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Is the denominator 10! or 6^10?

Hi Daniel, it is good that you gave feedback but it would have been a lot nicer if you had said thank you to all the people who allocated their time to your question.

How many positive integers are there whose digits strictly increase from left to right?
(For example, 28, 13589, and  4 are all such integers. "Strictly" means no two digits can be equal, so 15668 wouldn't count.)

e.g. 

How many ways can 5 digits can be chosen from 9

9C5 = 126

Each of these will have only 1 number in the order needed.

So that is 126 in 5 digit numbers

Do 1 digit numbers and/or 0 count? 

there are 10 of those.

 ------

1 digit numbers 10  (includes 0)

2 digit numbers  9C2 = 36

3 digit numbers  9C3 =84

4 digit numbers 9C4=126

5 digit number 9C5 = 126

6 digit numbers 9C6=84

7 digit numbers 9C7=36

8 digit numbers 9C8=9

9 digit numbers = 1

10+36+84+126+126+84+36+9+1 = 512 ways   

I will attempt to count them too ...

I know that six of the rumbers are 

1,1,3,3,5,5, and I am assuming that the others four must be 2,4 or 6 each.

So I get the probability as       \(\frac{1530900}{6^{10} }= \frac{175}{6912} \approx 0.025318287037037\)       Note: this has been edited (1st edit)

That is if i have made no mistakes.

There would be a number of more sensible ways to do this.     

That is far too simple Azun. The 6 on the top is definitely not correct.   

It is good that you gave it a go though. 

Since 10 fair six-sided dice are rolled, the total outcomes is \(6^{10}=60466176\).

And, the pairs are: 11,33

11,55

So, 2*3=6 pairs....

Thus, the probability is \(\frac{6}{60466176}=\frac{3}{3023308}\).

The age of John?  This question doesn't make sense?

The answer was 125. I got it wrong.

Since no repeating digits 9 digits available

Use these formulas:  m² + n²,  m² - n²,  2·m·n

For example:  If  m = 2  and  n = 1, then you have:  2² + 1²  =  5     2² - 1²​  =  3     2·2·1  =  4.

One place to start is to  look at the divisors of 162  and see which ones are possible legs for  "primitive" Pythagorean Triple  right triangles

The  divisors  of  162  are   1 | 2 | 3 | 6 | 9 | 18 | 27 | 54 | 81 | 162

  3   will  work  because   3-4-5  is a primitive  triple

We need to divide  162  by each factor....and then  multiply  both 4 and 5 by this factor to find one triple

So

162/3  =   54

So  54 * 4  = 216

And 54 * 5 = 270

So    162 -  216 - 270  is a   "triple"

And

9  will work  because

9 - 40 - 41 is a primitive triple

So 162/9  = 18

And  18 * 40  = 720

And 18 * 41  =  738

So  162 - 720 - 738  is a  "triple'

There are two more which are a lttle hard to find...the next is

27 - 364 - 365

So 162/27  = 6

And 364 * 6  = 2184

And 365 * 6  = 2190

So 162 - 2184  -2190   is  a "triple'

The last  is 162 - 6560- 6562

So....there are 4    which have one leg  = 162

But what is the upper limit? up to 10,000, 100,000, 1,000,000......etc.

-0.1x^2  - 0.6x  + 0.4    =  0      

First off....the easiest thing to  do  is to multiply  everything through  by  -10....this gives us

x^2  + 6x - 4  = 0        take  1/2 of 6  = 3....square it  =  9....add and subtract it

x^2   + 6x  + 9  - 4  - 9  = 0        factor the first three terms and simplify  the rest

(x + 3)^2  - 13  = 0       add 13 to both sides

(x + 3)^2  = 13           take both roots

x + 3  = ±√13       subtract 3  from  both sides

x  = ±√13  - 3    or    -3 ±√13   if you prefer!!!!

\(f(x)=\frac{2x-7}{\sqrt{x^2-5x+6}} \)

We only need to worry about the denominator, since all real numbers will "work" for the numerator

Note that  the function inside the square root in the denominator must be  >  0  

So...let's find this

x^2 - 5x  + 6  >  0     factor

(x - 3) ( x - 2)  >  0

Setting each factor to 0  and solving for x produces  x  = 2   and  x  = 3

So....we have   three  possible  intervals  for solutions

(-inf, 2)   and   [2, 3]   and (3, infinity)

Note that  any  value of x  in the middle interval will make the inequality  false

So.....the domain  is       x < 2  ∪  x > 3

Here's a graph that will confirm this :

https://www.desmos.com/calculator/f0cw1ybcym

It's been a joy  - and sometimes a real challenge  - to answer your questions, Tertre.....You have also provided some very good  solutions to many of the questions others have posted on the forum, as well.

So......Thanks for the kudos...and.....

See you on your return.....!!!!

The area  of the tirangle  = (1/2)side^2 * sin (60°)  = (1/2) 3^2 * √3/2 = (9/4)√3  units^2

We can  construct a circle at A  with a radius of 1......the area  bounded by the sides of the triangle and the circle will be  =

pi * (1)^2 * (60/360)  =   pi/6  units ^2

So....the probability that "X"  lies within one unit of "A"   is

[pi / 6]  / [ (9/4)√3]  =

pi /  6   *  4 / [9√3  ]  =

4 pi / [ 54 √3 ] =

 2 pi / [ 27√3]  ≈   .1344 ≈   13.44%

Here's a picture :

Look under your first post.

Look under your first post.

Look under your first post.

For each day, the expected amount of rain is (0.40)(0)+(.25)(4)+(.35)(10)=0+1+3.5=4.5. To find the total expected amount of rain for the days from Monday to Friday, we can add the amounts for each day, getting 5*4.5= 22.5 inches total.

We get 22.5 inches total as our final answer.

The chance of winning the jackpot is 1/10000. The value of the jackpot is $250000 . The expected winnings, minus the cost of the ticket is E=1/10000. $250000-$3=$2.50-$3.00=-$0.50

-$0.50 is our final answer.

The issue here is that the exterior doors are included in the count of how many doors a room has. Meaning, for example, if room 1 has some exterior doors, then they will be counted in n_1. Since i need to *prove* that the number of exterior doors in even, i can't assume that is even, and we're only told that every room has an even number of doors, so each n_i for i>0 is even. That means we have no way of knowing n_0 + n_1 + n_2 +...+n_r at this point that is even.