Question

Compute the value (8)
(6)

Answer 1

Or (8 over 6)
The answer is 28

Answer 2

maybe it is
\[\dbinom{8}{6}\]

Answer 3

Yes idk how to input that

Answer 4

\[\dbinom{8}{6}=\dbinom{8}{2}=\frac{8\times 7}{2}=4\times 7=28\]

Answer 5

How does the six change to a 2?

Answer 6

eight choose six is the same as eight choose two because \(6+2=8\)

Answer 7

plz kindly teach what it is all about..................i don't know even a bit

Answer 8

you can compute eight choose six too inf you like, but you will get the same answer, just more cancellation

Answer 9

I'm having trouble understanding. How did you know how to multiply?

Answer 10

\[\dbinom{8}{6}\] means the number of ways you can choose six items from a set of eight

Answer 11

ok ok i got it is permutations/////////////thanks...........

Answer 12

first of all you should understand that it is the same as \(\dbinom{8}{2}\) because choosing 6 people out of eight to invite to your party is the same as choosing 2 out of 8 not to invite

Answer 13

But then why multiply with 7?

Answer 14

so you have 8 choices for the first person, seven for the second, but order doesn't count, so you don't want to count choosing justin first and kesha second differently then kesha first and justin second.

by the counting principle you get
\[\dbinom{8}{2}=\frac{8\times 7}{2}\]

Answer 15

eight choices for the first person, seven for the second. it is an application of the counting principle
there is also a formula you can use, or you can use pascal's triangle

Answer 16

the general formula is
\[\dbinom{n}{k}=\frac{n!}{k!(n-k)!}\] but you don't really want to ever use it

Answer 17

for example you could write
\[\dbinom{8}{6}=\frac{8!}{6!(8-6)!}=\frac{8!}{6!2!}=\frac{8\times 7}{2}\]

Answer 18

I'm sorry I'm having trouble with how the 7 came into the equation

Answer 19

ok lets go slow

Answer 20

i have say 8 letters and i want to know how many ways i can choose any two of them
lets call the letters \(\{a,b,c,d,e,f,g,h\}\)

Answer 21

so have two slots to fill
( __ , __ ) in which to put the two chosen letters.

there are 8 choices for the first letter.
now that i have chosen one of the eight letters, there are only seven choices left for the second letter, because one has already been chosen

Answer 22

by the counting principle, if there are 8 ways to do one thing, and 7 ways to do another, there are \(8\times 7\) ways to do them together

Answer 23

but i do not want to count
(a, b) and (b, a) as two separate choices, because the order i choose them in makes no difference. in other words i can't tell the difference between (a,b) and (b,a)
so i have counted everything twice

therefore, i have to divide by 2 to get the answer i need, i.e.
\[\dbinom{8}{2}=\frac{8\times 7}{2}\]

Answer 24

is it clear where the 7 came from? eight choices for the first letter, then 7 for the second because one is already chosen

Answer 25

Yes I understand now!!!! But one last thing, how is 6 thee same as 2?

Answer 26

Is it from the n-k?

Answer 27

lets say you have six eggs to throw at cars in a parking lot, but there are 8 cars in the lot all together

you can either decide which six out of the eight cars to throw eggs at (the number of ways to do this is \(\binom{8}{6}\) or
you can choose two of the cars not to throw eggs at.
the number of ways to do that is \(\binom{8}{2}\)

Answer 28

it is the same problem exactly
this shows for example that \(\dbinom{12}{4}=\dbinom{12}{8}\)

Answer 29

and yes, if you want to appeal to the formula, it is from the \(k!(n-k)!\) in the denominator

Answer 30

for example
\[\dbinom{12}{4}=\frac{12!}{4!(12-4)!}=\frac{12!}{4!8!}=\frac{12!}{(12-8)!8!}=\dbinom{12}{8}\]

Answer 31

in math you can say
\[\dbinom{n}{k}=\dbinom{n}{n-k}\]

Answer 32

moral of the story is pick the one that is easiest

Answer 33

I understand it better! Thank you for taking the time to help me!

Answer 34

yw