# Combinatorics - HELP! ( FCP)

• Oct 26th 2013, 12:28 AM
Gurp925
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
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?
• Oct 26th 2013, 01:41 AM
acc100jt
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
• Oct 26th 2013, 01:57 AM
Gurp925
Re: Combinatorics - HELP! ( FCP)
Understood thankyou!