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by divya23 » Thu Jun 16, 2011 8:18 am
In a certain group of 10 members,4 teach only french and rest teach spanish or german.if the group is to chose 3 member comittee which must have atleast 1 member who teaches french,how many diff commitess can be chosen?


ans = 100
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by Frankenstein » Thu Jun 16, 2011 8:26 am
Hi,
From a group of 10 members, 3 members can be chosen in 10C3 = 120 ways
The committee can be formed without any French teachers in 6C3 ways = 20 ways
So, number of committees with at least 1 French teacher = 120-20 = 100
Cheers!

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by winniethepooh » Thu Jun 16, 2011 1:03 pm
Another way to solve this can be the probability of there being only 1 french teacher and 2 other teachers + 2 french teachers and 1 other teacher + 3 french teachers and none of the other teachers.
= 4C1 x 6C2 + 4C2 x 6C1 + 4C3 x 6C0
= 4 x 15 + 6 x 6 + 4 x 1
= 60 + 36 + 4
= 100
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by EddieU » Thu Jun 16, 2011 1:16 pm
sorry, how does 10C3 = 120?

Thanks
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by winniethepooh » Thu Jun 16, 2011 1:19 pm
It does
10C3 = 10x9x8x7x6x5x4x3x2x1/ 3x2x1x7x6x5x4x3x2x1
= 10x9x8/ 3x2x1
= 5x3x8
= 120.
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by amit2k9 » Thu Jun 16, 2011 7:44 pm
10c3 - 6c3 = 120- 20 = 100

6c3 for none of the people teach french.
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