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Original post deleted. I would have left it here but it just caused confusion.

Sorry guest. 

Oh I will explain that I misread the question so everyone can ignore my answer.  

Thanks for your answer and site reference guest  :)

What Melody said....!!!

Probability is a "curse" upon the world of mathematics!! Here is EXACTLY the same question and the 6 answers given, including one by a Professor of Mathematics, almost all of which differ from each other in their final answer!

https://www.quora.com/In-how-many-ways-can-4-distinct-balls-be-distributed-into-3-identical-boxes

Just wanted to try this one, too...

a4 / a2   = 3

a4  = 3a2

a4  = a2 + 2d   ...so...

a2 + 2d  = 3a2

2d  = 2a2

d  = a2

And   a5 /a3  =

[ a2 + 3d  ] / [ a2 + d  ] =

[d + 3d ] / [ d + d ]

4d / 2d  =

2

This is what Melody got, as well !!!

Let the sequence be

a, a+d, a+2d, a+3d, a+4d

\(\frac{a+3d}{a+d}=3 \qquad find \qquad \frac{a+4d}{a+2d}\\ 3a+3d=a+3d\\ 2a=0\\ a=0\\ so\\ \frac{a_5}{a_3}=\frac{a+4d}{a+2d}=\frac{4d}{2d}=2\)

In a roomful of 30 people, what is the probability that at least two people have the same birthday? 

What is the prob that 30 people in a room all have different birthdays?  (asssuming no one is born on 29th Feb)

365/365*(364/365)*(363/365) ...... [(366-30)]/365 =    0.2937

So the prob that 2 or more have the same birthday is   1-0.2937 = 0.7063 approximately.

No one else actually gave an answer, but if someone wants to compare my answer to the formula answer that would be interesting.

There is always plenty of room to go wrong with probablility but then I don't trust formulas if I do not understand how they are derived. So I would not automatically accept a formula answer or my own answer.  Either or both could be wrong.

Maybe they are the same and they are both right. That would almost be a first!    LOL

How many ways are there to put 4 balls in 3 boxes if the balls are distinguishable but the boxes are not?

The  3 boxes are all different and the 4 balls are identical

All 4 be in any box   That is 3 ways

3 in one box and 1 in another that would be   3*2 = 6 ways

2 in one box and 2 in another that would be   3*2 = 6 ways

2 in one box and 1 in each of the others   3 ways

That is a total of 18 ways I think.

Hi DS it is really good to see you again :)

We have two integers, \(x\) and \(y\).

\(x*y = -12\)

\(x + y = -3\)

\(y = -3-x\)

\(x*(-3-x) = -12\)

\(-x^2 - 3x + 12 = 0\)

Now, we can use the quadratic formula to find the two factors: \(\frac{-b ± \sqrt{b^2 - 4ac}}{2a}\).

\(\frac{3 ± \sqrt{57}}{-2}\), so the two roots are \(\frac{3 - \sqrt{57}}{-2}\) and \(\frac{3 + \sqrt{57}}{-2}\). These two numbers are the only numbers that have a product of \(-12\) and a sum of \(-3\), BUT, they are NOT integers.

- Daisy

We can make a chart for all the sums:

(The first row is the sum, and the second is the number of ways to get that sum)

So as you can see in the chart, the possibilities for rolling a sum of \(5\) is \(4\), and the number of possibilities for any sum is \(36\). This means that your probability would be \(\frac{4}{36} = \frac{1}{9}\).

- Daisy

Let's start with the price of the pants:

Both started at \($30\) but then were discounted by \(15%\)%. That means that the discounted price is \(85\)% of \(30\). Given this, we have \(30 \times 0.85 = 25.5\), which is our price. Now, we need to add a \(5\)% tax. We can get \(25.5 \times 1.05 = 26.775 ≈ 26.78\). Since it was asking for the total price, and Nona bought two pairs of pants, we would have \(26.78\times2 = $53.56\).

Next, let's move onto the discounts of a dress:

We have two cases;

Case 1: A \($50\) dollar dress that was discounted by \(20\)% and \(10\)%

Case 2: A \($50\) dollar dress that was discounted \(30\)%

Case 1: Both the coupons added would be \(0.90(0.80(50))\) which equals \(36\).

Case 2: We have \(0.70 \times 50\) which equals \(35\).

Wow, what a close call. The \(30\)% coupon saves \($1\) more than the \(20\)% and \(10\)% coupons. 

Finally, let's see how many possible outfits Nona can make:

Nona has \(9\) pairs of pants, \(12\) shirts and \(3\) belts, and an outfit can consist of a pair of pants, a shirt and a belt or no belt. 

We will start with the pants and shirts. In order to find all possible combinations, we can do \(9 \times 12 = 108\). We will save that number, and use it for our next step. In order to find the possible amount of outfits with a belt, we can do \(108 \times 3 = 324\). So \(324\) is the number of outfits WITH a belt, and \(108\) is the number of outfits WITHOUT a belt. When we add these, we get our answer, \(432\).

