1.How many license plates can be formed if every license plate has 2 different letters followed by 2 different digits?

3. How many 4-digit numbers have the last digit eequal to the sum of the first two digits?

Possibilities

11*2     10             10

20*2      10            20

12*3      20            40

30*3      10            50

13*4      20            70

22*4       10           80

40*4        10          90

14*5       20          110

23*5      20           130

50*5      10           140

15*6     20            160

24*6      20            180

33*6      10            190

60*6      10            200

16*7       20           220

25*7      20            240

43*7      20            260

70*7      10            270

17*8      20            290

26*8      20            310

35*8      20            330

44*8      10             340

80*8      10             350

18*9      20            370

27*9      20            390

36*9     20             410

45*9     20              430

90*9     10              440

There you go  -  I get 440

1.How many license plates can be formed if every license plate has 2 different letters followed by 2 different digits?

26*25*10*9

2.How many 3-digit numbers have the property that the first digit is twice the second digit?

21*        10

42*         10

63*         10

84*          10

There is a start anyway 

@above  i don't know what you are talking about

Why not?

Are you the person who asked the question?

I get  something a little different from Melody

Assuming that we have to start with a non-zero digit :

If we start with 1, we have 9 possibilites for the second digit  0-8   

Srarting with 2, we have 8 possibilities for the second digit, 0-7

Starting with 3, we have 7 possibilities for the second digit, 0-6

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.

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Starting with 9, we only have 1 possibility for the second digit, 0

So........the sum of all the possibilites for the second digit, with the beginning digit being 1-9,  is just the sum of the first 9 positive integers = (9)(10)/2  =  45

And since the 3rd digit can be 0-9, we have 10 possibilites.........so .......  45 * 10   = 450   4-digit numbers that have the last digit equal to the sum of the first two digits

Hi Chris,

Yes I could have miss a set of 10 easily enough.  I can't be bothered trying to find it though :)