Combinations/Permutations.
\[_{10}C _{4}\] @mathslover
\(_{n} C_{r} = \cfrac{n!}{r!(n-r)!}\)
Plug-in n = 10 and r = 4 here.
\[\frac{ 10! }{ 4!(10-4)!}\]
Right... So, it becomes : \(\cfrac{10!}{4!(6)!}\)
10! = 10 * 9 * 8* 7 *(6 *5 * 4 * 3 * 2 * 1 ) 6! = 6 * 5 * 4 * 3 * 2 *1 4! = 4 * 3 * 2 * 1 In the expansion of 10! , notice that (6*5*4*3*2*1) is common with 6! So, 10! = 10 * 9 * 8 * 7 * 6!
Didn't we already do this? :O
Yes. @tHe_FiZiCx99 I got really mad about it because I lost all the notes that i took on it. And asked mathslover to help. >_<
Oh :o don't feel bad :>
Well...I don't know what to do next after what mathslover said. I need it like in step by step.
might i suggest this not really the way anyone does this?
\[_{10}C_4=\frac{\overbrace{10\times 9\times 8\times 7}^{\text{ 4 terms}}}{4\times 3\times 2}\] cancel first multiply last
\[_{10}C_4=\frac{10\times\cancel 9^3\times \cancel8\times 7}{\cancel4\times \cancel3\times \cancel2}=10\times 3\times 7=210\]
Ok. That's what was confusing me.
don't get married to the formula \[\frac{n!}{k!(n-k)!}\] it is just a formula, not for consumption
I like \[ _{10}C_4 = \frac{_{10}P_4}{4!} \]
Ok.. Thank you so much!!! @satellite73!!! I will definitely take notes on this!:) And put them in a safe spot!!:)
Therefore, we can say : \(\cfrac{10!}{4! \times 6!} = \cfrac{10 * 9 * 8 * 7 * 6!}{4! \times 6!}\) Thus, 6! will cancel out and we get : \(\cfrac{10 * 9 * 8 * 7 \times \cancel{6!}}{(4!) \times \cancel{6!}}\) \(\cfrac{10*9*8*7}{4*3*2*1}\)
Sorry, my net was not working so, couldn't post the next step. :(
@wio hahaha...lol..
It's ok. @mathslover I though you just decided not to help me..But thanks though. At least I got some help.
I just can't simply deny someone to help . Unfortunately, due to weather conditions , the internet access went down .. :( anyways, you got better help by @satellite73 .
hmm...that's what caused it/...bad weather!>_< But anyways....Thanks to everybody who has tried to help!!!!!:)
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