permutation & combinations How many odd three digit numbers are there when tens digit is greater than units digit and hundreds digit is greater than tens digit?


*

*$225$

*$ 45$

*$ 50$

*$230$



My attempt:
The units digit can be $1$ or $3$ or $5$ so $3$ ways ($9$ cannot be taken)
units digit when = 1       _ _ _1  ten's digit = 2,3,4,5,6,7,8  
in this if ten's digit = 2  hundred's digit =  3,4,5,6,7,8,9..  -- 7 ways
in this if ten's digit = 3  hundred's digit =  4,5,6,7,8,9...     6 ways
so on upto ten's = 8 hundreds = 9 --1 way  i.e, nothing but  sum of first 7 terms  i.e, 28
similarly for unit's= 3  ten's digit = 4,5,6,7,8     hundred's digit =  ,5,6,7,8,9..  -- 5 ways
so on upto ten's = 8 hundreds = 9 --1 way  i.e, nothing but  sum of first 5 terms  i.e, 15
similarly for unit's = 5 the ways can be  6  total 28+15+6 = 49
one number remained is 987 with this it is 50 ways.
calculating this take more time can anyone reduce this or is there any formula for this type of ques...

For a number to be odd, it must end with $1$ or $3$ or $5$ or $7$
Ending with 1: Let us fix $1$ at one's place now we have to fill ten's and hundred's place such that tens digit is greater than units digit and hundreds digit is greater than tens digit.
Placing $2$ at ten's place leaves us with $[3,4,5,6,7,8,9]$ i.e, $7$ choices.
Placing $3$ at ten's place leaves us with $[4,5,6,7,8,9]$ -- $6$ choices.
Placing $4$ at ten's places leaves us with $[5,6,7,8,9]$ i.e, 5 choices... so on.
No. of odd three digit numbers with 1 at one's place = $7+6+5+4+3+2+1 = 28$
Ending with 3: Fixing $3$ at one's place and filling ten's place with $4,5,6,7,8,9$ respectively leaves us with $5,4,3,2,1$ ways that sum up to $15$
Ending with 5: Fixing $5$ at one's place and filling ten's place with $6,7,8,9$ respectively leaves us with $3,2,1,0$ ways that sum up to $6$
Ending with 7: Fixing $7$ at one's place and filling ten's place with $8$ and hundred's place with $9$ gives us number $987$ with counts $1$ to total sum.
Total numbers : $28+15+6+1 = 50$
