The subsets of the set {s, t, u} consisting of the three elements s, t

You are forgetting an empty set, which is also a subset of {s, u, w, x} and do not contain t.

I am a little bit confused by your solution can you please clarify.
I agree we need to find out how many subsets are possible without t, so 4 letter could have 1 set ({SUWX} 4!/4!=1), 3 letters could have 4 sets (4!/3!=4), 2 letters could have 6 sets (4!/2!x2!=6), and 1 letter could have 4 sets. So overall 15 sets and the answer is D. Where did i go wrong?[/quote]

You are forgetting an empty set, which is also a subset of {s, u, w, x} and do not contain t.[/quote]


Thanks Bunuel, i got it, but do you think the way of thinking was correct?

Yes, your approach is correct:
# of subsets with 4 elements is 1: \(C^4_4=1\);
# of subsets with 3 elements is 4: \(C^3_4=4\);
# of subsets with 2 elements is 6: \(C^2_4=6\);
# of subsets with 1 elements is 4: \(C^1_4=4\);
plus 1 empty set.

1+4+6+4+1=16.

I thought that the formula for number of subsets for n elements was 2^n -1

Can anybody explain why this formula does not apply in this case?

Thanks!
Cheers
J

The number of subsets of a set with n elements is 2^n, including an empty set.

Take the task of building subsets and break it into stages.

Stage 1: Determine whether or not to place "t" in the subset
The question tells us that "t" cannot be in the subset.
So, we can complete stage 1 in 1 way (that is, we DO NOT place "t" in the subset

Stage 2: Determine whether or not to place "s" in the subset
We can either HAVE "s" in the subset or NOT HAVE "s" in the subset
So, we can complete stage 2 in 2 ways

Stage 3: Determine whether or not to place "u" in the subset
We can either HAVE "u" in the subset or NOT HAVE "u" in the subset
So, we can complete this stage in 2 ways

Stage 4: Determine whether or not to place "w" in the subset
We can either HAVE "w" in the subset or NOT HAVE "w" in the subset
So, we can complete this stage in 2 ways

Stage 5: Determine whether or not to place "x" in the subset
We can complete this stage in 2 ways

By the Fundamental Counting Principle (FCP), we can complete the 5 stages (and thus build all possible subsets) in (1)(2)(2)(2)(2) ways (= 16 ways)

Answer: E

Note: the FCP can be used to solve the MAJORITY of counting questions on the GMAT. So, be sure to learn it.

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Dear Moderator,
Can we please update the OA for this question, Thank you.

________________
Done. Thank you.

When the answer choices are relatively small, we might also consider the straightforward strategy of listing and counting

We get:
{ }
{s}
{u}
{w}
{x}
{su}
{sw}
{sx}
{uw}
{ux}
{wx}
{suw}
{sux}
{swx}
{uwx}
{suwx}

Answer: E

Cheers,
Brent

Bunuel

When including an empty set, will the number of subsets not including one of the elements of the set always be (2^n)/2? Seems to work out each time.

And then when you don't include an empty set, you can do 2^n - 1 to get the number of subsets x, and then (x-1)/2 for the number of subsets not including one of the elements of the set.

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