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605-655 (Medium)|   Combinations|                              
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tickledpink001

Thanks. Please can you help me understand how would we solve something like- a set contains 6 elements, how many ways exist to choose a subset of 3 from this set?

The number of ways to select a 3-element subset from a set containing 6 different elements is simply 6C3 = 20.
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­here the "event" that produces outcomes to count is the presence or absence from the set of a letter
in the example
x might be present or absent (2 outcomes)
y might be present or absent (2 outcomes)
w might be present or absent (2 outcomes)
total = 2^3

same applies for xyzw --> 2^4 = 16
now how many sets can be formed without w?
we take into account how many sets can be formed where we surely have w and than subtract this value from the total.

let's assume w is in every set--> we would have to arrange only the other 3 letters --> xyz. so how many way can we arrange them?
x might be present or absent (2 outcomes) (we assume each one have already w inside)
y might be present or absent (2 outcomes) (we assume each one have already w inside)
z might be present or absent (2 outcomes) (we assume each one have already w inside)
total = 2^3 = 8 possible outcomes where we have always w
(example --> if x, y, and z are absent the resulting set will be {w})
how many sets DO NOT contain w?

total = 16
sets containing w = 8

16 - 8 = 8 sets do not contain w
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{w, x, y} has 2^3 subsets (including selecting none)

{w, x, y, z} will have 2^4 subsets.

Out of these half will have w, so subsets of the set {w, x, y, z} that contain w = 2^3 = 8
JDPB7
The subsets of the set {w, x, y} are {w}, {x}, {y}, {w, x}, {w, y}, {x, y}, {w, x,y}, and { } (the empty subset). How many subsets of the set {w, x, y, z} contain w ?


(A) Four
(B) Five
(C) Seven
(D) Eight
(E) Sixteen
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Deconstructing the Question

We need to count how many subsets of \(\{w,x,y,z\}\) contain w.

The key idea is that w is required, so it is already fixed in every valid subset.

That means the only real choices are about x, y, and z.

Step-by-step

Start with w already included.

So every valid subset begins with:

\(\{w\}\)

Now decide what to do with x, y, and z.

For each of these 3 elements, there are 2 choices: include it or do not include it.

So the total number of valid subsets is:

\(2 \cdot 2 \cdot 2 = 2^3 = 8\)

To see this concretely, the valid subsets are:

\(\{w\}\)

\(\{w,x\}\)

\(\{w,y\}\)

\(\{w,z\}\)

\(\{w,x,y\}\)

\(\{w,x,z\}\)

\(\{w,y,z\}\)

\(\{w,x,y,z\}\)

So there are 8 subsets that contain w.

Answer: D
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