Help please (combinatory & permutations)

[tropic sigil]

Question: a) In how many ways can you draw a straight line from A to one of the numbers, and another straight line from B to another number (I got 56)

b) How many of the configurations in (a) satisfy that the lines do not cross? (I got 28)

c) In how many ways can you draw straight lines from A, B and C to three different numbers so that no lines cross?
I got 56 on this one too but me and my friend got different answers. I need help with specifically this question

BTW It's 3 letters and 8 numbers.

[tropic sigil]

<@&286206848099549185>

[solid robin]

First, look for the number of ways you can draw straight lines from A, B and , to three different numbers regardless of the crossings

As a second step, from the answer to my first question, deduce the number of ways without lines crossing each other. Hint : this is the same as having no repetition in your pairings (example, seeing A1 B2 C3 as identical as B1 A2 C3 or C1 A2 B3)

[tropic sigil]

Ill show you how i did it when i get home

[tropic sigil]

this is for b)

Idk i might have done something wrong on c), 56 sounds like a lot

[solid robin]

You got both right!
56 might sound a lot but look : if you compare to question b) for example, where you get only 28, for example for the combination A1 B2 you have choices from 3 up to 8 for C

And the large number of choices of C "outweighs" the restrictions you have (for example you don't have the combination A7 B8 where has it appears in b) )

[tropic sigil]

ohhhh theres 6!!! i just drew it out haahhaah

or wait

[solid robin]

Right ! :D

[tropic sigil]

ohhhh!!!!

xd

[tropic sigil]

thxxx this helped a lot

[solid robin]

However the more you add letters, eventually the less choices you would have
This is reflected by the binomial coefficients being maximal at n/2 C n and then decreasing again

$\binom{n}{n/2}$

[tropic sigil]

ohhhhhhh

wow

[elfin grail]

Twenty

[tropic sigil]

tysm wow

[solid robin]

Cool right? 😁

[tropic sigil]

u rly helped me

YES

tysmm

i was so confused all night yesterday

[solid robin]

No problem 😌

[tropic sigil]

have a good day!

[solid robin]

Glad it helped

You too !

[tropic sigil]

.close

[minor smelt]

Solved

Post marked as solved by @tropic sigil.

Use .unsolved if this was a mistake.

[tropic sigil]

sry this is silly but how would i write thsi with permutation and combinations? since A always has to be the smallest, B middle and C highest. and they only use 6 numbers each of the total 8.

so i use the bionomal coefficient?

[solid robin]

Yes, one definition of $\binom{k}{n}$ is "the number of possible ways of picking k elements among n elements without repetition"

[elfin grail]

Twenty

[solid robin]

And there is no such thing as a silly question especially in maths 😌

In terms of permutation, I'm not sure 🤔
I'd see that counting exercise as an (injective) function rather than a permutation

Like, you can count the number of functions $f$ from ${A,B,C}$ to ${1,2, \dots, 8}$ such that $f$ is an increasing monotonic function. (This correspond to the conditio' you said about A being the smallest, C the highest)

[tropic sigil]

i wrote this in my text but they might ask me to write it in some way

[elfin grail]

Twenty

[tropic sigil]

also i wrote it this way but combination and premutation gives the same answer lol

[solid robin]

Yeah, I believe the method you wrote is equivalent to this ^

[tropic sigil]

yeah

maybe i could remove the one i just sent

[solid robin]

Oh no you can leave it

[tropic sigil]

ill try sending it in and ill see what they say

thanks so much!

[solid robin]

Okay ! I'll be out a bit, if you have other questions I'll check it later 👍

[tropic sigil]

ty ty ty!

[solid robin]

[tropic sigil]

Hi @solid robin So they just replied and told me to consider that there could be combinations like 1, 2, 4 or 3, 5, 7 and so on. To solve that do I use combinations? So I did (8, 3) = 8! / 3!(8-3)! = 336/6 = 56

[solid robin]

That's it yeah, you choose 3 among 8 without repetition so that's $\binom{3}{8}$

[elfin grail]

Twenty

[solid robin]

Basically, if you want to count them all, that would be something like
1 2 3
1 2 4
1 2 5
1 2 6
1 2 7
1 2 8

1 3 4
1 3 5
1 3 6
1 3 7
1 3 8

1 4 5
1 4 6
1 4 7
1 4 8
etc
And then
2 3 4
2 3 5
2 3 6
...

2 4 5
2 4 6
...

Up until 6 7 8

[tropic sigil]

thanks i got it!!