@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

pretty sure it's because you can do a block-walking argument:
we start from the top of the triangle and we can only move down. so this means that the number of ways to get to a place on the triangle is equal to the sum of the ways to get to the two spaces above it - which is the definition of pascal's triangle. this means that every number on pascal's triangle also represents the number of ways to get to that number: 6248ca08-f23e-43be-87a7-4b1806d034dc-image.png
however, we can find the number of ways to get to this number! the total number of steps we will take is equal to the row that the number is in, because we move one row down every single step (the tip of the triangle with just a 1 is the 0th row). the position in the row corresponds to the number of right moves that are needed. this is because if we were to only have left moves, we would end up on the 1 on the very left. let's call the very left 1 the 0th number. so if we want to get to the 1st number, we would need 1 right move and the rest being left moves. if we wanted the 2nd number, we would need 2 right moves, etc. therefore, the number of ways to get to the kth number in the nth row is equal to the number of ways to arrange k left moves and n-k right moves. this is equal to n choose k. but since the number of ways is equal to the number, this means that the number is also n choose k.
the binomial theorem has coefficients of n choose 0, then n choose 1, all the way up to n choose n, which would correspond with row n of pascal's triangle.

tl;dr: the numbers in pascal's triangle represent the number of ways to get there, which in turn is equal to n choose k. since the binomial theorem has coefficients of n choose k for each k from 0 to n, it will correspond to row n of pascal's triangle.

@Desolate_101 yep it is lol

@RZ923 yeah, i was thinking the same thing, maybe it is for the teachers for the class or smthn