I broke maths.

[Hutchoo]

Just wondering if you'd like to have my babies now or some other time?

LOL. LOVE YOU MAN <3
Tytyty for the help guys

[TrueTears]

now pls

shall i go book a room now?

[mmonn1906]

Just went through the whole thread, amazing!
This looks like a total different language, but most of you say it's a fun and enjoyable subject. Which is good as.
Hope I have as much fun as you guys when I do it, I'll definitely come to VN for help

[Hutchoo]

Hmm, yeah. I like to ask the basic/fundamental questions here because everyone goes in depth.

Btw, this isn't just methods work, some of it is GMA (General Math Advanced)/specialist stuff.

[Hutchoo]

Just went through the whole thread, amazing!
This looks like a total different language, but most of you say it's a fun and enjoyable subject. Which is good as.
Hope I have as much fun as you guys when I do it, I'll definitely come to VN for help

Looks like you speak a different language yourself. Are you having fun negging my posts?

[Hutchoo]

I don't really understand this type of probability :/

There are 4 awards (Best player, Most improved, Most Consistent and Best effort) to be presented the players in the U16 softball team. If there are 15 players in the squad, and no player may receive more than one award, find the number of different ways the awards can be distributed.


AND ANOTHER ONE:
The 100m dash at the athletic sports has 8 competitors. In how many different ways can the first, second and third place ribbons be distributed?

Dw, got it. Thanks.

[TrueTears]

combinations, (15, 4) and then (8, 3)

[Hutchoo]

I honestly don't know how to do this question :/
Probability makes no sense to me... sigh.

Thanks in advance.

[xZero]

so you have 6 main courses, find the number of ways you can arrange them (order doesn't matter). now for starters, there are 3 options, you can choose either 3 or none so thats 4 options in total, follow that chain of logic for entrees and desserts and just times everything together at the end

[kamil9876]

Sometimes it is easier to calculate the complement. Let's count how many different meals there are where you can choose at most one from each category (and there is no restriction on how many courses you want). This is simply . Now let's count the complement: ie how many meals there are where you choose less than 2 courses. There is one way where you choose no courses, then by choosing only one course there are 3+5+6+3 different ways. So there are 4*6*7*4-(3+5+6+3)-1 ways of choosing at least 2 courses.

edit: fixed mistake, thanks Mao for pointing it out.

[Mao]

Sometimes it is easier to calculate the complement. Let's count how many different meals there are where you can choose at most one from each category (and there is no restriction on how many courses you want). This is simply . Now let's count the complement: ie how many meals there are where you choose less than 2 courses. There is one way where you choose no courses, then by choosing only one course there are 3+5+6+3 different ways. So there are 3*5*6*3-(3+5+6+3)-1 ways of choosing at least 2 courses.

I don't quite agree with this answer. Let's have A=3, B=5, C=6 and D=3

My line of thinking:
2 courses: A*(B+C+D) + B*(C+D) + C*D
3 courses: A*B*(C+D) + A*C*D + B*C*D
4 courses: A*B*C*D

Total is 654.


Upon revisiting, I see what has happened. Your method is the same as xZero's method, which is more elegant than mine. There is a careless mistake made in that the choice of entree (for example) is 4. Thus the number of different meals is 4*6*7*4. Take away the ways to have only 1 course and zero course, 4*6*7*4-(3+5+6+3)-1=654, the same as my brute-force answer.

[Hutchoo]

- seriously i can't delete posts sigh

[Hutchoo]

Hi guys, bit confused with this question!
The answer is 1.

[Dangster]

Hi guys, bit confused with this question!
The answer is 1.

log₅2x + log₅6 = log₅12x

therefore, log₅12x = log₅(4x+8)
12x = 4x+8
8x = 8
x = 1

[REBORN]

LHS is log5(2x times 6) therefore log5(12x)

Get rid of the log5

12x=4x+8
x=1