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Counting

by bia » Wed May 28, 2008 6:14 am
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
Bia
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by gmatinjuly » Wed May 28, 2008 6:55 am
My answer is 72 . Here is explaination.

Odd numbers between 800 and 900
Start : 801
End : 899

801-809 : 5
811-819 : 4 (For odd tens digit we will have one lesser number like 811 has reperion then 833 will be same etc)
821-829 : 5
831-839 : 4
841-849 : 5
851-859 : 4
861-869 : 5
871-879 : 4
881-889 : 0 (All 88_)
891-899 : 4

Total numbers between 801-899 : 5*4 + 4* 5 = 40

Odd numbers between 900 and 1000
Start : 901
End : 999

901-909 : 4 (as In hundreds place is 9 and all odd number from 1-10 will have atleast one 9 this case 909)
911-919 : 3 (Here we need to account for two one coz of tens place as odd number and other for number ending with 9
two numbers elimnated are 911 and 919)
921-929 : 4
931-939 : 3
941-949 : 4
951-959 : 3
961-969 : 4
971-979 : 3
981-989 : 4
991-999 : 0


Total numbers between 901-999 : 4*5 + 3* 4 = 32


Total Number = 40+32 = 72
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by AleksandrM » Wed May 28, 2008 9:27 am
There has got to be a shorter way of doing this!
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by netigen » Wed May 28, 2008 10:45 am
Lets say the number is XYZ

Lets look at the possibilities

X = 8 or 9
Y = 0-9
Z = 1,3,5,7,9

case 1:

X = 8
Y = 0,2,4,6 or 1,3,5,7,9
Z = 1,3,5,7,9

Possibilities = 1 x 4 x 5 + 1 x 5 x 4 = 40

case 2:

X = 9
Y = 0,2,4,6,8 or 1,3,5,7
Z = 1,3,5,7

Possibilities = 1 x 5 x 4 + 1 x 4 x 3 = 32

total = 40+32 = 72

Whats the OA
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