View Full Version : Attention: Math Genious!!!!!!! HELP NEEDED!!!

2.2vtec

07-20-2003, 11:27 PM

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

07-20-2003, 11:32 PM

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

Dope Dealer

07-20-2003, 11:32 PM

Eww, permutations....

Dope Dealer

07-20-2003, 11:34 PM

1. 144

AquamosH

07-20-2003, 11:34 PM

Originally posted by 88CRX

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

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

2.2vtec

07-20-2003, 11:37 PM

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

:eek:

Stratus_Power

07-20-2003, 11:40 PM

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

07-20-2003, 11:40 PM

[ _ _ _ ] _ _ _ _

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

07-20-2003, 11:42 PM

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

Sorry..

Stratus is right. I read it wrong.

Stratus_Power

07-20-2003, 11:44 PM

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

Melinda

07-20-2003, 11:47 PM

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

07-20-2003, 11:49 PM

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

07-20-2003, 11:53 PM

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

07-20-2003, 11:53 PM

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

Dope Dealer

07-20-2003, 11:54 PM

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

07-20-2003, 11:55 PM

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

07-20-2003, 11:56 PM

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

07-20-2003, 11:57 PM

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

07-20-2003, 11:58 PM

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

07-20-2003, 11:58 PM

then it should've been negative

Zephyr

07-20-2003, 11:59 PM

2. i think is 7!/3!

4. is 24 i think, use pascal's triangle

1

1 1

1 2 1

1 3 3 1

1 4 6 4 1

so 6 x 2 x 3=24

5. 11 nPr 4

plug those in ur graphing calculator... i cant find mine at the moment

Melinda

07-21-2003, 12:00 AM

Originally posted by Stratus_Power

then it should've been negative Hmm you are very right :thumbsup: Shows how much I've forgotten when i've been out of school

Weapon_R

07-21-2003, 12:00 AM

Originally posted by Melinda

You isolate the x

Still wrong :). I dont have the right answer, but it has something to do with a triangle.

1

1 1

2 2 2

3 4 4 3

4 6 8 6 4

5 10 14 14 10 5

Ring a bell for anyone?

max_boost

07-21-2003, 12:00 AM

You don't isolate the X, you multiply the stupid thing out 5 times! :rofl:

Melinda

07-21-2003, 12:01 AM

Bah, i give up! :D

D'z Nutz

07-21-2003, 12:01 AM

Originally posted by Melinda

Ahh shite...okay here goes again

(2x+3)^5

= 32x+243

= 7.59375 (convert that to fraction if you want)

No, you can't just multiply the power inside the brackets. For example,

(x+2)^2 = (x+2)(x+2) = x^2 +4x + 4,

Not x^2 + 4

(2x+3)^5 = (2x+3)(2x+3)(2x+3)(2x+3)(2x+3)

Zephyr

07-21-2003, 12:02 AM

Originally posted by Weapon_R

Still wrong :). I dont have the right answer, but it has something to do with a triangle.

1

1 1

2 2 2

3 4 4 3

4 6 8 6 4

5 10 14 14 10 5

Ring a bell for anyone?

refer to my triangle. i just taught this stuff two days ago to a kid...

and i just took precalculus last year still fresh in my mind, all this combination and perminatation

Dope Dealer

07-21-2003, 12:02 AM

I'm still trying to figuire out #2. The answer could also be 3!x5!.

I dunno, I forgot how to do Premutations and Combo's.

Weapon_R

07-21-2003, 12:04 AM

Originally posted by Zephyr

refer to my triangle. i just taught this stuff two days ago to a kid...

and i just took precalculus last year still fresh in my mind, all this combination and perminatation

Exactly why that triangle was an example...I couldn't even remember the name of it, but the idea was there :). It's been a long time since I took my last calculus course

Stratus_Power

07-21-2003, 12:06 AM

number 2 is tricky.. either all 3 letter must be vowel to start w/ or just the 1st letter has to be vowel.. if all 3 must be. then

3X2X1X4X3X2X1 = 144

Zephyr

07-21-2003, 12:06 AM

yea i had to do a 30 pascal triangle.. so i have it pretty much memorized... plus i just had the same stuff last year...and i just got hired to tutor

Zephyr

07-21-2003, 12:07 AM

Originally posted by Stratus_Power

number 2 is tricky.. either all 3 letter must be vowel to start w/ or just the 1st letter has to be vowel.. if all 3 must be. then

3X2X1X4X3X2X1 = 144

its total amount which is 7! (7X6X5X4X3X2X1) divided by the three vowels 3! (3X2X1)

D'z Nutz

07-21-2003, 12:07 AM

For #2,

Isn't it just 3 x 6! ?

Unless I'm reading the question wrong, isn't it asking that the first letter be a vowel, and then after that, the rest of the letters can be put anywhere, including the remaining vowels?

