All Answers 358087 Answers

Thank you, I have a better understanding of what this problem meant. I first tried plugging in some values and then graphing it. I figured there was an easier way!

I'm not sure how you've calculated your permutations, but it looks to me as though your groups 1122xy are the same as groups 1122yx ((where x and y are different).

Certainly, if you remove those repeats (if indeed they are repeats) you get back to the 1440/6^6 result.

What is the sum of the factors of 1024?   

Not sure what you're wanting here, but 1024 is 2¹⁰ and ten 2's would total 20. 

_.

Hello Alan: That conclusion is extremely surprising in light of having just completed the individual permutations of each of the following 16 groups. Because I permuted each of the 16 groups separately, I don't think the possibility of duplication is likely. Finally, I assembled the entire list and the computer counted them as the total of these 16 individual permutations and both agreed as to the total of =2,520 permutations!
So, Alan, where could the overcounting possibly be? Any ideas where to look for possible duplications?
1 - 112233 = 90 permutations
2 - 112234 = 180p 
3 - 112235 = 180p 
4 - 112236 = 180p 
5 - 112243 = 180p 
6 - 112244 = 90p 
7 - 112245 = 180p 
8 - 112246 = 180p 
9 - 112253 = 180p 
10 -112254 = 180p 
11 -112255 = 90p 
12 -112256 = 180p
13 -112263 = 180p 
14 -112264 = 180p 
15 -112265 = 180p 
16 -112266 = 90p
Here is a complete list of all 90 permutations of the first group:112233. See if you can find anything wrong with it. Thanks.
{1, 1, 2, 2, 3, 3}, {1, 1, 2, 3, 2, 3}, {1, 1, 2, 3, 3, 2}, {1, 1, 3, 2, 2, 3}, {1, 1, 3, 2, 3, 2}, {1, 1, 3, 3, 2, 2}, {1, 2, 1, 2, 3, 3}, {1, 2, 1, 3, 2, 3}, {1, 2, 1, 3, 3, 2}, {1, 2, 2, 1, 3, 3}, {1, 2, 2, 3, 1, 3}, {1, 2, 2, 3, 3, 1}, {1, 2, 3, 1, 2, 3}, {1, 2, 3, 1, 3, 2}, {1, 2, 3, 2, 1, 3}, {1, 2, 3, 2, 3, 1}, {1, 2, 3, 3, 1, 2}, {1, 2, 3, 3, 2, 1}, {1, 3, 1, 2, 2, 3}, {1, 3, 1, 2, 3, 2}, {1, 3, 1, 3, 2, 2}, {1, 3, 2, 1, 2, 3}, {1, 3, 2, 1, 3, 2}, {1, 3, 2, 2, 1, 3}, {1, 3, 2, 2, 3, 1}, {1, 3, 2, 3, 1, 2}, {1, 3, 2, 3, 2, 1}, {1, 3, 3, 1, 2, 2}, {1, 3, 3, 2, 1, 2}, {1, 3, 3, 2, 2, 1}, {2, 1, 1, 2, 3, 3}, {2, 1, 1, 3, 2, 3}, {2, 1, 1, 3, 3, 2}, {2, 1, 2, 1, 3, 3}, {2, 1, 2, 3, 1, 3}, {2, 1, 2, 3, 3, 1}, {2, 1, 3, 1, 2, 3}, {2, 1, 3, 1, 3, 2}, {2, 1, 3, 2, 1, 3}, {2, 1, 3, 2, 3, 1}, {2, 1, 3, 3, 1, 2}, {2, 1, 3, 3, 2, 1}, {2, 2, 1, 1, 3, 3}, {2, 2, 1, 3, 1, 3}, {2, 2, 1, 3, 3, 1}, {2, 2, 3, 1, 1, 3}, {2, 2, 3, 1, 3, 1}, {2, 2, 3, 3, 1, 1}, {2, 3, 1, 1, 2, 3}, {2, 3, 1, 1, 3, 2}, {2, 3, 1, 2, 1, 3}, {2, 3, 1, 2, 3, 1}, {2, 3, 1, 3, 1, 2}, {2, 3, 1, 3, 2, 1}, {2, 3, 2, 1, 1, 3}, {2, 3, 2, 1, 3, 1}, {2, 3, 2, 3, 1, 1}, {2, 3, 3, 1, 1, 2}, {2, 3, 3, 1, 2, 1}, {2, 3, 3, 2, 1, 1}, {3, 1, 1, 2, 2, 3}, {3, 1, 1, 2, 3, 2}, {3, 1, 1, 3, 2, 2}, {3, 1, 2, 1, 2, 3}, {3, 1, 2, 1, 3, 2}, {3, 1, 2, 2, 1, 3}, {3, 1, 2, 2, 3, 1}, {3, 1, 2, 3, 1, 2}, {3, 1, 2, 3, 2, 1}, {3, 1, 3, 1, 2, 2}, {3, 1, 3, 2, 1, 2}, {3, 1, 3, 2, 2, 1}, {3, 2, 1, 1, 2, 3}, {3, 2, 1, 1, 3, 2}, {3, 2, 1, 2, 1, 3}, {3, 2, 1, 2, 3, 1}, {3, 2, 1, 3, 1, 2}, {3, 2, 1, 3, 2, 1}, {3, 2, 2, 1, 1, 3}, {3, 2, 2, 1, 3, 1}, {3, 2, 2, 3, 1, 1}, {3, 2, 3, 1, 1, 2}, {3, 2, 3, 1, 2, 1}, {3, 2, 3, 2, 1, 1}, {3, 3, 1, 1, 2, 2}, {3, 3, 1, 2, 1, 2}, {3, 3, 1, 2, 2, 1}, {3, 3, 2, 1, 1, 2}, {3, 3, 2, 1, 2, 1}, {3, 3, 2, 2, 1, 1} = 90 permutations

