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# From a set of 6 questions, in how many ways can a student

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SVP
Joined: 29 Aug 2007
Posts: 2473
From a set of 6 questions, in how many ways can a student [#permalink]

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24 Oct 2007, 21:53
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From a set of 6 questions, in how many ways can a student solve 1 or more than 1 questions?

a. 6
b. 12
c. 63
d. 64
e. 65
VP
Joined: 08 Jun 2005
Posts: 1145

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24 Oct 2007, 22:16
total ways = 6C1+6C2+6C3+6C4+6C5+6C6 = 6+15+20+15+6+1= 63

SVP
Joined: 29 Aug 2007
Posts: 2473

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24 Oct 2007, 22:18
KillerSquirrel wrote:
total ways = 6C1+6C2+6C3+6C4+6C5+6C6 = 6+15+20+15+6+1= 63

can you tell, in how many way can these questions be solved in total?
VP
Joined: 08 Jun 2005
Posts: 1145

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24 Oct 2007, 22:21
GMAT TIGER wrote:
KillerSquirrel wrote:
total ways = 6C1+6C2+6C3+6C4+6C5+6C6 = 6+15+20+15+6+1= 63

can you tell, in how many way can these questions be solved in total?

64 ways

you need to add 6C0 = 6!/6! = 1 ---> solving none

Director
Joined: 11 Jun 2007
Posts: 915

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24 Oct 2007, 22:37
KillerSquirrel wrote:
total ways = 6C1+6C2+6C3+6C4+6C5+6C6 = 6+15+20+15+6+1= 63

nice work KS.. but seems too time consuming!
SVP
Joined: 29 Aug 2007
Posts: 2473

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24 Oct 2007, 23:25
KillerSquirrel wrote:
GMAT TIGER wrote:
KillerSquirrel wrote:
total ways = 6C1+6C2+6C3+6C4+6C5+6C6 = 6+15+20+15+6+1= 63

can you tell, in how many way can these questions be solved in total?

64 ways

you need to add 6C0 = 6!/6! = 1 ---> solving none

thanks. i was looking for a direct way to find the total.

for ex: what is the prob of getting 2 boys and 3 girls from 5 boys and 6 girls?

2 boys = 5c2 = 10
3 girls = 6c3 = 20
2 boys x 3 girls = 10x20 = =200

prob = 200/11c5 = 200/462

i am looking for a way to find 11c5 = 11!/{(5!)(11-5)1}
24 Oct 2007, 23:25
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