Attention: Math Genious!!!!!!! HELP NEEDED!!!

[2.2vtec]

I need someone on here to answer the following math 30 questions. This is really important cause it's worth 30 percent of my girlfriends mark and she's stuck on these questions. First person to answer correctly I will personally take out for however many beers they want!!!

2. If all the letters in the word DIPLOMA are used, then the number of different 7 letter arrangements that can be made beginning with the 3 vowels is....


3. An exective consisting of a president, vice-president, and treasurer and secretary must be formed from a group of 11 people. Calculate the number of executives possiable.

4.In the expansion of (2x+3)to the power of 5, what is the coefficient of the term that contains x to the power of 3

5. Algebraically solve for n in nP2=30

6. In a group of people, there are 10 females and 12 males. Determine the number of 4-member committiees consisting of at least 1 female that can be formed.

Please even if you can only answer a couple it will be greatly appreciated!!

[88CRX]

holy shit.... only been out of hs for a year and i already forget everything oh well

[Dope Dealer]

Eww, permutations....

[Dope Dealer]

1. 144

[AquamosH]

Originally posted by 88CRX
holy shit.... only been out of hs for a year and i already forget everything oh well

I remember those kinds of q's were so easy too!

[2.2vtec]

Originally posted by Dope Dealer
1. 144

sweet man thanks!! Can you do anymore of them and possiable write how to do them?? She needs to show her work.

THANK YOU GUYS SO MUCH!!! THIS IS DUE TOMARROW AT NOON

[Stratus_Power]

Originally posted by 2.2vtec
I need someone on here to answer the following math 30 questions. This is really important cause it's worth 30 percent of my girlfriends mark and she's stuck on these questions. First person to answer correctly I will personally take out for however many beers they want!!!

2. If all the letters in the word DIPLOMA are used, then the number of different 7 letter arrangements that can be made beginning with the 3 vowels is....


3. An exective consisting of a president, vice-president, and treasurer and secretary must be formed from a group of 11 people. Calculate the number of executives possiable.

4.In the expansion of (2x+3)to the power of 5, what is the coefficient of the term that contains x to the power of 3

5. Algebraically solve for n in nP2=30

6. In a group of people, there are 10 females and 12 males. Determine the number of 4-member committiees consisting of at least 1 female that can be formed.

Please even if you can only answer a couple it will be greatly appreciated!!

im just guessing here.. been too long
2. 7 letters = __, __, __, __, __, __, __
first letter can be 1 of the 3 vowels.. so its 3 different combo
2nd letter can be any of the 6 remaining letter.. .3rd letter can eb any of the 5 remaining.. etc.. so
3X6X5X4X3X2X1 = 2160 different arrangment.
3. same idea, 11X10X9X8 = 7920
4. i know there is way to do this.. but its been WAYYY too long, cant remeber.
5. whats nP2 haha
6. 4 member.. so __, ___ , ___, ___
let say first spot HAS to a female. so it can have 10 different combo.. so its 10X
for the 2nd spot, it can either be male or female.. minus the one female picked.. so its 11.. etc..
10X11X10X9 = 9900 different combo..

[Dope Dealer]

[ _ _ _ ] _ _ _ _
3 2 1 4 3 2 1

First 3 letters are vowels.

The rest are the other letters.

3 x 2 x 1 x 4 x 3 x 2 x 1 = 144

I'm not really sure though. I can't remember the exact way, but I think it might be the answer.

[Dope Dealer]

Woops, I thought the vowels all had to be at beginning.

Sorry..

Stratus is right. I read it wrong.

[Stratus_Power]

oh i see wat u mean.. actually u might be right too..

[Melinda]

Originally posted by 2.2vtec
4.In the expansion of (2x+3)to the power of 5, what is the coefficient of the term that contains x to the power of 3

5. Algebraically solve for n in nP2=30

4. (2x+3)5
= 10x=15
= x=1.5

5. nP2=30
= nP=15
= n=15/P

[Stratus_Power]

5. nP2= 30
so
n! / (n-2)!= 30

so n (n-1) (n-2)/ (n-2) (n-3).... = 30.. bottom cancel out
so n (n-1) = 30
n^2 -n = 30
n^2 - n - 30 = 0
(n+5) (n-6) = 0

so n= -5 or n = 6
n must be real number . so n = 6

[Weapon_R]

Originally posted by Melinda


4. (2x+3)5
= 10x=15
= x=1.5

No.


Its not (2x+3)5, but (2x+3)^5.

[max_boost]

That is basically the entire Permutations and Combinations unit. Someone must of been sleeping through class.

[Dope Dealer]

Originally posted by Melinda


4. (2x+3)5
= 10x=15
= x=1.5

It's to the power of, you can't foil. And it says in relation to the power of 3.

I've been trying to do it, but can't.

[Melinda]

Originally posted by Weapon_R


No.


Its not (2x+3)5, but (2x+3)^5.

Ahh shite...okay here goes again

(2x+3)^5
= 32x+243
= 7.59375 (convert that to fraction if you want)

[Melinda]

Originally posted by Dope Dealer


It's to the power of, you can't foil. And it says in relation to the power of 3.

I've been trying to do it, but can't.

In relations to the power of 3??? What the??? Ahh I give up! I did one of them for ya

[Stratus_Power]

Originally posted by Melinda

Ahh shite...okay here goes again

(2x+3)^5
= 32x+243
= 7.59375 (convert that to fraction if you want)

how did u manage to come up w/ a real number answer w/ an unknown in the question?!

[Melinda]

Originally posted by Stratus_Power


how did u manage to come up w/ a real number answer w/ an unknown in the question?!

You isolate the x
x= the number

[Stratus_Power]

then it should've been negative