Originally Posted by
Soroban Hello, prasum!
Let the four digits be: $\displaystyle \{a,\,b,\,c,\,d\}$
Select one of each digit: .$\displaystyle a\,b\,c\,d\,\_\,\_$
There are two cases to consider.
[1] The other two digits are the same: .$\displaystyle 4$ choices.
. . .The number is of the form: $\displaystyle a\,a\,a\,b\,c\,d$
. . .The digits can be permuted in: $\displaystyle {6\choose3} = 20$ ways.
. . .Hence, there are: .$\displaystyle 4 \times 20 \:=\:80$ such numbers.
[2] The other two digits are different: .$\displaystyle {4\choose2} \,=\, 6$ choices.
. . .The number is of the form: $\displaystyle a\,a\,b\,b\,c\,d$
. . .The digits can be permuted in $\displaystyle {6\choose2,2} = 180$ ways.
. . .Hence, there are: .$\displaystyle 6 \times 180 \:=\:1080$ such numbers.
The answer is: .$\displaystyle 80 + 1080 \:=\:1060$ numbers.