數學需簡單解題過程，謝謝！

(a)1C1 * 6C3 = 120
1C1 :
Since the thousand digit must be zero, therefore you can only select one (ie. 0) from the seven number.

6C3 :
As 0 is used to occupy thousand digit, you can select three number from the remaining six number. (ie. 1~6)
------------------------------------------------------------------------------------
(b)6C1 * 6C3 = 720
6C1 :
Since the thousand digit can be anything except zero, therefore you can select one number among 1~6.

6C3 :
You can select three number from the remaining six numbers for the other digits.

2012-05-17 14:19:26 補充：
Forgot to tell C means combination.
For example: 6C3 = 6!/3! = 6*5*4*3*2*1/3*2*1 = 6*5*4 = 120

2012-05-17 14:28:07 補充：
For this question, you can check whether you calculated right or wrong.
Since there are 7 numbers to form 3-digit or 4-digit number,
(It cannot be less than 3-digit as there is only one "0")
therefore the total combination = 7C4 = 7!/(7-4)! = 7!/3! (=7*6*5) = 840, which is the sum of (a) & (b)

2012-05-17 14:30:28 補充：
sorry 7!/3! = 7*6*5*4, not 7*6*5

唔多解釋，001講哂，不過佢計錯咗係7個數字唔係6個。

A)1*6*7*7=294
因為(When the thousands digit is 0, the number formed is treated as a 3-digit number.) 所以千位得一個0可以放，千位係1百位唔可以放0，所以6，十個位由0到6，7個數字都放得。

B)6*7*7*7=2058

千位唔可以放，1-6，6個數字都放得，所以千位係6，百十個都可以係7。

2012-05-17 14:26:35 補充：
B)6*7*7*7=2058

千位唔可以放(0)，1-6，6個數字都放得，所以千位係6，百十個都可以係7。

2012-05-17 14:36:14 補充：
咁我明喇，要4 different numbers

a) 1*6*5*4
千(一定係0)，百(除0外)，十(P(百-1)，個(P(十-1)
=120

b) 6*6*5*4
千(除0外)，百(P(除千外)，十(P(百-1)，個(P(十-1)
=720

(a) Consider the hundred digit, we can choose one number except 0 (6 choices). Then, for the ten digit, we also have 6 choices (including 0). For the unit digit, there are 5 choices remaining.
no. of possible 3-digit numbers = 6x6x5 = 180

(b) Similarly, the thousand digit can be any number except 0 (6 choices). Then, for the hundred digit, we also have 6 choices (including 0). For the ten digit, there are 5 choices remaining. There will be 4 choices remained for the unit digit.
no. of possible 4-digit numbers = 6x6x5x4 = 720