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M12-23

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Math Expert
Joined: 02 Sep 2009
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16 Sep 2014, 00:47
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35% (medium)

Question Stats:

67% (00:32) correct 33% (00:43) wrong based on 126 sessions

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How many subsets of $$\{a,b,c,d\}$$ including both $$a$$ and $$c$$ (order of elements does not matter) are there?

A. 3
B. 4
C. 5
D. 6
E. 7

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Posts: 46297

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16 Sep 2014, 00:47
Official Solution:

How many subsets of $$\{a,b,c,d\}$$ including both $$a$$ and $$c$$ (order of elements does not matter) are there?

A. 3
B. 4
C. 5
D. 6
E. 7

Here are the four subsets: $$\{a, c\}, \ \{a, b, c\}, \ \{a, c, d\}, \ \{a, b, c, d\}$$.

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Joined: 11 May 2014
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15 Jan 2015, 11:22
I think this question is good and helpful.
Is there a way to solve with combinatorics?
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Joined: 14 May 2014
Posts: 44
Schools: Broad '18 (WA)
GMAT 1: 700 Q44 V41
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24 Jul 2015, 09:27
Bunuel wrote:
Official Solution:

How many subsets of $$\{a,b,c,d\}$$ including both $$a$$ and $$c$$ (order of elements does not matter) are there?

A. 3
B. 4
C. 5
D. 6
E. 7

Here are the four subsets: $$\{a, c\}, \ \{a, b, c\}, \ \{a, c, d\}, \ \{a, b, c, d\}$$.

hi, Bunuel,

can we count {a,b,c,d} as subset of {a,b,c,d}..?
Math Expert
Joined: 02 Sep 2009
Posts: 46297

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24 Jul 2015, 09:31
riyazgilani wrote:
Bunuel wrote:
Official Solution:

How many subsets of $$\{a,b,c,d\}$$ including both $$a$$ and $$c$$ (order of elements does not matter) are there?

A. 3
B. 4
C. 5
D. 6
E. 7

Here are the four subsets: $$\{a, c\}, \ \{a, b, c\}, \ \{a, c, d\}, \ \{a, b, c, d\}$$.

hi, Bunuel,

can we count {a,b,c,d} as subset of {a,b,c,d}..?

____________________________
Yes.
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Joined: 14 Oct 2015
Posts: 36
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29 Oct 2015, 08:54
Bunuel wrote:
riyazgilani wrote:
Bunuel wrote:
Official Solution:

How many subsets of $$\{a,b,c,d\}$$ including both $$a$$ and $$c$$ (order of elements does not matter) are there?

A. 3
B. 4
C. 5
D. 6
E. 7

Here are the four subsets: $$\{a, c\}, \ \{a, b, c\}, \ \{a, c, d\}, \ \{a, b, c, d\}$$.

hi, Bunuel,

can we count {a,b,c,d} as subset of {a,b,c,d}..?

____________________________
Yes.

That does not make sense. How can a "sub" set be the same as the set?? {a,b,c,d} is not a subset of {a,b,c,d}. Agree to disagree.
Math Expert
Joined: 02 Sep 2009
Posts: 46297

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29 Oct 2015, 11:42
1
danjbon wrote:
Bunuel wrote:
riyazgilani wrote:

hi, Bunuel,

can we count {a,b,c,d} as subset of {a,b,c,d}..?

____________________________
Yes.

That does not make sense. How can a "sub" set be the same as the set?? {a,b,c,d} is not a subset of {a,b,c,d}. Agree to disagree.

Mathematically B is a subset of A if every member of B is a member of A. So, a set is a subset of itself.
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19 Jan 2016, 09:21
Buenel,

I have a (naive) question. Why (D,A,C,B) is not a subset of (A,B,C,D) - when order of elements doesn't matter ?
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19 Jan 2016, 09:25
1
cricketer wrote:
Buenel,

I have a (naive) question. Why (D,A,C,B) is not a subset of (A,B,C,D) - when order of elements doesn't matter ?

A set, by definition, is a collection of elements without any order. (While, a sequence, by definition, is an ordered list of terms.)
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19 Jan 2016, 09:31
Bunuel wrote:
cricketer wrote:
Buenel,

I have a (naive) question. Why (D,A,C,B) is not a subset of (A,B,C,D) - when order of elements doesn't matter ?

A set, by definition, is a collection of elements without any order. (While, a sequence, by definition, is an ordered list of terms.)

Thanks a lot. Very helpful/useful concept.
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10 Oct 2016, 22:28
I think this is a poor-quality question. Useless question
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15 Jul 2017, 12:40
Can this question be re-worded to say "include both a & c"?
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23 Jul 2017, 09:36
Why we don't consider singletons? (Set of one number here).
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23 Jul 2017, 20:49
Evgart wrote:
Why we don't consider singletons? (Set of one number here).

Doesn't the question say that the subsets must include at least both a and c?
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14 Aug 2017, 08:19
why not {a,d} and {b,d}?
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14 Aug 2017, 08:24
frndabu1 wrote:
why not {a,d} and {b,d}?

Because the subsets must include both a and c. Neither {a, d} nor {b, d} include both a and c.
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14 Aug 2017, 08:45
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07 Nov 2017, 12:11
the number of subsets from k elements is 2^k
here if we consider a and c together as one element (since they are always present together), won't the number of subsets: 2^3 = 8?
now a&c will always need to be present, therefore the only remaining subsets: 2^2 = 4

is above understanding correct?
Re: M12-23   [#permalink] 07 Nov 2017, 12:11
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