Question

From 5 employees at a company, a group of 3 employees will
be chosen to work on a project. How many different groups
of 3 employees can be chosen?

Answer 1

\(\dbinom{5}{3}\) or \(_5C_3\) read "five choose 3" and computed via
\[\binom{5}{3}=\frac{5\times 4}{2}\]

Answer 2

I did n!/r!(n!-r!) and got the wrong answer. I might be doing it wrong.

Answer 3

don't get married to the formula

Answer 4

first of all, it should be clear that 5 choose 3 is the same as 5 choose 2
is that clear or no?

Answer 5

oh, i see, your formula is wrong, but no matter you should use it anyway

Answer 6

*SHOULDN'T

Answer 7

I'm sorry I haven't studied math for over a year and trying to refresh my math skills for placement testing hha

Answer 8

ok lets go slow

Answer 9

because if you think, rather than use a formula, this is really easy

Answer 10

ok:D

Answer 11

you have five people (or whatever) and you want to select 2 of them
make a fraction, you have two slots to fill up top as \(5\times 4\) and in the denominator you put \(2\times 1\) which really only need to write \(2\)

Answer 12

so you have \(_5C_2=\frac{5\times 4}{2}=5\times 2=10\)

Answer 13

wanna do another one that is a bit harder?

Answer 14

yes

Answer 15

ok lets compute \(\binom{10}{7}\)

Answer 16

the first thing to realize is that choosing 7 out of 10 is identical to choosing 3 out of 10
is that clear or no?

Answer 17

yes that's clear.

Answer 18

ok so instead we will compute
\[\binom{10}{3}\] now we have 3 slots to fill in the numerator starting at 10 to give a numerator of \(10\times 9\times 8\) and a denominator starting at 3 giving \(3\times 2\)

Answer 19

so
\[\binom{10}{7}=\binom{10}{3}=\frac{10\times 9\times 8}{3\times 2}\] cancel first multiply last

Answer 20

you get
\[\frac{10\times 9\times 8}{3\times 2}=10\times 3\times 4=120\]

Answer 21

ohh it's getting back to me now, this is paired with sigma as well right?

Answer 22

I remember doing something like this back in high school but i just totally forgot the name of it

Answer 23

if you want to write a formula, it is
\[\binom{n}{k}=\frac{n!}{k!(n-k)!}\] but the formula is silly to use because it just tells you all you need to cancel

Answer 24

there are formula that involve sigmas, which is about addition, and some are related to combinations, but it is not really the same

Answer 25

ok, now I understand how the answer was 10. Thank you so much! I don't want to place in a prerequisite class at my college because I already took pre-calc back in high school..plus the classes that come before pre-calc in college don't even count towards graduation.

Answer 26

hahah, so thank you very much!!

Answer 27

yeah it is easy to get stuck in those annoying low math courses, and then they make them harder than they need to
good luck
yw