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Probability

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Source: — Problem Solving |

by earth@work » Thu Oct 30, 2008 10:17 am
Total ways=8c3=56 ways
now, let us take ways in which a couple is always present in committee, for this let us take couples as one object
i.e. Ways when couple present in committee = 4c1*6c1 =24
Ways when no couple present = 56-24 =32
IMO E=32
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by mbaapplicant2008 » Thu Oct 30, 2008 7:18 pm
Yes, the answer is E.
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by cramya » Thu Oct 30, 2008 8:23 pm
Can u explain a little more please on the 4c1*6c1?

now, let us take ways in which a couple is always present in committee, for this let us take couples as one object
i.e. Ways when couple present in committee = 4c1*6c1 =24
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by earth@work » Sun Nov 02, 2008 6:18 pm
cramya wrote:Can u explain a little more please on the 4c1*6c1?

now, let us take ways in which a couple is always present in committee, for this let us take couples as one object
i.e. Ways when couple present in committee = 4c1*6c1 =24
Hi cramya, sorry for delayed reply.
4c1*6c1 .....
the 4c1 parts comes from considering couples as one object i.e instead of 8 people we consider 4 people only... ways to selects 1 out of 4 is 4c1 (this will give us couple in committee of 3
we are left with 3rd place in the committe to be filled by any remaining 6 people = 6c1
therefore , total ways when atleast one couple is present =4c1*6c1 =24
hope its clear now
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by cramya » Sun Nov 02, 2008 8:48 pm
Thanks Earth@work!
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