their digits are different

[sanju09]

How many odd three-digit integers greater than 800 are there such that all their digits are different?
A. 40
B. 56
C. 72
D. 81
E. 104

OA C

[moutar]

Total number of odd numbers between 800 and 999 = 100

Think which numbers do have a repeated digit.

Between 800 and 899
Numbers in the 80s: 881, 883, 885, 887, 889 (5)
Numbers wth repeated tens and units: 811, 833, 855, 877, 899 (5)
Total = 10

Between 900 and 999 (inclusive):
Numbers with a nine at the end: 909, 919, 929, ..., 999 (10)
Numbers in the nineties: 991, 993, 995, 997 (4)
Numbers wth repeated tens and units: 911, 933, 955, 977 (4)
Total = 18

Gives total = 100 - 10 - 18 = 72 = Answer C

[cubicle_bound_misfit]

Well I am li'l stuck here.

thosand place can be 8 or 9

for case 8 unit place can be 1, 3,5, 7, 9

for any unit place (say 1) and thosand place ( say 8) the tenth place can be filled up in 8 ways

duplice 881,883,885,887,889,991,993,995,997,999

I am getting 5*2*8 -10 = 70 ?

whats wrong?

[shidoshide]

Well I am li'l stuck here.

thosand place can be 8 or 9

for case 8 unit place can be 1, 3,5, 7, 9

for any unit place (say 1) and thosand place ( say 8) the tenth place can be filled up in 8 ways

duplice 881,883,885,887,889,991,993,995,997,999

I am getting 5*2*8 -10 = 70 ?

whats wrong?

According to your calculation, 5*2*8, there are no 881, 883, 885, 887, and 889. More specifically,
for 8X1: 8 ways to fill the tenth place should be 0, 2, 3, 4, 5, 6, 7, 9
for 8X3: 8 ways to fill the tenth place should be 0, 1, 2, 4, 5, 6, 7, 9
for 8X5: 8 ways to fill the tenth place should be 0, 1, 2, 3, 4, 6, 7, 9
for 8X7: 8 ways to fill the tenth place should be 0, 1, 2, 3, 4, 5, 6, 9
for 8X9: 8 ways to fill the tenth place should be 0, 1, 2, 3, 4, 5, 6, 7.

Also, there are no 991, 993, 995, 997, 999. More specifically,
for 9X1: 8 ways to fill the tenth place should be 0, 2, 3, 4, 5, 6, 7, 8
for 9X3: 8 ways to fill the tenth place should be 0, 1, 2, 4, 5, 6, 7, 8
for 9X5: 8 ways to fill the tenth place should be 0, 1, 2, 3, 4, 6, 7, 8
for 9X7: 8 ways to fill the tenth place should be 0, 1, 2, 3, 4, 5, 6, 8.
However, for 9X9, this format is not desired.

So, in conclusion, there are totally 5*8 + 4*8 = 72 numbers. Hope this helps!

[cubicle_bound_misfit]

Thanks. But pure premutation anyone?

[sanju09]

Thanks. But pure premutation anyone?

Sure! First of all there is no thousand's place in the required numbers; refer to your last post. Although there is a hundred's place for all such numbers that can either be filled by a '8' or a '9' here.

When '8' fills the 100's place, 1's place can be filled in 5 ways and 10's in 8 ways; totalling to 5 * 8 = 40.

Or when '9' fills the 100's place, 1's place can be filled in 4 ways and 10's in 8 ways; totalling to 4 * 8 = 32.

Hence such numbers 40 + 32 = 72 in total.

[shargaur]

we have only two scenario here

odd number 1,3,5,7,9
even number 0,2,4,6,8

Case I : first place filled by 1 (only 8)
Third place filled by 5 options (1,3,5,7,9)
Second place (8)
= 1*8*5 = 40

Case II first place filled by 1(only digit 9)
second place by 8
third place by 4 (1,3,5,7)
=1*8*4=32
total 72