Question

@dontre

Answer 1

this is for you\[\Huge C(7,4)= > \frac{7!}{4!3!}=>\frac{7 \times 6 \times 5 }{3 \times 2}\]

Answer 2

so I only do two numbers down on top?

Answer 3

did u get the step i missed? leme do it properly
\[\Large C(7,4) = \frac{7!}{4!3!}= \frac{7 \times 6 \times 5 \times \cancel{4!}}{\cancel{4!} 3!}\]

Answer 4

property used : \[\Huge n! = n(n-1)!\]

Answer 5

I think I see it

Answer 6

\[\Huge C(N,R) = \frac{N!}{R! (N-R)!}\]
\[\Huge C(7,4) = \frac{7!}{4! 3!} = \frac{ 7 \times 6 \times 5 \times 4!}{4! 3!}\]

Answer 7

extending the property i stated earlier,
n!=n(n-1)!
n!=n(n-1)(n-2)!...etc

Answer 8

what if its two problems together...for instance...₇C₄/₇C₁

Answer 9

same procedure

Answer 10

solve both nd divide them

Answer 11

\[\Huge C(n,1) = n\] always

Answer 12

7

Answer 13

and 7C4 we just solved :)

Answer 14

35/7 = 5