Let xyz is 3 digits number.

Given that sum of digits = 10

$$ \therefore $$ x + y + z = 10 ......(1)

Also x can't be 0 as if x = 0 then it will become 2 digits number.

So, x $$ \ge $$ 1, y $$ \ge $$ 0, z $$ \ge $$ 0

As x $$ \ge $$ 1

$$ \Rightarrow $$ x $$-$$ 1 $$ \ge $$ 0

Let x $$-$$ 1 = t

$$ \therefore $$ t $$ \ge $$ 0

From equation (1)

(x $$-$$ 1) + y + z = 9

$$ \Rightarrow $$ t + y + z = 9

Now this problem becomes, distributing 9 things among 3 people t, y, z.

Number of ways we can do that

= $${}^{9 + 3 - 1}{C_{3 - 1}} = {}^{11}{C_2} = 55$$

Now when 3 digit number is 900 then t = 9, y = 0, z = 0.

And when t = 9, then

x $$-$$ 1 = 9

$$ \Rightarrow $$ x = 10

But we can't take x = 10 in a 3 digits number. So, we have to remove this case.

$$ \therefore $$ Total number of 3 digit numbers = 55 $$-$$ 1 = 54.