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CAN someone help here? PLEASE....!

Problem Solving — algebra and arithmetic (GMAT Focus Edition)
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CAN someone help here? PLEASE....!

by svbla » Sat Mar 19, 2016 2:06 pm
Of the 5 distinguishable wires that lead into an apartment, 2 are for cable television service, and 3 are for telephone service. Using these wires, how many distinct combinations of 3 wires are there such that at least 1 of the wires is for cable television?

I know it may be very simply for you, but it kills me!!!
I can figure it out that using combination formula to solve this issue, but why subtract 1 in the end? why? I don't get it!
What if it says at least 2of the wires for cable? what to do then?

I hate math so much, please help here if you could!!!
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by MartyMurray » Sat Mar 19, 2016 3:21 pm
Case 1: 2 Cable Wires Used

C1 C2 T1

C1 C2 T2

C1 C2 T3

2C2 x 3C1 = 1 x 3 = 3

Case 2: 1 Cable Wire Used

C1 T1 T2

C1 T1 T3

C1 T2 T3

C2 T1 T2

C2 T1 T3

C2 T2 T3

2C1 x 3C2 = 2 x 3 = 6

Total Number Of Combinations: 3 + 6 = 9
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by [email protected] » Sat Mar 19, 2016 4:10 pm
Hi svlba,

IF you were allowed to use any 3 wires (from the group of 5), then you could use the Combination Formula without having to do anything else:

Combinations = N! / K!(N-K)! = 5! / 3!2! = (5)(4) / (2)(1) = 10 possible combinations of 3 wires

HOWEVER, the prompt tells us that AT LEAST one cable wire must be included in the group of 3. There is ONE option that has 0 cable wires though (the one with all 3 telephone wires), so we have to subtract that one option from the total.

10 - 1 = 9

The number of possible combinations of wires is relatively small, so you could also have 'brute forced' this prompt - and just list out all of the options (as Marty pointed out).

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by Matt@VeritasPrep » Fri Apr 15, 2016 1:50 pm
We've got two possibilities:

1: We only use one cable TV wire;
2: We use BOTH cable TV wires.

In the first case, we'd have (2 choose 1) * (3 choose 2), or 6. In the second, we'd have (2 choose 2) * (3 choose 1), or 3. So our total is 6 + 3, or 9.
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