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# a set of 15 different word are given . In how many ways is it possible

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Intern
Joined: 07 Sep 2010
Posts: 5
a set of 15 different word are given . In how many ways is it possible [#permalink]

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23 Sep 2010, 06:55
00:00

Difficulty:

65% (hard)

Question Stats:

47% (02:27) correct 53% (01:28) wrong based on 15 sessions

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a set of 15 different word are given . In how many ways is it possible to choose a subset of not more than 5 words

a 4944
b 4^15
c 15^4
d 4943
[Reveal] Spoiler: OA
Math Expert
Joined: 02 Sep 2009
Posts: 39753
Re: a set of 15 different word are given . In how many ways is it possible [#permalink]

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23 Sep 2010, 07:00
holy wrote:
a set of 15 different word are given . In how many ways is it possible to choose a subset of not more than 5 words

a 4944
b 4^15
c 15^4
d 4943

Hi, and welcome to Gmat Club.

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1 (empty subset) + 15C1 (choosing 1 word) + 15C2 (choosing 2 words) + 15C3 (choosing 3 words) + 15C4 (choosing 4 word)+ 15C5 (choosing 5 words)
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Intern
Joined: 07 Sep 2010
Posts: 5
Re: a set of 15 different word are given . In how many ways is it possible [#permalink]

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23 Sep 2010, 16:49
thanks Bunuel... sure willl keep in mind while posting next time
Intern
Joined: 10 Oct 2010
Posts: 23
Location: Texas
Re: a set of 15 different word are given . In how many ways is it possible [#permalink]

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11 Oct 2010, 04:23
holy wrote:
a set of 15 different word are given .

Bag of 15 choices.

holy wrote:
In how many ways is it possible to choose a subset

Combo box arrangement
(_)(_)(_)(_)(_)/5! OR xC5 OR x!/(x-5)!*5! OR xP5/5!

holy wrote:
of not more than 5 words

Table

# of words: Events
0: 1
1: (15)/1! = 15
2: (15)(14)/2! = 105
3: (15)(14)(13)/3! = 455
4: (15)(14)(13)(12)/4! = 1365
5: (15)(14)(13)(12)(11)/5! = 3003
---------------------------------------
Total = 4944
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Re: a set of 15 different word are given . In how many ways is it possible [#permalink]

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16 Dec 2010, 07:31
cabelk wrote:
holy wrote:
a set of 15 different word are given .

Bag of 15 choices.

holy wrote:
In how many ways is it possible to choose a subset

Combo box arrangement
(_)(_)(_)(_)(_)/5! OR xC5 OR x!/(x-5)!*5! OR xP5/5!

holy wrote:
of not more than 5 words

Table

# of words: Events
0: 1
1: (15)/1! = 15
2: (15)(14)/2! = 105
3: (15)(14)(13)/3! = 455
4: (15)(14)(13)(12)/4! = 1365
5: (15)(14)(13)(12)(11)/5! = 3003
---------------------------------------
Total = 4944

This is pretty much the same exact way I did it.

Is there a faster way?
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Re: a set of 15 different word are given . In how many ways is it possible   [#permalink] 16 Dec 2010, 07:31
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