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Victor has 8 associates

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by sanju09 » Sat Mar 19, 2011 4:21 am
Victor has 8 associates. In how many ways can he call one or more of them to banquet?
(A) 8
(B) 63
(C) 64
(D) 255
(E) 256



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by vineeshp » Sat Mar 19, 2011 5:40 am
Is the OA D?

Assuming the question is just to find out how many ways can atleast one person be called out of 8,

8C1 + 8C2 + 8C3 + 8C4 + 8C5 + 8C6 + 8C7 + 8C8

255?
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Just telling you what I know and think. I am not the expert. :)
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by srcc25anu » Sat Mar 19, 2011 2:42 pm
ways to call atleast 1 = (2^n) - 1
in this case its 2^8 - 1 = 256-1 = 255
hence D
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by manpsingh87 » Sun Mar 20, 2011 8:22 am
sanju09 wrote:Victor has 8 associates. In how many ways can he call one or more of them to banquet?
(A) 8
(B) 63
(C) 64
(D) 255
(E) 256



Made up
for every associate there are two options either he/she will be called or won't be called at banquet.
as there are 2 options for every associate therefore total no. of ways of inviting 8 associates will be 2^8. now here i have to subtract one case at which none of them will be called. so total no. of ways of inviting the 8 associates will be 2^8-1 that is 255.
hence [spoiler]D
[/spoiler]
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by force5 » Sun Mar 20, 2011 9:36 am
yes should be 255

(2^n)-1 ( where n=8)
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