FUT 2-1 GX

GX 2-0 KC

FUT 2-0 TH

GEN 2-1 RRQ

PRX 2-0 GEN

GEN 2-1 RRQ

TS 2-1 TLN

TLN 2-0 DFM

DRX 2-0 TS

TS 2-1 TLN

DRX 2-0 DFM

T1 2-0 ZETA

T1 2-0 GE

BLD 2-0 ZETA

ZETA 2-0 GE

BLD 2-1 T1

EG 2-0 KRÜ

G2 2-0 EG

G2 2-0 KRÜ

LEV 2-0 LOUD

LOUD 2-0 100T

SEN 2-1 LEV

LEV 2-0 100T

C9 2-1 FUR

FUR 2-0 MIBR

NRG 2-0 C9

C9 2-0 MIBR

NRG 2-0 FUR

SEN 2-0 LOUD

:(

Unfortunately, many people are not on this platform to "Learn more about Valorant Esports while maintaining a positive presence in the community." I find many of the interactions on this website to consist of either someone flaming a player, someone flaming a team, someone impolitely disagreeing with someone or flaming someone for liking a certain team. Many people on this forum are here to argue with other people as a way to make themselves feel better. I think many of those people dislike your positivity.

TLDR: many people dislike geospliced because geospliced is a good source of positivity, and many people aren't on the forum for positivity.

I suspect people dislike geospliced simply because of how nice he is, however, I believe he is a friendly source of positivity on this platform.

geospliced is correct here, I don't think that vlr users are the best representation of the average EMEA fan. The majority of FNC fans probably do know that Derke is Finnish.

honestly im down, just rip the band-aid off now

Notice how i didn't critique how the season wasn't boring, I explained the reason for the boring offseason. My point can exist with your point. And no, it's not Leo Faria.

There won't be

They also needed a longer break last year to get franchising up and running. Different reason.

We had to have a longer break to get VCT China up and running, I don't understand complaining. There will be less of a break in future seasons.

go to sleep, easy

the moose is on the loose

i have three PCs, i play solo q on one, duo q on the other, and 5 stack on the last PC all at the same time

pop a caffeinated beverage of your choice and play val, you're not sleeping tonight

YES

PLEASE I WANT TO DO MY PACIFIC PICK EM'S PLEASE I BEG PLEASE

21816

To solve this problem, we can use a combinatorial approach. We'll break it down into cases:

Choose 1 red ball and 1 blue ball from each of the 4 boxes.
Choose 2 red balls and 1 blue ball from one box, and 1 red ball and 2 blue balls from each of the remaining 3 boxes.
Choose 1 red ball and 2 blue balls from one box, and 2 red balls and 1 blue ball from each of the remaining 3 boxes.
Let's calculate the number of ways for each case:

Choosing 1 red ball and 1 blue ball from each of the 4 boxes:
(
3
1
)
×
(
2
1
)
(
1
3
​
)×(
1
2
​
) ways from each box
Total ways =
(
3
1
)
×
(
2
1
)
(
1
3
​
)×(
1
2
​
)^4

Choosing 2 red balls and 1 blue ball from one box, and 1 red ball and 2 blue balls from each of the remaining 3 boxes:
(
3
2
)
×
(
2
1
)
(
2
3
​
)×(
1
2
​
) ways to choose from one box
(
3
1
)
×
(
2
2
)
(
1
3
​
)×(
2
2
​
) ways to choose from each of the remaining 3 boxes
Total ways =
(
3
2
)
×
(
2
1
)
×
(
3
1
)
×
(
2
2
)
(
2
3
​
)×(
1
2
​
)×(
1
3
​
)×(
2
2
​
)^4

Choosing 1 red ball and 2 blue balls from one box, and 2 red balls and 1 blue ball from each of the remaining 3 boxes:
(
3
1
)
×
(
2
2
)
(
1
3
​
)×(
2
2
​
) ways to choose from one box
(
3
2
)
×
(
2
1
)
(
2
3
​
)×(
1
2
​
) ways to choose from each of the remaining 3 boxes
Total ways =
(
3
1
)
×
(
2
2
)
×
(
3
2
)
×
(
2
1
)
(
1
3
​
)×(
2
2
​
)×(
2
3
​
)×(
1
2
​
)^4

Now, we sum up the total number of ways from each case to get the final answer. Let's calculate it.

