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subset query !

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subset query ! [#permalink] New post 29 Jun 2011, 03:51
00:00
A
B
C
D
E

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100% (03:05) correct 0% (00:00) wrong based on 1 sessions
kindly explain...

to form subset of ( 10, 14, 17, 24 ) containing an odd element , istn't it ok to take 4p1 ? which gives answer as 4.

please guide on probability. i am realy week on this probability.permutation and combination !!!
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Re: subset query ! [#permalink] New post 29 Jun 2011, 03:57
Could you please post the full question ?
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Re: subset query ! [#permalink] New post 29 Jun 2011, 04:06
how many different subset of the set (10, 14, 17, 24) contain an odd umer of elements?
a.) 3
b.) 4
c.) 8
d.) 10
e.) 12

pls help!
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Re: subset query ! [#permalink] New post 29 Jun 2011, 04:13
kashishh wrote:
how many different subset of the set (10, 14, 17, 24) contain an odd number of elements?
a.) 3
b.) 4
c.) 8
d.) 10
e.) 12

pls help!


{10}
{14}
{17}
{24}
{10,14,17}
{10,14,24}
{10,17,24}
{14,17,24}

\(C^{4}_{1}+C^{4}_{3}=8\)

Ans: "C"
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Re: subset query ! [#permalink] New post 29 Jun 2011, 04:17
4C1 + 4C3 = 4 + 4 = 8

Answer - C
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Re: subset query ! [#permalink] New post 29 Jun 2011, 04:17
doesn't this quest mean 17 as odd element ?
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Re: subset query ! [#permalink] New post 29 Jun 2011, 04:19
how should i get through probabilty questions? i am not able to make any improvement ... any read to recommend?
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Re: subset query ! [#permalink] New post 29 Jun 2011, 04:22
Ans 1 - This question means that, if you select the elements of the set in all possible ways, how many of them would comprise odd numbers. So you can select 0,1,2,3,4 elements at a time. fluke has shown how many combinations are there for 1 and 3 elements.
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Re: subset query ! [#permalink] New post 29 Jun 2011, 04:23
Ans 2 - Read the Permutations and Combinations + Probability chapters in GMAT Math Book.

math-combinatorics-87345.html

math-probability-87244.html
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Re: subset query ! [#permalink] New post 29 Jun 2011, 04:27
thanks to you both !!
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Re: subset query ! [#permalink] New post 29 Jun 2011, 14:44
odd number of elements.

As we have 4 elements , to form odd number of elements we can either form 1 element sets or 3 element sets.

so total possibilities = 4c1+4c3 = 4 +4 =8

Answer is C.
Re: subset query !   [#permalink] 29 Jun 2011, 14:44
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