Of the 5 distinguishable wires that lead into an apartment, 2 are for

Yes, it is. It's a direct approach in contrast to {total}-{restriction} approach used in the posts above.

Thanks Bunuel. I always have my fingers crossed when I solve combinations.

Hi,

In these Q, where atleast is used, the best is to find none and then subtract from total..
here total ways= 5C3=5!/3!2!=10..
ways where no cable television is used=3c3=1..
therefore ways containing atleast one wire of cable tele=10-1=9..
ans D

Thanks to the GMAT Club Maths Book for explaining Combinations. Hope, I've done justice to the same (with a correct answer)

Given:
Total number of wires = 5
Number of cable wires = 2
Number of telephone wires = 3

To find:
Number of combination which has at least one cable wires

Solution:
No of ways of selecting 'at least' 1 cable wire means, we can select more than one as well. The minimum we can select is one and the maximum we can select, given the constraints that 3 wires need to be selected in total and there are 2 cable wires, is 2

Since it is a combination of wires, the arrangement is not important

Approach 1:

Number of ways of selecting at least one cable wire in a selection of 3 wires from 5 wires = Selection 1 (Number of ways of selecting one cable wire and two telephone wires )+ Selection 2 (Number of ways of selecting two cable wires and 1 telephone wire)

Selection 1
Number of ways of selecting one cable wire = 2C1 = 2
Number of ways of selecting 2 telephone wires = 3C2 = 3
Total = 2C1 * 3C2 = 6 ( m ways of doing something and n ways of doing something else together give m*n ways of doing - the holy grail rule in Combinatorics)

Selection 2
Number of ways of selecting one cable wire = 2C2 = 1
Number of ways of selecting 2 telephone wires = 3C1 = 3
Total = 2C2 * 3C1 = 3 ( m ways of doing something and n ways of doing something else together give m*n ways of doing - the holy grail rule in Combinatorics)

Selection 1 + Selection 2 = 9 ways of selecting 3 wires out of 5 such that at least one is a cable wire


Approach 2

Number of ways of selecting 3 wires out of 5 such that at least one is a cable wire = Selection X (Total number of ways of selecting 3 wires from the 5) - Selection Y (total ways of selecting 3 wires such that none is a cable i.e all the three are telephone wires)

Total number of ways of selecting 3 wires out of 5 = 5C2 = 10
Number ways of selecting 3 wires such that none is a cable i.e all the three are telephone wires = 3C3 ( 3 telephone wires and we are selecting all the three at once) = 1

Selection X - Selection Y = 9

Answer is Option D

GMAT Club Math Book https://gmatclub.com/forum/gmat-math-book-in-downloadable-pdf-format-130609.html

So I suck at combinations/permutations/whichever one this question was. When there aren't many possibilities I have better luck just listing the combinations.

CTT
TCT
TTC
CCT
TCC
CTC

I am apparently missing three though? According to the other explanations which seem a lot more correct then what I did...

Hi,
why you are going wrong is because you are missing out on an important word..distinguishable..
so take C as C1 and C2... T as T1,T2,T3..
now try, you will get the answer

Hmm...it seems that will give me way more than 9? Way, way more. Even just adding C1 and C2 to the first 3 gets me to 9-not even considering that there is T1, T2...

C1TT
TC1T
TTC1
C2TT
TC2T
TTC2
CCT
TCC
CTC

hi,
one of the C1 and C2 is must and rest three are T1,T2,T3..
the ways are
C1,C2,T1
C1,C2,T2
C1,C2,T3
C1,T1,T2
C1,T3,T2
C1,T1,T3
C2,T1,T2
C2,T1,T3
C1,T3,T2..
there are 9 ways..
and any one above is just one way..
you can't take C1,C2,T1 as
C1,C2,T1..
C2,C1,T1
T1,C2,C1
you are picking up three, order does not matter as The Q is about combinations..
hope it is clear now..

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