Let's calculate the number of ways for each case:

Choosing 1 red ball and 1 blue ball from each of the 4 boxes:
(
3
1
)
(
1
3
​
) ways to choose 1 red ball from each box
(
2
1
)
(
1
2
​
) ways to choose 1 blue ball from each box
Total ways =
(
3
1
)
×
(
2
1
)
(
1
3
​
)×(
1
2
​
)^4

Choosing 2 red balls and 1 blue ball from one box, and 1 red ball and 2 blue balls from each of the remaining 3 boxes:
(
3
2
)
(
2
3
​
) ways to choose 2 red balls from one box
(
2
1
)
(
1
2
​
) ways to choose 1 blue ball from one box
(
3
1
)
(
1
3
​
) ways to choose 1 red ball from each of the remaining 3 boxes
(
2
2
)
(
2
2
​
) ways to choose 2 blue balls from each of the remaining 3 boxes
Total ways =
(
3
2
)
×
(
2
1
)
×
(
3
1
)
×
(
2
2
)
(
2
3
​
)×(
1
2
​
)×(
1
3
​
)×(
2
2
​
)^4

Choosing 1 red ball and 2 blue balls from one box, and 2 red balls and 1 blue ball from each of the remaining 3 boxes:
(
3
1
)
(
1
3
​
) ways to choose 1 red ball from one box
(
2
2
)
(
2
2
​
) ways to choose 2 blue balls from one box
(
3
2
)
(
2
3
​
) ways to choose 2 red balls from each of the remaining 3 boxes
(
2
1
)
(
1
2
​
) ways to choose 1 blue ball from each of the remaining 3 boxes
Total ways =
(
3
1
)
×
(
2
2
)
×
(
3
2
)
×
(
2
1
)
(
1
3
​
)×(
2
2
​
)×(
2
3
​
)×(
1
2
​
)^4

Now, let's calculate each of these expressions.

Let's first calculate the number of ways for each case:

Choosing 1 red ball and 1 blue ball from each of the 4 boxes:
(
3
1
)
(
1
3
​
) ways to choose 1 red ball from each box
(
2
1
)
(
1
2
​
) ways to choose 1 blue ball from each box
Total ways =
(
3
1
)
×
(
2
1
)
(
1
3
​
)×(
1
2
​
)^4

Choosing 2 red balls and 1 blue ball from one box, and 1 red ball and 2 blue balls from each of the remaining 3 boxes:
(
3
2
)
(
2
3
​
) ways to choose 2 red balls from one box
(
2
1
)
(
1
2
​
) ways to choose 1 blue ball from one box
(
3
1
)
(
1
3
​
) ways to choose 1 red ball from each of the remaining 3 boxes
(
2
2
)
(
2
2
​
) ways to choose 2 blue balls from each of the remaining 3 boxes
Total ways =
(
3
2
)
×
(
2
1
)
×
(
3
1
)
×
(
2
2
)
(
2
3
​
)×(
1
2
​
)×(
1
3
​
)×(
2
2
​
)^4

Choosing 1 red ball and 2 blue balls from one box, and 2 red balls and 1 blue ball from each of the remaining 3 boxes:
(
3
1
)
(
1
3
​
) ways to choose 1 red ball from one box
(
2
2
)
(
2
2
​
) ways to choose 2 blue balls from one box
(
3
2
)
(
2
3
​
) ways to choose 2 red balls from each of the remaining 3 boxes
(
2
1
)
(
1
2
​
) ways to choose 1 blue ball from each of the remaining 3 boxes
Total ways =
(
3
1
)
×
(
2
2
)
×
(
3
2
)
×
(
2
1
)
(
1
3
​
)×(
2
2
​
)×(
2
3
​
)×(
1
2
​
)^4

Now, let's calculate each of these expressions.

