A committee consists of four women and three men. The committee will randomly select two people to attend a conference in Hawaii. Find the probability that both are women.
Okay the solution is..... 4C2=6 and 7C2=21.. 6/21=.2857...
I have gone through this chapter and have not seen when C2 stands for. Another thing is, where is the 6 AND 21 coming from?
Some help, pleasssse.
Okay the solution is..... 4C2=6 and 7C2=21.. 6/21=.2857...
I have gone through this chapter and have not seen when C2 stands for. Another thing is, where is the 6 AND 21 coming from?
Some help, pleasssse.
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4C2 = 4!/(2!2!) = 6
7C2=7!/(2! 5!) = 21
(by the definition of mCn)
7C2=7!/(2! 5!) = 21
(by the definition of mCn)
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In case you need a little more explanation, the 7C2 = 21 on the bottom is because there are 21 ways of choosing 2 people from 7 - these are called Combinations, when the order of selection doesn't matter. You might have an nCr button on your calculator that does it for you without factorials? cont..
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then there are 4 C 2 = 6 ways of choosing 2 women from 4 (and you could say multiplied by 3 C 0 = 1 way of picking 0 men from the 3, but this doesn't matter obviously). so in all there are 6 ways of choosing 2 women, out of the 21 possible selections.
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