3 x 6 x 5 x 4 x 3 x 2 x 1

Zephyr

07-21-2003, 12:08 AM

Originally posted by D'z Nutz

For #2,

Isn't it just 3 x 6! ?

Unless I'm reading the question wrong, isn't it asking that the first letter be a vowel, and then after that, the rest of the letters can be put anywhere, including the remaining vowels?

3 x 6 x 5 x 4 x 3 x 2 x 1

its basically a combination with limits just like i have MOM how many combos can i make? but M repeats twice so its

3!/2!

Weapon_R

07-21-2003, 12:09 AM

Originally posted by Stratus_Power

number 2 is tricky.. either all 3 letter must be vowel to start w/ or just the 1st letter has to be vowel.. if all 3 must be. then

3X2X1X4X3X2X1 = 144

I think that is the right way. It asks for the number of combinations that can be made beginning with the three vowels.

Stratus_Power

07-21-2003, 12:09 AM

i did alll the other questions ( except 5 ) in page 1

Dope Dealer

07-21-2003, 12:09 AM

Originally posted by Zephyr

its total amount which is 7! (7X6X5X4X3X2X1) divided by the three vowels 3! (3X2X1)

John,

If all three letters had to be values, couldn't it be 3! x 5! also?

With the packaging method..

[ V V V ] L L L L

3 combo's to start, 4 to follow.

5 elements in total.

Zephyr

07-21-2003, 12:12 AM

yea maybe that way, i rarely have done this limitation to this extreme, usualy its just like

IDIOT, how many ways can i arrange it?

5!/2!

edit: o wait nm...maybe some other way i keep forgetting SHIT NOW IM LOST

2.2vtec

07-21-2003, 12:12 AM

nP2, that 2 beside the P is a small subsribt below the P...same with the n.

Ok last question...kinda long..

standard licsence plates consist of 3 letters and 3 didgets.

1. determine the maximum number of standard licsence plates avaliable. state any assumption(s) you make.

2. determine the number of license plates that include the letter Z

3. assume it becomes necessary to increase the number of avaliable license plates. Provide 2 alternative proposals in which a maximum of 6 characters consisting of letters, and/or didgits can be used. Include any assumptions you make and determine the number of license plates for each proposal

Zephyr

07-21-2003, 12:14 AM

uhh 1. im gonna take a wacking at it but is it 10x6x26x6 (assuming 26 letters in the alphabet, 10 digits and 6 spaces on the plate)

Dope Dealer

07-21-2003, 12:15 AM

1. License Plates

L L L - D D D

26 x 26 x 26 - 10 x 10 x 10

= 17576000 Possible Plates Available.

Zephyr

07-21-2003, 12:16 AM

Originally posted by Dope Dealer

1. License Plates

L L L - D D D

26 x 26 x 26 - 10 x 10 x 10

= 17576000 Possible Plates Available.

OHHHH ASSUMING 3 LETTERS 3 DIGITS.... yea then that would work... assuming that the first is a letter not a digit cuz i never seen a plate with zero as first thing

Dope Dealer

07-21-2003, 12:17 AM

Originally posted by Zephyr

OHHHH ASSUMING 3 LETTERS 3 DIGITS.... yea then that would work... assuming that the first is a letter not a digit cuz i never seen a plate with zero as first thing

Then it would be 9 x 10 x 10 x 26 x 26 x 26.

But I think it's talking standard license's in Alberta.

L L L D D D

2.2vtec

07-21-2003, 12:19 AM

ok i'm getting alot of answers...could somone(stratus power) put all the answers on one post so it's more clear. Thanks guys and PM your number and I'll take you out for beers!!

Dope Dealer

07-21-2003, 12:19 AM

2.

_ _ _ = _ _ _

L L L D D D

1 x 26 x 26 x 10 x 10 x 10

676 000

Zephyr

07-21-2003, 12:20 AM

the hardest combination question i seen is "how many possible combinations of 5 card straights can u make out of a 52 card deck?"

Zephyr

07-21-2003, 12:21 AM

3. is 11nPr4 = 7920

Dope Dealer

07-21-2003, 12:23 AM

3.

Proposal 1

4 Letters, 2 Digits

_ _ _ _ - _ _

L L L L D D

26 x 26 x 26 x 26 x 10 x 10

= 45 697 600

Proposal 2

All Digits

6 Letters 0 Digits

_ _ _ _ _ _

L L L L L L

26 x 26 x 26 x 26 x 26 x 26

= 308 915 776

Dope Dealer

07-21-2003, 12:23 AM

Originally posted by Zephyr

3. is 11nPr4 = 7920

Yes.

Zephyr

07-21-2003, 12:24 AM

evan try doing "how many possible combinations of 5 card straights including random suit combos can u make out of a 52 card deck?" :rofl:

Stratus_Power

07-21-2003, 12:25 AM

Originally posted by 2.2vtec

nP2, that 2 beside the P is a small subsribt below the P...same with the n.

