Math problem

[joonkimdds]

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it's been so long since i took math classes so i forgot how to solve this simple problem. so plz be generous to teach me 😀
a house has 4 rooms, and a painter has 9 colors. if the rooms r a,b,c, and d, how many diff ways can he paint the rooms?

solution says 9P4= 9!/(9-4)! = 9!/5!=9*8*7*6*5!/5! = 3024

I don't know what P stands for and what ! stands for.

there is a similar problem.

there r 6 people with 4 diff positions. how many diff committtes can consist of these offices?
this time, solution says 6P4 and blah blah and answer is 3601.

I don't know why 6 comes infront of P and 4 comes afterward.
Why can't I write 4P6?



oh and 1 more q uestion if u don't mind 🙂
the book i read says there are 81 numbers between 10 and 90. I think it should be 79... 😕
number between 1 and 3 would be 2 so why would they include those two boundaries as the number between?

 

[plooberman]

1. well for the first couple problems, you can think of it as simply having 9 different typs of paint to choose for the first house "a". choose one, and that leaves 8 differnt paints for house "b", 7 for "c", 6 for "d".

9x8x7x6 is your answer. This is the easiest way to think about that problem.

2. "!" means factorial. when "!" appears infront of a number "n", you have to multiply all whole numbers between "n" and "zero".

E.g. 5! = 5 factorial = 5x4x3x2x1

9P4 or 9!/(9-4)! are fancier ways of solving that problem. you learn in precalculous or algebra that n!/(n-#)! is a formula to solve this kind of problem. (9x8x7x6x5!)/5! => 9x8x7x6 (5! cancel out)

I forgot what the P stands for but just memorize that you have to include the # of options (paints) before the "P" and the # of objects (houses to be painted) after the P

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not sure about the 81 numbers one...where are these problems coming from? there could be a mistake in ur answer key. good luck

it's been so long since i took math classes so i forgot how to solve this simple problem. so plz be generous to teach me 😀
a house has 4 rooms, and a painter has 9 colors. if the rooms r a,b,c, and d, how many diff ways can he paint the rooms?

solution says 9P4= 9!/(9-4)! = 9!/5!=9*8*7*6*5!/5! = 3024

I don't know what P stands for and what ! stands for.

there is a similar problem.

there r 6 people with 4 diff positions. how many diff committtes can consist of these offices?
this time, solution says 6P4 and blah blah and answer is 3601.

I don't know why 6 comes infront of P and 4 comes afterward.
Why can't I write 4P6?

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oh and 1 more q uestion if u don't mind 🙂
the book i read says there are 81 numbers between 10 and 90. I think it should be 79... 😕
number between 1 and 3 would be 2 so why would they include those two boundaries as the number between?

 

[Chga]

woah what book is this?

 

[joonkimdds]

woah what book is this?

It's from DAT destroyer. I got it for free from this website.
I went to DAT destroyer website and saw a sample problem. then i came back here and wrote that sample from the website seems good to me and the person who works in that website saw what I posted. He appreciated what i have done(what did i do...saying the book is good ) and he sent me this huge book that has so many good problems and very good explanations. It includes org chem, gen chem, bio and quantitative reasoning.
Written by Dr. Jim Romano.

oh and now i understand why it's 81. I just had to reread the problem 😍

 

[Law0830]

Hi i was just looking at the math questions that were posted and its the same ones that i have been working on through the DAt Destroyer. Heres what i think, first off P is permutation and it is used for probability questions concerning arrangements. Now C is combinations its also used for probability questions concerning how many different ways you can do something but your forming groups of things, so you have to factor in the amount of what group your making and how large it is. This would cause the possibile amount of arrangements to decrease in value because it is now more specific. So 6P4 is just basically 6 * 5 * 4 * 3. and 9 P 4 is 9* 8 * 7 * 6. Now 6 C 4 would equal to 6 * 5 * 4 * 3 divided by 4 * 3 * 2 *1.

Now for the numbers between 10 and 90 you have to take into effect the 10 and 90 giving you a total of 81 numbers.

 

[joonkimdds]

Hi i was just looking at the math questions that were posted and its the same ones that i have been working on through the DAt Destroyer. Heres what i think, first off P is permutation and it is used for probability questions concerning arrangements. Now C is combinations its also used for probability questions concerning how many different ways you can do something but your forming groups of things, so you have to factor in the amount of what group your making and how large it is. This would cause the possibile amount of arrangements to decrease in value because it is now more specific. So 6P4 is just basically 6 * 5 * 4 * 3. and 9 P 4 is 9* 8 * 7 * 6. Now 6 C 4 would equal to 6 * 5 * 4 * 3 divided by 4 * 3 * 2 *1.

Now for the numbers between 10 and 90 you have to take into effect the 10 and 90 giving you a total of 81 numbers.

TY. I wonder how you got DAT destroyer(did u buy it?).
Now I understand what 6p4 stands for but then if i see 6 ppl with 4 positions, how do i know if it's 6p4 or 4p6? aren't 6 ppl with 4 position and 4 ppl with 6 position supposed to have the same probability?

Same thing with the painting. aren't 4 rooms with 9 colors supposed to have the same probability as 9 rooms with 4 colors?

 

[Law0830]

I understand why you would think that but no thats not true. Yes it would give you the same outcome but the answer you woudl get would differ in units. What i mean is that you would get not how many ways the rooms could be painted but how many rooms can fit into 4 different paints!!! So you have to use the analogy of what is being done to something the what is before the P and what you will be doing is placed on the right.

I actually purchased the DAT Destroyer. I see you got it for free howd you get it. One of my friends just reccommended it to me so i got it and its been my best investment in my life so fdar. Very challenging problems with new twists that i would have ordinarily have thought of. I feel a lot more confident.

 

[aud336]

it's been so long since i took math classes so i forgot how to solve this simple problem. so plz be generous to teach me 😀
a house has 4 rooms, and a painter has 9 colors. if the rooms r a,b,c, and d, how many diff ways can he paint the rooms?

solution says 9P4= 9!/(9-4)! = 9!/5!=9*8*7*6*5!/5! = 3024

I don't know what P stands for and what ! stands for.

there is a similar problem.

there r 6 people with 4 diff positions. how many diff committtes can consist of these offices?
this time, solution says 6P4 and blah blah and answer is 3601.

I don't know why 6 comes infront of P and 4 comes afterward.
Why can't I write 4P6?



oh and 1 more q uestion if u don't mind 🙂
the book i read says there are 81 numbers between 10 and 90. I think it should be 79... 😕
number between 1 and 3 would be 2 so why would they include those two boundaries as the number between?

reply of the second question
subtract these two numbers,10 and 90, won't give you the answer you want. For example, there are 11 number between 10 an 20 because you are counting two "0s", 10 and 20. (10,11,12....20). I hope, I answer them correctly.