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# 4 GMATPrep Qs

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Joined: 29 Aug 2007
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Re: 4 GMATPrep Qs [#permalink]  02 Sep 2009, 19:35
If a side of isosceles triangle is x, then hypotenuse is x sqrt2.

Perimeter of an isosceles traingle= 16 + 16 sqrt2
x + x + x sqrt2 = 16 + 16 sqrt2
2x + x sqrt2 = 16 + 16 sqrt2
x sqrt2 (1+sqrt2) = 16 (1+sqrt2)
x sqrt2 = 16
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Joined: 29 Jul 2009
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Re: 4 GMATPrep Qs [#permalink]  06 Sep 2009, 09:58
LenaA wrote:
I did not get your example with meals.
But I will give you mine...also food related...

You are going to have a dessert and you have a selction of 4 chocolate cakes and 3 cheese cakes. You are going to eat 2 types of cake...how many desserts options can you have?
1) you are going to have one chocolate and one cheese cake then you have $$4\times 3=12$$ dessert options
2) there is no limitation you can have 2 chocolate or 2 cheese or one chocolate and one cheese...
then you have 7 types of cake and you can have $$7C2=\frac{7!}{5!2!}$$dessert options...

ok now it's clear, but how can I calculate the following in the exam? $$7C2=\frac{7!}{5!2!}$$

thanks,
Manager
Joined: 10 Aug 2009
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Re: 4 GMATPrep Qs [#permalink]  06 Sep 2009, 18:09
LenaA wrote:
I did not get your example with meals.
But I will give you mine...also food related...

You are going to have a dessert and you have a selction of 4 chocolate cakes and 3 cheese cakes. You are going to eat 2 types of cake...how many desserts options can you have?
1) you are going to have one chocolate and one cheese cake then you have $$4\times 3=12$$ dessert options
2) there is no limitation you can have 2 chocolate or 2 cheese or one chocolate and one cheese...
then you have 7 types of cake and you can have $$7C2=\frac{7!}{5!2!}$$dessert options...

ok now it's clear, but how can I calculate the following in the exam? $$7C2=\frac{7!}{5!2!}$$

thanks,

The following way: $$\frac{7\times 6\times 5!}{5!\times 2\times 1}=\frac{7\times 6}{2}=21$$
Re: 4 GMATPrep Qs   [#permalink] 06 Sep 2009, 18:09

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