@genuineopossum to clarify, 9 is supposed to be (10-x) for 1 and 8is supposed to be (10-x) for 2 and so on.

@audrey said in Module 0 Week 3 Challenge Q20:

@fantasticcrow Yay, I'm glad that helped!! And as for counting versus probability, it all depends on what the question asked for. To give a simple example:

How many ways are there to make a three digit number where all the digits are even? [This would be a counting problem, because you want to count how many 3-dig #'s there are where all the digits are even]

Vs.

If you choose any three digit number at random, what is the probability that all its digits are even? [This would be a probability problem, because you're asked what the chance of something happening is-- so in this case, you would take the answer from #1, and divide by how many three digit numbers there are!]

In the context of this problem, we can't use simple counting because not all of the options are equally likely to be chosen. In other words, you should use counting when all the options are equally likely ("weighted")-- for example, choosing a three digit number at random. But in this case, they're not, so we have to actually multiply out probabilities, if that made sense.

Very helpful. Thank you!

@lucky_ducky1 thanks! you gave a great explanation.

-1x-2+3-4+(5x6-7^2)- -18=?

@RZ923 wait... What???? Prof.Loh = Married ???

@excitedarmadillo-0 No, it is just the same test, it doesn't change

@Desolate_101 yes

@daringhorse i think you meant 3^0+3^1 and 5^0+5^1 right, i just want to check

Thank you

So sorry about that @professionalbronco !!

But yes, your solution is absolutely right. Another way of looking at is the "algorithm" Professor Loh mentioned in the video: by sorting alphabetically, you'll get that there are 7 options for person A to make a pair, 5 options for the second pair, 3 for the third, and 1 for the last. $$7\cdot5\cdot3\cdot1=\boxed{105}$$ as well.

Also from the lesson, using the expression you had but expanding out the 4!:
$$\frac{\binom{8}{2}\cdot\binom{6}{2}\cdot\binom{4}{2}\cdot\binom{2}{2}}{4\cdot3\cdot2\cdot1}$$
and then expanding the binomial coefficients and cancelling:
$$\frac{\frac{8\cdot7}{2}\cdot\frac{6\cdot5}{2}\cdot\frac{4\cdot3}{2}\cdot\frac{2\cdot1}{2}}{4\cdot3\cdot2\cdot1} \implies \frac{(\cancel{4}\cdot7)\cdot(\cancel{3}\cdot5)\cdot(\cancel{2}\cdot3)\cdot(\cancel{1}\cdot1)}{\cancel{4}\cdot\cancel{3}\cdot\cancel{2}\cdot\cancel{1}}$$

which leaves us the same expression for \( 7!! \) 🙂

also !!
did you know that the exclamation in the factorial is kind of like the "index" of how much you're shifting each number (double factorial shifts by 2, triple by 3, and so on). Soooo... technically we could get to triple factorials (or higher) maybe:
\(8!!!=8\cdot5\cdot2=80\)
\(11!!!!=11\cdot7\cdot3=231\)

@RZ923 agreeed

@Todymaster 2017 macbook air 13' 8gb ram 1600mhz ddr3 1.8 chz dual core intel i5 with intel hd graphics 6000 1536mb

@debbie he's done 1,000 exams? That's a sheesh moment

@nastya Do you have a list of each Greek letter's uses? That would be very helpful.

RIP its been 2 years and its still on debbie's todo list

@inspirationaljackrabbit pretty cool shape ngl

@The-Blade-Dancer (long leg)^2 = (3^2) - (3 / 2)^2 which simplifies to 3/2√3