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...
 A: Each number must end with $1$ or $3$ or $5$ or $7$:


*

*In order to generate numbers that end with $1$, choose $2$ digits from $[2\dots9]$

*In order to generate numbers that end with $3$, choose $2$ digits from $[4\dots9]$

*In order to generate numbers that end with $5$, choose $2$ digits from $[6\dots9]$

*In order to generate numbers that end with $7$, choose $2$ digits from $[8\dots9]$


So the total amount of numbers is $\binom82+\binom62+\binom42+\binom22=28+15+6+1=50$.
A: How many solutions does $a>b>c$ have for $a,b,c\in\{0,1,2,3,4,5,6,7,8,9\}$?
Hint: for a given choice of $b$, how many choices do you have for $a$ and $c$? And how could it possibly be related to this strange sum?
A: 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 : $28+15+6+1 = 50$
A: 
the units digit can be 1or 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...
