Lottery game

[Vemuri]

Entries in a particular lottery game are made up of 3 digits, each 0 through 9. If the order of digits in the entries matters, how many different possible enteries exist in which all 3 digits are not equal?

A. 516
B. 720
C. 989
D. 990
E. 1321

[cramya]

Hope I dint miss something.


First digit 10 choices including 0 since the problem says 0 can be the first digit

2nd digit->9 choices

3rd digit-> 8 choices


10*9*8 = 720

[cramya]

I did miss something.


2 digits can be equal but not all 3.

10*9*2 =180

But ech of the above picks can be arranged in 3 ways

So 180*3 but we need to divide by 2 since some pairs of arrangement are the same


I pick 1 2 1 and then 2 1 1

3 different numbers can be formed by either of these 2 picks but the resulting unique numbers is just half. Out of 540 only 270 are unique.(hope I dint mess up here, Ian or some one else please correct me)

180*3/2 = 270

So 720+270 = 990


Nice prob.

Regards,
CR

[m&m]

alternatively we could solve this by saying total choices - number of choices with all 3 digits same.

Total choices = 10*10*10
Choices with all 3 digits same = 10 (incld 000, 111, 222, 333)

1000 - 10 = 990

Voila

[cramya]

alternatively we could solve this by saying total choices - number of choices with all 3 digits same.

Total choices = 10*10*10
Choices with all 3 digits same = 10 (incld 000, 111, 222, 333)

1000 - 10 = 990

Well done....

[Vemuri]

alternatively we could solve this by saying total choices - number of choices with all 3 digits same.

Total choices = 10*10*10
Choices with all 3 digits same = 10 (incld 000, 111, 222, 333)

1000 - 10 = 990

Voila

Cool approach !!!