Question

question below

Answer 1

\[nC4 + nC5\]

Answer 2

use d formula for n c r and do d simplification

Answer 3

nCr= n!/ r! (n-r)!

Answer 4

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

Answer 5

n!/ 4! (n-4)! + n!/ 5! (n-5)!

Answer 6

n(n-1)(n-2)(n-3)/24 + n(n-1)(n-2)(n-3)(n-4)/ 120

Answer 7

it's ok?

Answer 8

i still can't see it

Answer 9

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

Answer 10

right?

Answer 11

yes

Answer 12

n!/ (n-4)! = 1.2.3....(n-4)(n-3)(n-2)(n-1)n / (1.2.3....(n-4))

Answer 13

yeah...

Answer 14

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

Answer 15

let me try

Answer 16

n(n-1)(n-2)(n-3)/24 + n(n-1)(n-2)(n-3)(n-4)/ 120

Answer 17

i got that, but how do you simplify further?

Answer 18

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

Answer 19

@virtus i dont think so lol

Answer 20

16Cn +16C9 =17C9

Answer 21

FIND "n"

Answer 22

why not @hba

Answer 23

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

Answer 24

OH LOL ok then

Answer 25

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

Answer 26

use the identity nCr + nC(r-1) = (n+1)Cr