[Math] permutation & combinations

combinationspermutations

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…

Best Answer

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

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