- Daisy

By the chart, you can see that there are:

Fifteen \(1\)'s

Fifteen \(2\)'s

Fifteen \(3\)'s

Fifteen \(4\)'s

Five \(5\)'s

Five \(6\)'s

Five \(7\)'s

Five \(8\)'s

Five \(9\)'s    and 

Four \(0\)'s (Of course, the zero's would not matter, because they are asking for the sum).

So we are trying to find \(s\), the sum of the digits. We can get the equation \(s = 15(1) + 15(2) + 15(3) + 15(4) + 5(5) + 5(6) + 5(7) + 5(8) + 5(9) \), which is equal to \(s = 15 + 30 + 45 + 60 + 25 + 30 + 35 + 40 + 45\), which we can finally get \(s = 325\)

- Daisy

Thanks!

For \(a23\), \(n = 22\). We can set up the equation \({2}^{22}\times a22 = 2p\). We can see that \(a22 = 2^{21}\times a21\). We can keep doing this until we get: \(2^{22+21+20...+3+2+1} \times 1 = 2^p\), so \(p = 22+21+20...+3+2+1 = 253\). 

- Daisy

Number  of band members in first fromation  is

(n -2)(n + 8)

Number of band members in second formation  =  n (2n - 3)

But...the difference in the numbers will be  ≥ 4...since at least 4 drummers are excluded in the second grouping....so we have this inequality

(n - 2) ( n + 8)  - n (2n - 3) ≥ 4  simplify

n^2 + 6n - 16 - 2n^2 + 3n - 4 ≥  0

-n^2 + 9n - 20 ≥ 0    multiply through  by  -1....change the inequality sign direction

n^2 - 9n + 20 ≤ 0   (1)

The easiest  way to solve this is to change it to an equality

n^2 - 9n + 20  = 0       factor

(n - 5) ( n - 4)  = 0

Setting each factor to 0  and solving for  n produces  n = 5  or n  = 4

We have 3 possible intervals that will solve (1)

(-inf, 4)  U [4 ,5 ]  U (5 ,inf)

If we let  n   =  4.5...then note that

(4.5)^2 - 9(4.5) + 20   = -.25     which makes  (1)  true

So since n  can only assume integer values...... n  = 4  or  n  = 5   will make  (1) true

And the sum of these values  = 9

Here's a graph of the solution :  https://www.wolframalpha.com/input/?i=n%5E2+-+9n+%2B+20%C2%A0%E2%89%A4+0

Since  AB  =AC....then triangle  ABC  is isosceles

And...in an isosceles triangle the altitude bisects the base

So...let BC  be the base and we have that

So...DC = 1/2 BC

And area of  triangle ABC  = (1/2) BC (AD)

180  = (1/2) (12) (AD)

180  = 6 AD

30  = AD

And  since  AB and DE are parallel...then

angle BAC  = angle DEC

angle ACB  =angle ECB

So...by AA congruency

 ΔABC  ~ ΔEDC

And DC = 1/2 *BC  = 1/2 *  12   = 6

So....the altitude of Δ EDC  = 1/2 AD =  1/2 * 30  = 15

So...the area of ΔEDC = (1/2)(6)(15)  = 45

So...the are of ABDE  = area of  ΔABC  - area of ΔEDC  =

180  - 45   =

135  units^2

x^2 + 6x + y^2 - 4y   =  -c     complete the square on x  and  y

x^2 - 6x  + 9  + y^2 - 4y + 4  =  -c + 9 + 4

Note that if the radius is 4.....then the right side must  =  r^2  = 4^2  = 16

So

-c + 9 + 4   =16

-c + 13   =16

-c  = 3

c  = -3

Yep...you can complete the square and you will still get the same  "circle" equation that I got on the other problem....

No prob, Mathtoo....!!!

Thanks, CPhill! You can also complete the square, riight?

Using my earlier answer....we had  a circle  with a center  of   (7, 24)  and a radius of 25

So....the minimum x value  is at the point  ( 7 - 25 , 24)   = (-18, 24) 

So....the minimum x value is  -18

Thanks, guest....here's another way

(1/3) are red  =  30 *1/3  = 10 are red

So...20  are white

50% are 4 door   = 15  are 4 door

So    ....15 are 2 door

8 are 2 door  and white.....so 12 are 4 door and white  since we have 20 white total

And since 15 are 4 door  and 12 of these are white...then 3 must be 4 door and red

Understanding Sterling numbers isn't too difficult. But going to the site is very difficult, and it's impressive you went there without a link.

Here's my best attempt

The sum of the elements on row n of Pascal's triangle  = 2^n

So....in  base 10, we have

Log 2^n  =  

n Log 2  =  f(n)

And

f(n) / log 2   =    

n log 2  / log 2  =   

n