Question

HELP

Answer 1

In how many ways can a committee consisting of 3 faculty members and 5 students be formed if there are 8 faculty members and 11 students eligible to serve on the committee?

Answer 2

i have the answer
but i have no idea how to get it.

Answer 3

so do you know anything about "picking" and "choosing" in terms of probability?

Answer 4

8 choose 3 times 11 choose 5
does that sound familiar?

Answer 5

here are the answers but i have no idea how they get them and ive been plugging in numbers for the past half hour and no luck

Answer 6

the only answer i could figure out was the 8!/3!(5!)
9 did 8*5*3 which gave me 120 then divided it by 3 and that gave me 40 then i added 8^2 and got 56 .
don't know if thats how they got the answer but i tried the same out for the second one and it didnt work.

Answer 7

@phi are you going to explain this or should I? I think it would be confusing if two people do it/

Answer 8

I could keep up
if both of you explained.

Answer 9

\[ \frac{8\cdot 7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1}{3 \cdot 2\cdot 1\cdot 5\cdot 4\cdot 3\cdot 2\cdot 1} \\
\]
notice the 5! cancels with the top
\[ \frac{8\cdot 7 \cdot 6}{3 \cdot 2\cdot 1} \]

Answer 10

the bottom is 3*2= 6 which cancels the 6 up top
that just leaves 8*7 = 56

Answer 11

Is that what you are asking?

Answer 12

@Beautykillsszz, I'll let @phi explain it, since he is already doing so lol. probability can be a bit difficult to learn as it's all about seeing things in a particular way, having two people explain something in a different way can cause confusion.

Answer 13

i dont understand how you got that.

Answer 14

First, you know
\[ 8! = 8\cdot 7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1 \]
right ?

Answer 15

why did you put after 3 2 1
5 4 3 2 1 ?

Answer 16

8C5 is short for
\[ \frac{8!}{5! (8-5)!} \]
or
\[ \frac{8!}{5!\ 3!} \]
right ?

Answer 17

okay i get that part now .

Answer 18

next, you need to know how fractions "work"
when you multiply them, you multiply top times top and bottom times bottom
\[ \frac{2}{3}\cdot \frac{3}{4} = \frac{2\cdot 3}{3\cdot 4} \]
but we can change the order of multiplying so we could say
\[ \frac{2\cdot 3}{3\cdot 4}= \frac{2\cdot 3}{4\cdot 3} \]
and we can re-write that by "undoing the multiply"
\[ \frac{2\cdot 3}{4\cdot 3} = \frac{2}{4}\cdot \frac{3}{3} \\= \frac{2}{4} \cdot 1 \\= \frac{2}{4} \]

Answer 19

that is a long-winded explanation of why you can "cancel" a number that is in the top and bottom

Answer 20

\[ \frac{8\cdot 7 \cdot 6 \cdot \cancel{5 \cdot 4 \cdot 3 \cdot 2 \cdot 1}}{3 \cdot 2\cdot 1\cdot \cancel{5 \cdot 4 \cdot 3 \cdot 2 \cdot 1}} \]

Answer 21

\[\frac{ 11 }{ 5 }\] = \[\frac{ 11! 10! 9! 8! 7! 6! 5! 4 3! 2! 1! }{ 5!4!3!2!1! }\]

Answer 22

or not ?

Answer 23

11! means 11 * 10 * 9 .... *1
notice no ! except for 11!

Answer 24

or is the 5 going to be 5 4 3 2 1 11 10 9 8 7 6

Answer 25

and we say 11C5 because it's more complicated than what you wrote
the top is 11!
the bottom has 2 numbers
5! (from the 5) and (11-5)! which shortens to 6!
so all together
\[ \frac{11!}{5! \ 6!} \]

Answer 26

462 i mean

Answer 27

how do they get that number

Answer 28

First, do you see how to change
11C5 to \( \frac{11!}{5! \ 6!} \)

Answer 29

yess 11 at the top and take 11-5 which gives you 6

Answer 30

and the 5! in the bottom also

Answer 31

so you get 11/ 5!6!

Answer 32

11! / 5! 6!

be careful

Answer 33

cause if you add 6 and 5 you get 11

Answer 34

yes

Answer 35

now to get the number that represents, I would pick the biggest number in the bottom (the 6!)
and cancel the 6! up top

remember 11! means
11*10*9*8*7* (6*5*4*3*2*1)
notice the 6! in the bottom will "cancel" the (6*5*4*3*2*1) up top

Answer 36

okay

Answer 37

in other words
\[ \frac{11\cdot 10 \cdot 9 \cdot 8 \cdot 7 \cdot\cancel{6\cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1} }{5! \cancel{6\cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1} }\]

Answer 38

do i take the 5! and put 5! 4 3 2 1

Answer 39

?

Answer 40

next rewrite 5! as 5*4*3*2*1
\[ \frac{11\cdot 10 \cdot 9 \cdot 8 \cdot 7 }{ 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1} \]

Answer 41

yaaayyy i got my answer!!!!

Answer 42

you sure like !. when you expand 5! the ! goes away.

Answer 43

thank youuuuuuuuu

Answer 44

can we do another one ?

Answer 45

ok

Answer 46

l give you another question and ill try to solve it and you tell me if i am wrong or write

Answer 47

please and thank you

Answer 48

post some of your work so we can make sure you stay on the right path.

Answer 49

okay

Answer 50

here is the question
An employer interviews 7 people for 4 management positions in a company, and 3 of the 7 people are men. If all 7 are qualified, in how many ways can the employer fill the management positions if exactly 2 are men?

Answer 51

here are my euations\[\frac{ 7 }{ 4 } and \frac{ 3 }{ 7 }\] i want to say 7/2 but i don t thintk that is correct its a trick one right ?

Answer 52

i think only the 7/4 is correct.

Answer 53

you should use C (for choose)

But yes, it's a bit tricky (I often seemed to get "wrapped around the axle" when I learned this stuff).

I would break it down into 3 men and 4 women

there are 4 jobs.

exactly 2 are men

now many ways can we pick 2 men out of 3 ?
and once we have 2 jobs filled, we are left with 2 more jobs to fill using the 4 women
how many ways are there to choose 2 women out of 4 ?

Answer 54

7 Choose 4 or 7C4 would mean we have to pick 4 people out of 7

that would work if we did not care about men/women.

but if you have subcategories i.e. men and women, we break the 7 into
3 and 4
and pick *separately* from each category

Answer 55

For the men we would do 3 choose 2 (i.e. out of 3 men, choose 2 of them)
Then for the women, we would do 4 C 2 (out of the 4 women, choose 2 of them)
we end up with filling all 4 jobs (2 men and 2 women)