive then the third book can be chosen from $6$ different books.

Complete step-by-step answer:
Here we are given that there are $10$ different books on the shelf. So let them be
${A_1},{A_2},{A_3},{A_4},{A_5},{A_6},{A_7},{A_8},{A_9},{A_{10}}$ books on the shelf. Now we are said to select $3$ books from these ten books such that two of them are consecutive.
Case (1): If two consecutive books are chosen from the end.
${A_1}{A_2},{A_2}{A_3},...........,{A_9}{A_{10}}$
So here two are the possibility that either we chose ${A_1}{A_2}$ or ${A_9}{A_{10}}$
If we chose${A_1}{A_2}$, then ${A_3}$ cannot be the third book because the question is saying that only two are consecutive. So we can choose the third book only in $7$ ways.
So the number of ways of choosing two consecutive books$= 2$
Number of ways of choosing third book$= 7$
So for the case (1)
Total number of ways$= 2 \times 7 = 14$
Case (2):
If consecutive books are not at both the ends.
\\_\\_\\_\\_\\_{A_4}{\text{ }}{A_5}\\_\\_\\_\\_\\_
Here we cannot take either ${A_1}{A_2}$ nor ${A_9}{A_{10}}$
So we can take ${A_2}{A_3},{A_3}{A_4},{A_4}{A_5},{A_5}{A_6},{A_6}{A_7},{A_7}{A_8},{A_8}{A_9}$
So the total number of pair$= 7$
Number of ways to choose two consecutive books in middle$= 7$
Number of ways to choose third book, here we cannot take ${A_3}{\text{ or }}{A_6}$
So the number of books that can be selected$= 6$
So total number of ways$= 6(7) = 42$
Now from case (1) and case (2), we got the total number of ways of choosing the third book in which exactly two of them are consecutive$= 14 + 42 = 56$

Note: We have different cases, now we are confused where we have to multiply and where to add. So if we have $a$ ways of doing something and $b$ ways of doing other things, then there are $a + b$ ways to choose one of them and we have $a \times b$ number of ways to do both of them.