<s>[INST] The number of 3 digit numbers of the form xyz such that $x
The number of 3 digit numbers of the form $xyz$ such that $x<y$ and $z\leq y$ is N such that n is a 3 digit number of the form $abc$ then find $a+c-b$     
My approach:
Case 1:
$x<y$
$z<y$
$x,y,z\neq 0$
Choosing 3 digits out of 9 numbers: $\binom 93$, greatest of them will always be $y$, hence ways of arranging is (exchanging x and z) $2\cdot\binom 93=54$. 
Case 2:
$x,y$ can't be $0$. Taking $z=0$, number of ways of selecting the other 2 digits are $\binom 92=36$. There is only one way to arrange them.      
case 3:
$y=z$,
_99, values for $x=8$
_88, values for $x=7$
_77, values for $x=6$
.
.
.
_11, values for $x=0$
Total possibilities=$8+7+6+\cdots+1=36$       
Hence total permutations=$54+36+36=126$     
This gives $a+c-b=5$ which is not the right answer. Why is this logic wrong?
 [/INST] Your case 1 needs supplementing with case 1a) $x=z$, which gives $\binom 92$ additional cases, with the larger choice $=y$ and the smaller $=x=z$ -- another $36$ cases.
Then $$\overset{\text{case 1}}{2\cdot \binom 93} + \overset{\text{case 1a}}{\binom 92} + \overset{\text{case 2}}{\binom 92} + \overset{\text{case 3}}{\binom 92}
= 2\cdot 84 +3\cdot 36 = 168+108 = 276$$
giving your final calculation of $2+6-7 = 1$
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