Ok last question...kinda long..

standard licsence plates consist of 3 letters and 3 didgets.

1. determine the maximum number of standard licsence plates avaliable. state any assumption(s) you make.

2. determine the number of license plates that include the letter Z

3. assume it becomes necessary to increase the number of avaliable license plates. Provide 2 alternative proposals in which a maximum of 6 characters consisting of letters, and/or didgits can be used. Include any assumptions you make and determine the number of license plates for each proposal

1. 26X26X26X10x10X10 = 17576000 assumption: letters/digits can be repeated

2. 1 X 26 X 26 X 10 X 10 X 10 = 676000 assumption: letters/digits can be repeatd

3. replace one of the digit with letters

so

26 X 26 X 26 X 26 X 10 X 10 = 45697600

OR

for # 1 and 2... use the assumpetion of letters/digits CANNOt be repeats.. so if u want more combo, you can make it so they are allowed to repeat itself

Stratus_Power

07-21-2003, 12:26 AM

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.

or either all 3 letter must be vowel to start w/ then

3X2X1X4X3X2X1 = 144

3. same idea, 11X10X9X8 = 7920

4. i know there is way to do this.. but its been WAYYY too long, cant remeber.

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

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

07-21-2003, 12:26 AM

Originally posted by Stratus_Power

1. 26X26X26X10x10X10 = 17576000 assumption: letters/digits can be repeated

Also assume that a number cannot be the first character on the plate.

Stratus_Power

07-21-2003, 12:28 AM

Originally posted by Dope Dealer

Also assume that a number cannot be the first character on the plate.

doesnt really matter.. if its 3 lette rand 3 digits.. then even if u go first " letter" as number..it iwll stil be

10X26X26X26X10X10..

Dope Dealer

07-21-2003, 12:29 AM

Originally posted by Stratus_Power

doesnt really matter.. if its 3 lette rand 3 digits.. then even if u go first " letter" as number..it iwll stil be

10X26X26X26X10X10..

Yeah, but it's illegal to have a plate starting with 0....

I was just doing it logically :)

Dope Dealer

07-21-2003, 12:30 AM

Originally posted by Zephyr

evan try doing "how many possible combinations of 5 card straights including random suit combos can u make out of a 52 card deck?" :rofl:

Ugh....

I don't even know anything about cards in the first place :D

Do you know how to do it?

Stratus_Power

07-21-2003, 12:31 AM

Originally posted by Dope Dealer

Yeah, but it's illegal to have a plate starting with 0....

I was just doing it logically :)

:thumbsup: i agree w/ you

Stratus_Power

07-21-2003, 12:34 AM

Originally posted by Zephyr

the hardest combination question i seen is "how many possible combinations of 5 card straights can u make out of a 52 card deck?"

13312?

kaput

07-21-2003, 10:56 AM

.

V6-BoI

07-21-2003, 09:27 PM

Originally posted by kaput

I should be able to do all of these right now... I'm taking a university stats class thats pretty much based on perms and combs. I'm not reading all the shit to see what has and hasn't been solved though, so if someone reposts the questions that need to be done I'll give them a try.

:werd:

same here, I'm taking math 271 right now and it involves this kind of shit too, so I'll try to give it a shot of someone reposts the questions that need to be done.

kaput

07-21-2003, 09:41 PM

.

James

07-22-2003, 05:30 AM

All the Wrong answers mixed in are funny, so many different ones! :rofl:

Originally posted by Zephyr

evan try doing "how many possible combinations of 5 card straights including random suit combos can u make out of a 52 card deck?" :rofl:

1,392,230....there ya go.

Akagi Redsuns

07-22-2003, 07:51 AM

Originally posted by 2.2vtec

snip........

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

........snip

Originally posted by D'z Nutz

No, you can't just multiply the power inside the brackets. For example,

(x+2)^2 = (x+2)(x+2) = x^2 +4x + 4,

Not x^2 + 4

(2x+3)^5 = (2x+3)(2x+3)(2x+3)(2x+3)(2x+3)

Yep that's correct, it works out to,

X^5+10X^4+40X^3+80X+32

So the answer is the X^3 coeffecient which is 40.

This is about the only one I will do since I don't have time do it all. There are textbooks that explains all this and with examples you know? Hell the instructor should have been explaining it. Good luck to the GF, better start studying for the finals...it's still worth 50% of the final grade?

V6-BoI

07-22-2003, 05:02 PM

Originally posted by kaput

I'm in engg 319, but I bet either of us could solve it with a quick look through our math 30 notes.

Haha yeah that's true. Uni math is so much different.

Seanith

07-22-2003, 07:57 PM

Originally posted by kaput

I'm in engg 319, but I bet either of us could solve it with a quick look through our math 30 notes.

yeah i did that class last year, and those questions look easy. I'm too lazy to check my text book though :D

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