help

[feral blaze]

Help

[eternal lake]

1. Wait patiently for a helper to come along.
2. Once someone helps you, say thank you and close the thread with:

+close

3. Feel free to nominate the person for helper of the week in #helper-nominations
4. Do not ping the mods, unless someone is breaking the rules.
5. If you're happy with the help you got here, and the server overall, you can contribute financially as well:

[muted moon]

how did u get 28 @feral blaze

[feral blaze]

Idk

I solved it last year

[muted moon]

oh lol well

[feral blaze]

How do I solve it

??

[muted moon]

for these type of questions

usually its exponentiation

[silent badger]

i think u need to find the number of non zero positive integral solutions of x + y+ z = 6

which will be 28

[muted moon]

no

there are three types of chocolate

if u have 3 types of chocolate

and youre buying 6

u can buy it in a total of $3 \times 3\times 3\times 3\times 3\times 3$

[heady lintel]

wolfqz

[silent badger]

huh, i think this is wrong but it seems correct

@feral blaze what is the answer?

[feral blaze]

3⁶

I am not able to understand why tho

[silent badger]

well assume u have 6 places to put the chocolates into

in each place, u have 3 ways to put a chocolate because u have 3 types of them

[feral blaze]

Thankss

Makes sense

Thanks

Thankyou

Thank you

How does that point thing come up

Thank you @silent badger @muted moon

[fierce dome]

@feral blaze has given 1 rep to @silent badger @muted moon

[silent badger]

btw @muted moon i am trouble understanding why my solution was wrong, i seemed to have missed some cases but which ones?

[muted moon]

a lot

3^6 is astronomically large compared to 28

💀

[silent badger]

yea but what went wrong

[feral blaze]

Oh I was gonna close it but you guys have discussion

I'll read itt

[muted moon]

just count some cases

u will reach 29 in no time

[silent badger]

yea i get that but x + y +z = 6 where x y z are types of chocolates seems to intuitvely make sense

i wanna know why that isnt right

im having a feeling the answer is actually 28

wait

it is 28

3^6 is counting all the permutations and that too without considering repititions, we need selections, not permuatations

so 28 will be correct

we prolly should get some third person to recheck this

[feral blaze]

@silent badger @muted moon

Help with 48

Don't give the answee

Just help me a little

[silent badger]

sry i accidentally wrote the whole queation

[feral blaze]

I didn't read it

Yet

[silent badger]

number of games between men will be 2 * mc2

[feral blaze]

The games b/w the men will be

[feral blaze]

Yehh

[silent badger]

and number of games between women and men will be 2 * 2m

[feral blaze]

I got this much

[feral blaze]

That is thiss

[silent badger]

?

[feral blaze]

The 2m

Part

I don't get it

[silent badger]

oh, well there are 2 women and each plays 2 games with m men, so 2m * 2 games

[feral blaze]

Still don't get it

We're do we get the 2m

From

[silent badger]

well ok lets start with a small example

lets say there 2 women, and each women has to play 1 game with 2 men, how many games will there be?

2*2 = 4

[feral blaze]

I get that the games b/w the men will be mc2

But the games b/w men and women should be m+2c2 right???

[feral blaze]

Yehh

[silent badger]

or, think of it like this

u need to select 1 man so mC1 and 1 women so 2C1 so 2C1 * mC1 = 2m

[feral blaze]

Yehh

This makes

Sense

To mee

Thankss

[silent badger]

this will include cases where there are no women selected and only men, but those cases r already in mC2 and we are only looking for men-women matches, we need to specify what we select

[feral blaze]

Omg I am so stupid

Sorry

[silent badger]

@feral blaze i asked on another maths server, the answer is 28

[feral blaze]

Well again

How is it 28

[silent badger]

i'll think of a decent explanantion and be back to you in a few minutes

[feral blaze]

Ok

[silent badger]

assume that u have 6 identical coins

and u have to distribute all of them among three beggers A B C

let number of coins given to A,B,C be a,b,c respectively

=> a + b+ c = 6

now if u look carefully, the number of non-0 integral solutions of this equation is the answer to ur question, we just used beggers instead of types of chocolates

now, each time u give the coins, 2 partitions will be created b/w the 6 coins

now lets add 2 dummy coins to our 6 coins

all we need to do is select 2 coins out of the now 8 coins, and each time consider that selection to be the dummy coins and thus the partitions and thus a solution to the qeuation

so the number of solutions to the equation and hence ur question will be 8C2

i hope that is clear, for more details of this method, u can search "begger's method" on google, it's a really interesting way to solve such questions

u can ask if u have any doubts in this

[feral blaze]

I gof

It

I remember

How I got 28 last time

I'll ask my teacher

[feral blaze]

Too