Question

Help Asap! I am having so much trouble with this subject and I have multiple problems! Please help! lol

Answer 1

so you have 23 students, 20 of which wont hold office

Answer 2

yes

Answer 3

So take 23 and set that to your combinations of 1st office

Answer 4

Now, take that number down to 22 and that is how many can hold 2nd office * 23.

Answer 5

combinations as in ncr=n!/((n-r)! r!) ????

Answer 6

Huh?

Answer 7

I don't know the equation.

Answer 8

lol okay. I think I have an idea of what to do. Still a little lost tho. lol

Answer 9

You now have the combinations for 1st and 2nd office.

Answer 10

So 23 *22

Answer 11

Now you want to multiply that by 21 for how many students can hold 3rd office.

Answer 12

@Pixel

Answer 13

Is that right?

Answer 14

10626

Answer 15

i believe so

Answer 16

Okay! Thank you!

Answer 17

ill help

Answer 18

ok

Answer 19

mmmmmmm

Answer 20

\[_{n}C _{k} = \frac{ n! }{ k! (n-k)! }\]

Answer 21

dam

Answer 22

Where,

\[_{23}C _{3} = \frac{ 23! }{ 3! (20 - 3)! }\]

Answer 23

\(\color{#0cbb34}{\text{Originally Posted by}}\) @Shadow
Where,

\[_{23}C _{3} = \frac{ 23! }{ 3! (20 - 3)! }\]
\(\color{#0cbb34}{\text{End of Quote}}\)
Thanks!

Answer 24

(Missed a 23)
Where,

\[_{23}C _{3} = \frac{ 23! }{ 3! (23 - 3)! }\]

Answer 25

Think my mind was already thinking 20 lol

Answer 26

lollll I got it. lolll

Answer 27

But basically you get:

\[\frac{ 23! }{ 3! \times 20! } \rightarrow \frac{ 23 \times 22 \times 21 \times 20! }{ 3! \times 20! }\]

Answer 28

Becoming:
\[\frac{ 23 \times 22 \times 21 }{ 3 \times 2 \times 1} = 1771\]

Answer 29

Thank you! I have another problem posted if you're able to help on it! :D