Q. The number of three-digit numbers having only two consecutive digits identical is
NTA AbhyasNTA Abhyas 2022
Solution:
Let us consider two cases which will occur according to the given question.
Case $1$ : First two digits will be identical
Case $2$ : Last two digits will be identical
Take case $1$ ,
Number of ways we can select a digit for hundredth place $ \rightarrow $ $9$ (Given that it is a three-digit number so $0$ cannot occur at this place)
Number of ways we can select a digit for ten's place $ \rightarrow $ $1$ ( $\because $ It has to be same as the digit at hundredth place)
Number of ways we can select a digit for unit place $ \rightarrow $ $9$ ( $\because $ It is different from digit at tenth place)
$\therefore $ Total number of three-digit numbers $=9\times 1\times 9=81$ .
Take case $2$ ,
Number of ways we can select a digit for hundredth place $ \rightarrow $ $9$ (Given that it is a three-digit number so $0$ cannot occur at this place)
Number of ways we can select a digit for ten's place $ \rightarrow $ $9$ ( $\because $ It is different from digit at hundredth place)
Number of ways we can select a digit for unit place $ \rightarrow $ $1$ ( $\because $ It has to be same as the digit at tenth place)
$\therefore $ Total number of three-digit numbers $=9\times 9\times 1=81$ .
Number of three-digit numbers having only two consecutive digits identical is $=81+81=162$ .
Case $1$ : First two digits will be identical
Case $2$ : Last two digits will be identical
Take case $1$ ,
Number of ways we can select a digit for hundredth place $ \rightarrow $ $9$ (Given that it is a three-digit number so $0$ cannot occur at this place)
Number of ways we can select a digit for ten's place $ \rightarrow $ $1$ ( $\because $ It has to be same as the digit at hundredth place)
Number of ways we can select a digit for unit place $ \rightarrow $ $9$ ( $\because $ It is different from digit at tenth place)
$\therefore $ Total number of three-digit numbers $=9\times 1\times 9=81$ .
Number of ways we can select a digit for hundredth place $ \rightarrow $ $9$ (Given that it is a three-digit number so $0$ cannot occur at this place)
Number of ways we can select a digit for ten's place $ \rightarrow $ $9$ ( $\because $ It is different from digit at hundredth place)
Number of ways we can select a digit for unit place $ \rightarrow $ $1$ ( $\because $ It has to be same as the digit at tenth place)
$\therefore $ Total number of three-digit numbers $=9\times 9\times 1=81$ .
Number of three-digit numbers having only two consecutive digits identical is $=81+81=162$ .