Answer
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Hint: As we see it is a basic question of permutation and combination and we have to make a four digit number by using digits 0,1,2,3,4,5. And we see digits contain 0 so we can not fill 0 at a thousand places and rest we can arrange in such a manner so that number gets repeated.
Complete step-by-step answer:
The thousand's place cannot be filled by 0, so the number of ways to fill the thousand's place is 5.
The remaining 3 places can be filled by 6 numbers in $6 \times 6 \times 6$ ways (as numbers can be repeated).
Thus the number of 4 digit number = $5 \times 6 \times 6 \times 6$
Number of 4 digit number is $1080$
The number of 4 digit numbers that do not contain any repeated digits = ${}^5{P_1} \times {}^5{P_3}$ --- (1)
(a thousand's place is to be filled by one of 1,2,3,4,5 and the remaining 3 places are to be filled by 3 of the remaining 5 digits including 0).
But equation 1 contains numbers which contain no repeated digits as well as which contain at least one repeated digit.
The number of 4 digit numbers which contain at least one repeated digit =$1080 - {}^5{P_1} \times {}^5{P_3}$
Now how to solve permutation
${}^n{P_r} = \dfrac{{n!}}{{\left( {n - r} \right)!}}$
${}^5{P_1} = \dfrac{{5!}}{{\left( {5 - 1} \right)!}}$ = $\dfrac{{5 \times 4 \times 3 \times 2 \times 1}}{{4 \times 3 \times 2 \times 1}}$ = $5$
Similarly
${}^5{P_3} = \dfrac{{5!}}{{\left( {5 - 3} \right)!}}$ = $\dfrac{{5 \times 4 \times 3 \times 2 \times 1}}{{2 \times 1}}$ = $5 \times 4 \times 3 = 60$
So number of 4 digit numbers which contain at least one repeated digit = $1080 - \left( {5 \times 60} \right)$
$ \Rightarrow 1080 - 300$
$ \Rightarrow 780$
So number of 4 digit numbers which contain at least one repeated digit is $780$
So option C is the correct answer.
Note: As I say in hint we can note take 0 at thousand place because when we take 0 at thousand place it is a three digit number example $0123$ is a three digit number. We can also see that the question talks about at least one digit repeat not all digits repeat these both terms are different.
Complete step-by-step answer:
The thousand's place cannot be filled by 0, so the number of ways to fill the thousand's place is 5.
The remaining 3 places can be filled by 6 numbers in $6 \times 6 \times 6$ ways (as numbers can be repeated).
Thus the number of 4 digit number = $5 \times 6 \times 6 \times 6$
Number of 4 digit number is $1080$
The number of 4 digit numbers that do not contain any repeated digits = ${}^5{P_1} \times {}^5{P_3}$ --- (1)
(a thousand's place is to be filled by one of 1,2,3,4,5 and the remaining 3 places are to be filled by 3 of the remaining 5 digits including 0).
But equation 1 contains numbers which contain no repeated digits as well as which contain at least one repeated digit.
The number of 4 digit numbers which contain at least one repeated digit =$1080 - {}^5{P_1} \times {}^5{P_3}$
Now how to solve permutation
${}^n{P_r} = \dfrac{{n!}}{{\left( {n - r} \right)!}}$
${}^5{P_1} = \dfrac{{5!}}{{\left( {5 - 1} \right)!}}$ = $\dfrac{{5 \times 4 \times 3 \times 2 \times 1}}{{4 \times 3 \times 2 \times 1}}$ = $5$
Similarly
${}^5{P_3} = \dfrac{{5!}}{{\left( {5 - 3} \right)!}}$ = $\dfrac{{5 \times 4 \times 3 \times 2 \times 1}}{{2 \times 1}}$ = $5 \times 4 \times 3 = 60$
So number of 4 digit numbers which contain at least one repeated digit = $1080 - \left( {5 \times 60} \right)$
$ \Rightarrow 1080 - 300$
$ \Rightarrow 780$
So number of 4 digit numbers which contain at least one repeated digit is $780$
So option C is the correct answer.
Note: As I say in hint we can note take 0 at thousand place because when we take 0 at thousand place it is a three digit number example $0123$ is a three digit number. We can also see that the question talks about at least one digit repeat not all digits repeat these both terms are different.
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