Hi Ron,

What is wrong if I do it as indicated below ?

Asking because this method came to my mind first in the mock and I could not get to the correct answer.

I have 4 places to fill.

_ _ _ _

I start with the last digit(one's place), I have 10 numbers to chose from. Similarly for 3rd digit I have 9 numbers to chose from, 8 numbers for the 2nd place.

Now for the first place I cannot include 0, so I have 6 numbers to chose from(7 numbers if 0 were allowed) for the first place (the thousand's place).

I get 6*8*9*10. which gives 4320.

RonPurewal wrote:ankitmisri wrote:1. Each number is to consist of four different digits from 0 to 9

Hence no number can repeated

2. the first digit cannot be 0

hence you can only choose 1 of the 9 numbers(1-9) for the 1st place i.e. is you have 9 options for 1st digit

For the 2nd digit again you can again choose 1 out 9 numbers (now u have 0 available) in 9c1 ways =9

For the 3rd digit only 8 numbers are available coz of "1."...u can choose that in 8c1 ways = 8

For the 4th place only 7 numbers are available coz of "1."...u can choose that in 7c1 ways = 7

Since all four selected individually, you have to multiply them 9*9*8*7 = 4536

yep. this is the most direct method, and thus the best.

if you have a problem in which

ORDER MATTERS, then you should be able to solve the problem by

multiplying numbers together.

you shouldn't have to bother with factorials, etc., unless you have a problem on which order DOESN'T matter.