m12#23 - Number System : Retired Discussions [Locked]
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# m12#23 - Number System

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12 Jun 2010, 05:24
How many subsets of $$(a,b,c,d)$$ including both $$a$$ and $$c$$ and (order of elements does not matter) are there?

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

OA is B and OE is:
Here are the four subsets: $$(a,c)(a,c,b)(a,c,d)(a,c,b,d)$$

I want to know whether $$(a)$$ and $$(c)$$ are valid subsets or not?
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Re: m12#23 - Number System [#permalink]

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12 Jun 2010, 07:51
ykaiim wrote:
How many subsets of $$(a,b,c,d)$$ including both $$a$$ and $$c$$ and (order of elements does not matter) are there?

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

OA is B and OE is:
Here are the four subsets: $$(a,c)(a,c,b)(a,c,d)(a,c,b,d)$$

I want to know whether $$(a)$$ and $$(c)$$ are valid subsets or not?

Generally $$(a)$$ and $$(c)$$ are valid subsets of the set $$(a,b,c,d)$$, but for this question we are looking for subsets that include both $$a$$ and $$c$$.

Hope it's clear.
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Re: m12#23 - Number System [#permalink]

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12 Jun 2010, 07:56
Thanks for clearing my doubt.
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Re: m12#23 - Number System [#permalink]

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17 Jun 2011, 23:15
Is (a,b,c,d) is a subset of (a,b,c,d) ?
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Re: m12#23 - Number System [#permalink]

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17 Jun 2011, 23:23
Every set has 2^n subsets where n is the number of elements in the set, and every set is a subset of itself.

Note: Null set is a subset of every set!!!
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Re: m12#23 - Number System [#permalink]

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18 Jun 2011, 00:34
thanks sudhir, so that would mean,

for a set {a,b}, number of subsets = 2^2 = 4.

{null},
{a}
{b}
{a,b}

for the question at hand in this thread : we will have 2^3 = 8 subsets, from which we will remove the null subset,

thus we are left with 7.now order does not matter, so out of the 6 sets formed with{(ac),b,d}, we discard 3 ?
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Re: m12#23 - Number System   [#permalink] 18 Jun 2011, 00:34
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# m12#23 - Number System

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