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\[nC4 + nC5\]
use d formula for n c r and do d simplification
nCr= n!/ r! (n-r)!

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Other answers:

is it ok if you show me, kinda new to this
n!/ 4! (n-4)! + n!/ 5! (n-5)!
n(n-1)(n-2)(n-3)/24 + n(n-1)(n-2)(n-3)(n-4)/ 120
it's ok?
i still can't see it
see... n!= 1.2.3.4....n
right?
yes
n!/ (n-4)! = 1.2.3....(n-4)(n-3)(n-2)(n-1)n / (1.2.3....(n-4))
yeah...
so now u can solve ur problem....:)
let me try
n(n-1)(n-2)(n-3)/24 + n(n-1)(n-2)(n-3)(n-4)/ 120
i got that, but how do you simplify further?
LOL @hba looks like you have no choice but to help this damsel in distress
  • hba
@virtus i dont think so lol
16Cn +16C9 =17C9
FIND "n"
why not @hba
  • hba
Its been along time ive studied it and i dont want to get you into another confusion :(
OH LOL ok then
16Cn +16C9 =17C9 16Cn+16C9=16C9 + 16C8 16Cn=16C8 n=8
use the identity nCr + nC(r-1) = (n+1)Cr

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