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virtus
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\[nC4 + nC5\]

akash123
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use d formula for n c r and do d simplification

akash123
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nCr= n!/ r! (nr)!

virtus
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is it ok if you show me, kinda new to this

akash123
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n!/ 4! (n4)! + n!/ 5! (n5)!

akash123
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n(n1)(n2)(n3)/24 + n(n1)(n2)(n3)(n4)/ 120

akash123
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it's ok?

virtus
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i still can't see it

akash123
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see... n!= 1.2.3.4....n

akash123
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right?

virtus
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yes

akash123
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n!/ (n4)! = 1.2.3....(n4)(n3)(n2)(n1)n / (1.2.3....(n4))

virtus
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yeah...

akash123
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so now u can solve ur problem....:)

virtus
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let me try

virtus
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n(n1)(n2)(n3)/24 + n(n1)(n2)(n3)(n4)/ 120

virtus
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i got that, but how do you simplify further?

virtus
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LOL @hba looks like you have no choice but to help this damsel in distress

hba
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@virtus i dont think so lol

virtus
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16Cn +16C9 =17C9

virtus
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FIND "n"

virtus
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why not @hba

hba
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Its been along time ive studied it and i dont want to get you into another confusion :(

virtus
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OH LOL ok then

akash123
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16Cn +16C9 =17C9
16Cn+16C9=16C9 + 16C8
16Cn=16C8
n=8

akash123
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use the identity nCr + nC(r1) = (n+1)Cr