Let's start by calculating the number of ways for each case:

Choosing 1 red ball and 1 blue ball from each of the 4 boxes:
(
3
1
)
(
1
3
​
) ways to choose 1 red ball from each box
(
2
1
)
(
1
2
​
) ways to choose 1 blue ball from each box
Total ways =
(
3
1
)
×
(
2
1
)
(
1
3
​
)×(
1
2
​
)^4

Choosing 2 red balls and 1 blue ball from one box, and 1 red ball and 2 blue balls from each of the remaining 3 boxes:
(
3
2
)
(
2
3
​
) ways to choose 2 red balls from one box
(
2
1
)
(
1
2
​
) ways to choose 1 blue ball from one box
(
3
1
)
(
1
3
​
) ways to choose 1 red ball from each of the remaining 3 boxes
(
2
2
)
(
2
2
​
) ways to choose 2 blue balls from each of the remaining 3 boxes
Total ways =
(
3
2
)
×
(
2
1
)
×
(
3
1
)
×
(
2
2
)
(
2
3
​
)×(
1
2
​
)×(
1
3
​
)×(
2
2
​
)^4

Choosing 1 red ball and 2 blue balls from one box, and 2 red balls and 1 blue ball from each of the remaining 3 boxes:
(
3
1
)
(
1
3
​
) ways to choose 1 red ball from one box
(
2
2
)
(
2
2
​
) ways to choose 2 blue balls from one box
(
3
2
)
(
2
3
​
) ways to choose 2 red balls from each of the remaining 3 boxes
(
2
1
)
(
1
2
​
) ways to choose 1 blue ball from each of the remaining 3 boxes
Total ways =
(
3
1
)
×
(
2
2
)
×
(
3
2
)
×
(
2
1
)
(
1
3
​
)×(
2
2
​
)×(
2
3
​
)×(
1
2
​
)^4

Now, let's calculate each of these expressions.

Let's calculate the number of ways for each case:

Choosing 1 red ball and 1 blue ball from each of the 4 boxes:
(
3
1
)
(
1
3
​
) ways to choose 1 red ball from each box
(
2
1
)
(
1
2
​
) ways to choose 1 blue ball from each box
Total ways =
(
3
1
)
×
(
2
1
)
(
1
3
​
)×(
1
2
​
)^4

Choosing 2 red balls and 1 blue ball from one box, and 1 red ball and 2 blue balls from each of the remaining 3 boxes:
(
3
2
)
(
2
3
​
) ways to choose 2 red balls from one box
(
2
1
)
(
1
2
​
) ways to choose 1 blue ball from one box
(
3
1
)
(
1
3
​
) ways to choose 1 red ball from each of the remaining 3 boxes
(
2
2
)
(
2
2
​
) ways to choose 2 blue balls from each of the remaining 3 boxes
Total ways =
(
3
2
)
×
(
2
1
)
×
(
3
1
)
×
(
2
2
)
(
2
3
​
)×(
1
2
​
)×(
1
3
​
)×(
2
2
​
)^4

Choosing 1 red ball and 2 blue balls from one box, and 2 red balls and 1 blue ball from each of the remaining 3 boxes:
(
3
1
)
(
1
3
​
) ways to choose 1 red ball from one box
(
2
2
)
(
2
2
​
) ways to choose 2 blue balls from one box
(
3
2
)
(
2
3
​
) ways to choose 2 red balls from each of the remaining 3 boxes
(
2
1
)
(
1
2
​
) ways to choose 1 blue ball from each of the remaining 3 boxes

​

The time complexity of searching an element in a red-black (RB) tree is O(log n), where n is the number of nodes in the tree.

Ask me anything

I also wanted to say that I have wanted to become more positive on this website because of you.

Thank you for your positivity on this website, Geospliced.
I am sorry that my fellow vlr users are downvoting you.