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by samirpandeyit62 » Mon Oct 29, 2007 4:18 am
2 exprs from 4 can be selected in 4P2 = 12 ways

now only possibities where the product is of the form

x- (by)^2

are (x+y)(x-y) & (x-y)(x+y) = 2 possibilties

so reqd P = 2/12 =1/6

Hope this is clear
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by Jimat » Mon Oct 29, 2007 7:56 pm
2 expresion picked from another 4 = 4!/[(4-2)!*2!] = 6

The only expresion that results in a x^2-(by)^2 form is (x+y)*(x-y), there is no way to say that (x-y)*(x+y) is another way since multiplication is a conmutative operation.
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by ri2007 » Tue Oct 30, 2007 6:14 pm
Samir
Why do u use permutation instead of combination formula in this case?
Thanks
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by samirpandeyit62 » Tue Oct 30, 2007 11:59 pm
Well both will do here even if u use Combinations, then also u will get the same result.
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by ri2007 » Wed Oct 31, 2007 5:56 am
Hey Samir

Sorry but this is not clear to me.

4P2 = 12 ways but 4C2 = 6

Now you have said that possibities where the product is of the form

x- (by)^2

are (x+y)(x-y) & (x-y)(x+y) = 2 possibilties

so reqd P = 2/12 =1/6

If you use combination then the ans would be 2/6 = 1/3 instead of 1/6

Aorry for the repeated question. As always thanks in advance for your help
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by samirpandeyit62 » Wed Oct 31, 2007 6:03 am
If we use Combinations then favorable outcome will be only

(x+y)(x-y) = 1 outcome as order would not be important, we can use it here coz multiplication works both ways

so it will be 1/4C2 = 1/6
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Samir
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by yuri » Fri Nov 02, 2007 12:32 am
I figured this out today as well (semi-guessed, but got it right). Anyway, I apologize for probably stupid question, but what does this mean??? What are P and C in these:

4P2 = 12 ways but 4C2 = 6
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by beeparoo » Sat Nov 03, 2007 12:23 pm
yuri wrote:I figured this out today as well (semi-guessed, but got it right). Anyway, I apologize for probably stupid question, but what does this mean??? What are P and C in these:

4P2 = 12 ways but 4C2 = 6
Yuri, it is short-form notation for PERMUTATION and COMBINATION calculations.

For example: 4C2 = "Four choose 2" = 4!/(2!2!)
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by yuri » Sun Nov 04, 2007 12:06 am
thank you beeparoo!
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