Question

. Evaluate: 9C4

126
36
13
20

Answer 1

hint: \[nCr \implies \frac{n!}{r!(n-r)!}\]
so \[9C4 \implies \frac{9!}{4!(9-4)!}\]
does that help?

Answer 2

whats the choice

Answer 3

what is 9! @black12 ?

Answer 4

just tell me in expanded form don't calculate that

Answer 5

would it be 36

Answer 6

hmn i think you are slow with your basics in this topic:

\[\large{\color{red}{n! = 1*2*3*4* .... *(n-1)*n}}\]
so :
\[\large{\color{violet}{9!=1*2*3*4*5*6*7*8*9}}\]

got my point?

Answer 7

so what would the final answer be then

Answer 8

sorry we don't provide direct answers, but if you want step by step understanding and soln I can help you with this

Answer 9

\[\large{\frac{9!}{4!(9-4)!}=\color{violetred}{\frac{9!}{4!*5!}}=\color{red}{\frac{1*2*3*4*5*6*7*8*9}{1*2*3*4*(1*2*3*4*5)}}}\]
can u continue further?

Answer 10

I got the 9/20 part so it would be 20 then right

Answer 11

No no .. you cant multiply 4! * 5! = 20
that is purely wrong

you know about factorial?

Answer 12

if u dont multiply then what?

Answer 13

expand 4!

4! = 1*2*3*4
got it?

Answer 14

ok

Answer 15

\[\large{\frac{9!}{4!(9-4)!}=\color{violetred}{\frac{9!}{4!*5!}}=\color{red}{\frac{1*2*3*4*5*6*7*8*9}{1*2*3*4*(1*2*3*4*5)}}}\]

now can u tell me what will you get after above multiplications?

Answer 16

multiply each one of those numbers

Answer 17

cancel the terms that are same and then multiply

Answer 18

this stuff is confusing can we just skip the teaching part. whats the final

Answer 19

sorry friend i can not do that..

direct answer not allowed here

Answer 20

Two ways to do it.
1. Use calculator.
2. As the above mentioned, use \[_nC_r = \frac{n!}{r!(n-r)!}\]
Which part are you confused at?

Answer 21

tired too use a calulator and got some big number

Answer 22

Hmm.. Actually, using calculator is faster.. But if you want to do it on your own, it's okay.
So, what have you got so far? Which part are you confused at?

Answer 23

so do i jus calculate all the top numbers toghter then the bottom numbers and reduce

Answer 24

Hmm..
For '!' , you need to 'expand', and then cancel the common factors.
\[_9C_4 = \frac{9!}{4!(9-4)!}\]\[_9C_4 = \frac{9\times 8 \times 7 \times 6 \times 5!}{4!(5)!}\]
So, can you cancel the common factor here first?

Answer 25

can u say that again your message looked like this on my screen
Hmm..
For '!' , you need to 'expand', and then cancel the common factors.
\[_9C_4 = \frac{9!}{4!(9-4)!}\]\[_9C_4 = \frac{9\times 8 \times 7 \times 6 \times 5!}{4!(5)!}\]
So, can you cancel the common factor here first

Answer 26

That's it!

Answer 27

i dont understand

Answer 28

Which part?

Answer 29

so would it be 126 then

Answer 30

Indeed it is!