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M12-23

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M12-23  [#permalink]

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New post 15 Sep 2014, 23:47
1
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A
B
C
D
E

Difficulty:

  35% (medium)

Question Stats:

66% (00:31) correct 34% (00:42) wrong based on 131 sessions

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Re M12-23  [#permalink]

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New post 15 Sep 2014, 23:47
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: \(\{a, c\}, \ \{a, b, c\}, \ \{a, c, d\}, \ \{a, b, c, d\}\).

Answer: B
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Re M12-23  [#permalink]

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New post 15 Jan 2015, 10:22
I think this question is good and helpful.
Is there a way to solve with combinatorics?
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Re: M12-23  [#permalink]

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New post 24 Jul 2015, 08:27
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: \(\{a, c\}, \ \{a, b, c\}, \ \{a, c, d\}, \ \{a, b, c, d\}\).

Answer: B


hi, Bunuel,

can we count {a,b,c,d} as subset of {a,b,c,d}..?
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Re: M12-23  [#permalink]

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New post 24 Jul 2015, 08:31
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: \(\{a, c\}, \ \{a, b, c\}, \ \{a, c, d\}, \ \{a, b, c, d\}\).

Answer: B


hi, Bunuel,

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

____________________________
Yes.
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PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

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Re: M12-23  [#permalink]

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New post 29 Oct 2015, 07:54
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: \(\{a, c\}, \ \{a, b, c\}, \ \{a, c, d\}, \ \{a, b, c, d\}\).

Answer: B


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.
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Re: M12-23  [#permalink]

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New post 29 Oct 2015, 10:42
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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Re: M12-23  [#permalink]

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New post 19 Jan 2016, 08:21
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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Re: M12-23  [#permalink]

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New post 19 Jan 2016, 08:25
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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Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

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Re: M12-23  [#permalink]

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New post 19 Jan 2016, 08:31
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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Re M12-23  [#permalink]

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New post 10 Oct 2016, 21:28
I think this is a poor-quality question. Useless question
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Re: M12-23  [#permalink]

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New post 15 Jul 2017, 11:40
Can this question be re-worded to say "include both a & c"?
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Re: M12-23  [#permalink]

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New post 23 Jul 2017, 08:36
Why we don't consider singletons? (Set of one number here).
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New post 23 Jul 2017, 19:49
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New post 14 Aug 2017, 07:19
why not {a,d} and {b,d}?
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New post 14 Aug 2017, 07:24
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New post 14 Aug 2017, 07:45
Great! careful reading!!!
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New post 07 Nov 2017, 11:11
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?
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Re: M12-23  [#permalink]

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New post 26 Oct 2018, 12:18
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: \(\{a, c\}, \ \{a, b, c\}, \ \{a, c, d\}, \ \{a, b, c, d\}\).

Answer: B


Dear Bunuel,

Is there any other way to approach alike questions. What if the subsets were asked for {a, b, c, d, e, f, g}?
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Re: M12-23  [#permalink]

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New post 04 Nov 2018, 09:34
Is this fair game on the real test? I haven't seen subsets covered anywhere. Any reference material would be helpful....
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Re: M12-23 &nbs [#permalink] 04 Nov 2018, 09:34
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