Question

Number of three digits numbers of the form xy with x < y, z < and x= 0,y,z will be

Answer 1

Case 1:-

$x<y$

$z<y$

$x,y,z\ne 0$

Choosing three digits out of 9 numbers then

$9\times 3=27$

Then, ways of arranging is

$2\times 27=54$.

Case 2:-

$x,y$ can’t be $0$. Taking $z=0$, number of ways of selecting the other two digits are

$2\times 2\times 9=36$

Only one way to arrange them.

Case 3:-

$y=z,$

$-99,valuesforx=8$

$-88,valuesforx=7$

$-77,valuesforx=6$

$................$

$-11,valuesforx=0$

Then, total possibilities $=8+7+5+-----+1=36$

Hence the total permutations is $54+36+36=126$

Hence, this is the answer.