Combinatorics - HELP! ( FCP)

Hi guys here is the question . An area code is the first 3 digits in a phone number and indicates the location of either the province or the city. In Canada, the following area codes exist:

Manitoba 204

Saskatchewan 306

Québec (Québec City) 418

Montreal 514

Newfoundland 709

Québec 450, 819

British Columbia 250, 604, 778

Alberta (south) 403

Ontario 519, 613, 705, 807

Yukon and NW Territories 867

Toronto (Ontario) 289, 647

Ontario (Toronto Metro) 416

New Brunswick 506

Alberta 780

Nova Scotia 902

Notice that there are 3 area codes for British Columbia: 250, 604 and 778. It will be necessary to

add another area code as the population increases. The new area code cannot be the same as an

existing code, it must begin with a 3 and end in an even number. Determine the number of

possible area codes to choose from.

A. 40

B. 44

C. 49

D. 50

What i attempted was to use the fundamental counting principle and see there is 3 spots first spot is taken so 3 _ _ second spot can be any one from 0-9 and the third one has to be even and cant be the same as any other area code meaning 4 possible choices 10*4 = 40 but the answer says 49 what did i do wrong?

Re: Combinatorics - HELP! ( FCP)

The new are code has to be 3 _ _

There are ten ways to fill the 2nd digit (0 - 9).

Ignoring the existing area codes, there are 5 ways to fill the last digit (0, 2, 4, 6, 8) *note that 0 is even

By multiplication principle, there are 10 x 5 = 50 codes to choose from.

However, the area code cannot be "306" as it is taken by Saskatchewan.

Hence, the number of possible area codes to choose from = 50 - 1 = 49

Re: Combinatorics - HELP! ( FCP)