Question

Please help ASAP!!

How many combinations of 3 students can a teacher choose from 32 students?


A.
96


B.
4960


C.
29,760


D.
32,768

Answer 1

I would say there are
\[\left(\begin{matrix}32 \\ 3\end{matrix}\right)\]
combos

Answer 2

It's not C for sure...Can you walk me through it??

Answer 3

oh oh It's D

Answer 4

32*32*32

Answer 5

so there are 32 students lets call them x1, x2, ..... , x32
and you want to make groups of 3 people out of those 32 people

it can be {x1, x2, x3}, or {x1, x3, x20}, or {x1, x3, x23} and much more
and you want to count all of those possible combos

do you know how to expand the equation what i wrote before?

Answer 6

Umm sort of... So my solution how i got to the answer is wrong im guessing.

Answer 7

just solve
\[\left(\begin{matrix}32 \\ 3\end{matrix}\right) = \frac{ 32! }{ 3!29! }\]

Answer 8

ok

Answer 9

you know how the factorial works right?
\[\frac{ 32! }{ 3!29! } = \frac{ (32)(31)(30)....(1) }{ [(3)(2)(1)] [(29)(28)(27)...(1)]}\]

Answer 10

I kind of know how it works it just using it that gets me stuck

Answer 11

if you write out the whole thing you can see you can cancel stuff out and you are left with

\[\frac{ (32)(31)(30) }{ (3)(2)(1) }\]

Answer 12

you can simplify it even more and you are left with
\[(32)(31)(5)\]
and solve that and its your answer

Answer 13

B

Answer 14

correct hope you understood what i did