# 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...

• What are your thoughts? a and d are very close together, but very far apart from b and c. Why do you think this is? – Sloan Mar 24 '15 at 14:35
• Please note that copy-pasting of homework problems is frowned upon here. You should at least tell us what you tried to solve it. – A.P. Mar 24 '15 at 14:36
• i had got answer 50 through my calculaton but it is too time taking and i cant conform it whether it is correct or not – sanjana Mar 24 '15 at 15:22
• just try with --- 3 digit where last can be filled by 1 or 3 or 5 only and so on – sanjana Mar 24 '15 at 15:24
• $987$ fulfils all conditions, but its unit digit is none of $1$, $3$ or $5$. – celtschk Oct 24 '16 at 22:57

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 $\binom82+\binom62+\binom42+\binom22=28+15+6+1=50$.

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?

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 : $28+15+6+1 = 50$

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...

• 50 is correct, as shown by String's answer. His hint made more explicit was that for a specific choice of $b$, there will be $\lfloor \frac{b}{2}\rfloor$ number of choices for $a$, and $9-b$ number of choices for $c$. By multiplication principle then for a specific $b$, you have $(9-b)\lfloor\frac{b}{2}\rfloor$ number of choices. Breaking this into cases for $b\in\{0,1,2,\dots,9\}$ and summing gives you $\sum_{b=0}^9(9-b)\lfloor\frac{b}{2}\rfloor$=50 – JMoravitz Mar 24 '15 at 16:30