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kouranjelika
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Combinatorics Question

by kouranjelika Wed Jun 18, 2014 7:45 pm

Hi,

I am making this up, because I wanted to try out different scenarios of selections.

Imagine there are 26 people are at a party. 5 women and the rest are men (so, 21 men). If we were to select a group of 3 people, how many different ways can we select a group where there are at least 2 women.

Here's what I have:

5 women; 21 men
SO: we can have 2 women and 1 man OR 3 women (using the 1 - trick here won't be much help I think because we would need to calculate no women and 1 woman, same story, just an extra step):


If it is 2 women and 1 man:
It is 5C2 * 21C1: so 10 * 21 = 210
If it is 3 women only:
Then it is 5C3 = 10

Together we have 210 + 10 = 220. Am I correct?

Thank you!
"A creative man is motivated by the desire to achieve, not by the desire to beat others."
-Ayn Rand
kouranjelika
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Re: Combinatorics Question

by kouranjelika Wed Jun 18, 2014 8:38 pm

Here's another one I came up with.
Say there is some sort of international conference where 8 countries are represented by 2 people from each country.

If we were to randomly select two speakers, what is the probability that these two speakers will be from a) same country b) different countries?

a) Same country.
There are 4 such possibilities (2 of each country are selected simultaneously). Total number of possibilities is 8 choose 2, which is 28. So 4/28 which reduces to 1/7 is the answer.
b) Different coutnries.
Denominator would still be the same, 28. Top would be different, 4 choose 1 * 3 choose 1 should be it, because I am mentally splitting the group into singulars from each country and then selecting one from the first group and one from the second group minus 1 for the country we already picked. But the total turns out to be 3/7 and for some reason a) and b) don't add up to 1. Can you find my error?

Thank you!
"A creative man is motivated by the desire to achieve, not by the desire to beat others."
-Ayn Rand
tim
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Re: Combinatorics Question

by tim Fri Jun 20, 2014 2:58 am

Your first example is just fine. As for the second part of the second one, your 4C1*3C1 is wrong. Honestly, it's best just to use the 1-x trick here; if the probability they are from the same country is 1/7, the probability they are from different countries is 6/7. To verify: pick anyone for the first seat. Then there are 7 other people to pair her with, 6 of whom are from a different country.
Tim Sanders
Manhattan GMAT Instructor

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