bloody tough no. prop sum

[vkb16]

How many odd three-digit integers greater than 800 are there such that all their digits are different?

40
56
72
81
104

OA is 72

How do u go about solving this kinda sum?

thanks

[xcusemeplz2009]

total digit =10...(0,1,2,3,....9)
total odd digit=5...(1,3,5,7,9)
let no be xyz
two cases first x can be 8 or 9 ( as no is > 800)

first case :fix x as 8(1 way)
z have any one value out of odd digits therefore 5 ways to select
now for y digit left 8 so ways
total ways=1*5*8=40

second case : fix x as 9=> 1 way
z can be selected out of 4 odd digit( 9 alraedy selected) in 4 ways
Y can be seleceted from 8 digits in 8 ways

total=1*8*4=32

total possible ways =40+32=72

[vkb16]

z have any one value out of odd digits therefore 5 ways to select
now for y digit left 8 so ways

why must be selected only out of odd digits, whereas Y can be selected out of all remaining numbers??

[BuckeyeT]

z have any one value out of odd digits therefore 5 ways to select
now for y digit left 8 so ways

why must be selected only out of odd digits, whereas Y can be selected out of all remaining numbers??

"How many odd three-digit integers..."

By his formula, xyz, z must be odd. So, it can only be selected out of odd integers. If it was "2", an even integer, the number would be xy2. And for all values, (xy), xy2 would be even. For example, 842 is even. Since we're only looking for odd numbers, z must be selected as an odd number.

[xcusemeplz2009]

z have any one value out of odd digits therefore 5 ways to select
now for y digit left 8 so ways

why must be selected only out of odd digits, whereas Y can be selected out of all remaining numbers??

the question is asking for a three digit odd no. and for any no. to be odd the digit at one's place has to be a odd

[vkb16]

thanks a lot guys! that made it clear...