Question

22c6

Answer 1

n C r = n!
---
r!(n-r)!

Answer 2

i get that part i just dont get how to solve it and my book doesnt really explain it very well

Answer 3

Do you know what `5!` means ?

Answer 4

yea its 5*4*3*2*1

Answer 5

\(\LARGE\color{blue}{ \rm{n~C~r}= \frac{n!}{r!\times(n-r)!} }\)


\(\LARGE\color{blue}{ \rm{22~C~6}= \frac{22!}{6!\times(22-6)!}=\frac{22!}{6!\times16!} }\)

Answer 6

\(\LARGE\color{blue}{ \frac{16!\times 17\times 18 \times 19 \times 20 \times 21 \times 22}{6!\times16!} }\)

Answer 7

\(\LARGE\color{blue}{ \frac{16!\times 17\times 18 \times 19 \times 20 \times 21 \times 22}{16!\times 1 \times2 \times 3 \times4 \times5 \times6} }\)

Answer 8

all you need is to simpify

Answer 9

\(\LARGE\color{blue}{ \frac{16!\times 17\times (3 \times 6) \times 19 \times (4 \times 5)\times 21 \times (2 \times 11)}{16!\times 1 \times2 \times 3 \times4 \times5 \times6} }\)

Answer 10

So you get ` 17 × 19 × 21 × 11`

Answer 11

and then just a calculator

Answer 12

ok i get it now thank you... if my book would have put it that way i would have actually understood but they pretty much just gave me the formula to use and said figure it out yourself

Answer 13

Yes... books often do this. I was recently reading a calculus 1 book, and barely understood how to integrate

Answer 14

G☼☼D LUCK :)