Im wondering if in the last exercise I can just expand both (x!) (x!) one to be negative and the other one to be positive? so I will end up with something like (x) (x-1)! (x)(x+1) / (x-1) (x+1) = (x)(x) = x ^2
So if I understand your question correctly, you can rewrite x! to be x(x-1)! but you cannot rewrite the other x! to be x(x+1)!, this is because the definition of factorial states that you multiply all the decreasing integers down to one (i.e. 4! = 4*3*2*1 or 4*3!), so we have to always decrease our values. That's why I hop over to the denominator where I have (x+1)! which I can rewrite as (x+1)*x! which is helpful to cancel out the x! in the numerator. As soon as we throw variables into these problems, it can get confusing. If you want to check your logic you can always temporarily choose a natural number to plug in for x, and test it out. Does it still make sense when you have numeric values? Does it adhere to all the rules and definitions? Sometimes this can be a helpful way to check your logic. Thanks for your question!
Hello there! Sorry, this response is so delayed, I just saw your question! 8! / 5!(8-5)! comes from the Combinations, or "choose", formula from combinatorics. If you're interested in learning more about this formula I talk about/derive it in my Combinations vs Permutations tutorial (ruclips.net/video/ssoTzwA_Aug/видео.html) 😊
Every factorial video is just so painfully simple… you did a good job in your explanations. But, how about something like. ((n-1)!/(n-k-1)!) + (k*(n-1)!)/((n-k)!*(n-1-k)!) And solving it algebraically not from a combinatorial proof perspective.
Brett is amazing! I share her videos with my classes.
Thank you for reasoning out loud, especially the relationship btw x in numerator and denominator. Very helpful in making sense of it all!
Oh good! So glad to hear that helped!
finally it all makes sense. really helpful!
Thank you very much ❤️ all the way from South Africa 🇿🇦
Cool, thanks for sharing the lesson about factorials. Video looks good!
Thank you so much for watching 🙂
this was so riveting
Thank you so much🥰🥰. It was helpful
Im wondering if in the last exercise I can just expand both (x!) (x!) one to be negative and the other one to be positive? so I will end up with something like (x) (x-1)! (x)(x+1) / (x-1) (x+1) = (x)(x) = x ^2
So if I understand your question correctly, you can rewrite x! to be x(x-1)! but you cannot rewrite the other x! to be x(x+1)!, this is because the definition of factorial states that you multiply all the decreasing integers down to one (i.e. 4! = 4*3*2*1 or 4*3!), so we have to always decrease our values. That's why I hop over to the denominator where I have (x+1)! which I can rewrite as (x+1)*x! which is helpful to cancel out the x! in the numerator. As soon as we throw variables into these problems, it can get confusing. If you want to check your logic you can always temporarily choose a natural number to plug in for x, and test it out. Does it still make sense when you have numeric values? Does it adhere to all the rules and definitions? Sometimes this can be a helpful way to check your logic. Thanks for your question!
@@MathHacks thx!
hey I have a question, when solving (8/5) how did u get the 5! (8-5)! I got lost in that part
Hello there! Sorry, this response is so delayed, I just saw your question! 8! / 5!(8-5)! comes from the Combinations, or "choose", formula from combinatorics. If you're interested in learning more about this formula I talk about/derive it in my Combinations vs Permutations tutorial (ruclips.net/video/ssoTzwA_Aug/видео.html) 😊
Compound Interest-Future Value please
But what is (n/2)!. I don't know how I can simplify this.
Every factorial video is just so painfully simple… you did a good job in your explanations. But, how about something like.
((n-1)!/(n-k-1)!) + (k*(n-1)!)/((n-k)!*(n-1-k)!)
And solving it algebraically not from a combinatorial proof perspective.
Lol