Question

Please Help. Permutations/combinations.

Answer 1

\[_{10}C _{4}\]

Answer 2

@mathslover Can you help me please?

Answer 3

Do you know the formula for \(_{n}C_r\) ?

Answer 4

Yes. It is \[_{n}C _{r}=\frac{ n! }{ r!(n-r)! }\]

Answer 5

\(_{n} C_r = \cfrac{n!}{r!(n-r)!}\)

Here you have : n = 10 and r = 4

So, you have : \(_{10} C_r = \cfrac{10!}{4!(10-4)!}\)

Answer 6

So, you have : \(_{10} C_r = \cfrac{10!}{4!(6!)}\)
Can you simplify this further?

Answer 7

I think so..Let me try and you tell me if I'm right...

Answer 8

Yep sure!

Answer 9

\[\frac{10\times9\times8\times7\times6\times5\times4\times3\times2\times1 }{ }\]

Answer 10

Kind of stuck on the bottom:(

Answer 11

\(\cfrac{10*9*8*7*6!}{4! \times 6!}\)

\(\cfrac{10*9*8*7 \times \cancel{6!}}{4! \times \cancel{6!}}\)

\(\cfrac{10\times 9\times 8\times 7}{4\times 3\times 2 \times 1}\)

Answer 12

Just simplify it now.

(I have to go for very urgent work. I am very sorry! @tHe_FiZiCx99 has his answer right and I am sure he will help you out if you get stuck at any step. good Luck! )

*Sorry again*

Answer 13

Alright..I guess @tHe_FiZiCx99 will help me. It's ok I forgive you!:)

Answer 14

I think I got it @tHe_FiZiCx99 Is it 210?

Answer 15

Apply the same logic! \(\ (4 * 3 * 2 * 1 = \color{green}{?} ) \) and (6 * 5 * 4 * 3 * 2 * 1 = \(\color{red}{?} \)

\(\color{red}{?} \)\(\ * ~ \color{green}{?} \)

Or you can easly apply the same idea that is when dealing with exponents

\(\ \sf \Large \dfrac{x^5}{x^2} = \dfrac{x * x * x * \cancel{x * x}}{\cancel{x * x}} = x * x * x = x^3 \)

\(\ \sf \Large \dfrac{x^5}{x^2} = x^{5-2} = x^3 \) :D get it?

Answer 16

Yes the overall answer is 210 :>

Answer 17

That is just totally confusing. Looks even a little challenging. Thanx for your help! @tHe_FiZiCx99 and @mathslover Yay!! I got it right!!

Answer 18

The only difference between them is that it HAS to be the same ones like mathslovers stated.

\(\ \cfrac{10*9*8*7 \times \cancel{6!}}{4! \times \cancel{6!}} \)

The 6! canceled out :)

Answer 19

Oh! Ok. makes sense now!!:) Thanx for explaining!:)

Answer 20

Yw

Answer 21

Thanks @tHe_FiZiCx99 for helping Yana.

Answer 22

Sure, you're welcome