This is not a hard question, for the y-axis (frequency density), the graph didn't show you the numbers like the x-axis, but you can figure out the numbers by looking at the table. For example, the table said, "when the time is 30-40 the frequency is 35." When you look at the graph and look for the time range of 30 to 40, you will find that the frequency density is at its highest point, which is 35. It is 7-blocks high, which means that one block on the y-axis is 5. 

Now for the answer:  For the first missing number -- 15, because it is 3-blocks high and 5*3=15. For the second missing number -- 25, because it is 5-blocks high and 5*5=25.

I might explain better if I am using a graph, if you don't understand, just ask.

Can someone help me?

Anyone got algebraic or system of equations challenge?!!

This site might help...

Here is a Monte Carlo program (in MathCad) that supports Guest #1's result:

Of course, since the Monte Carlo program depends on random numbers the result it returns may be slightly different each time it runs.  However, it does seem to provide strong support for Guest #1.

Happy New Year to you too CU

Happy New Year to all forum users.    

What is the smallest positive integer that is divisible by each of the first six positive integers?

LCM [1, 2, 3, 4, 5, 6] = 60

(1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60) >>Total = 12 Divisors.

4. How many ways are there to put 5 balls in 3 boxes if the balls are distinguishable but the boxes are not?

Assuming no restrictions  [ boxes may be empty ]  ..... we can compute  this with something known as "Stirling Numbers of the Second Kind"

We have

S(5, 1)  +  S(5,2)  + S (5,3)  =

    1      +    15       +     25     =

              41  ways 

Call S  the side of the  square

Using the Law of Sines

LC/ sin (LDC)  =  S / sin (67.5)

(S - 6)  /sin 22.5)  = S / sin (67.5)

(S - 6)  / S   =    sin (22.5)  / sin (67.5)

(S - 6) /S   =   sin (22.5) / cos(22.5)

(S - 6)  /S  =   tan (22.5)

(S - 6)  / S   =  tan (45/2)

(S - 6) / S  =  ( 1 - cos(45) )  / sin (45)

(S - 6) / S   =   ( 1 - √2/2 )  / ( √2/2)

(S - 6) / S  =  [ 2 - √2] / √2

(S - 6) / S  =    √2  - 1

S - 6  =  S [√2 - 1]

S  - S [ √2 -1]  =  6

S  [ 2 - √2 ]   =  6

S =   6  / [ 2 - √2]

S = 6 [ 2 +√2] / 2  =   3 [ 2 +√2]  =   6  + 3√2

And applying the Law of Sines again

S / sin (DKC)  = KC / sin(KDC)

S  / sin  (112.5)  =  KC  / sin (22.5)

KC  =   S* sin ( 22.5)  / sin (112.5)

KC  =  S * sin (22.5) / sin (67.5)

KC =  S * sin (22.5) / cos(22.5)

KC =  S *  tan (22.5)

KC =  S  *   [√2 - 1]

KC  = 3 [ 2 + √2]  [ √2 - 1]

KC  = 3 [ 2√2 + 2 - 1 - √2]

KC  =  3 √2

So   the length  of 1/2   of the diagonal  =  OC = 

 S√2  / 2     =  ( [ 6 + 3√2] * √2  ) /2  =     [ 6√2 + 6 ] / 2  =

3√2  +  3 

So

OK   =    OC  -  KC  =    [ 3√2 + 3 ]  - 3√2   =    3

3. How many ways are there to put 5 balls in 3 boxes if the balls are not distinguishable but the boxes are?

Let  n  =  number  of boxes

Let  k  =  number of balls

The number of  ways  [ with no restrictions....boxes may be empty]   is given by

C( k + n - 1 , n - 1)   =   C( 3 + 5 - 1 , 3 - 1)   =  C( 7, 2)   =  21 ways

2. How many ways are there to put 5 balls in 3 boxes if the balls are not distinguishable and neither are the boxes?

We  have

5    0    0

4    1    0

3    2    0

3    1    1

2    2    1

5  ways   

   [  The boxes are indistinguishable  so   5  0   0    =    0  5   0   =    0 0 5 ]

1. How many ways are there to put 5 balls in 3 boxes if the balls are distinguishable and the boxes are distinguishable?

Let  the number  of balls  =  k

Let the number of boxes  = n

The total number of ways   [ without restriction.....we can have empty boxes ]  is  given by

n^k  = 3⁵  =  243  ways

Guest's first answer is correct

LOL......!!!

THX.....( I do my best   ...  )

We have that

16 +  4x           16  + x      

______   =    _________                      cross-multiply

   12                12   -   x

( 16 + 4x) ( 12 - x)  =  12 ( 16 + x)     simplify

192 + 48x - 16x  - 4x^2   =  192  + 12x       

32x  -  4x^2  =  12x       rearrange as

4x^2  =  20x

4x^2 -  20x  +  0      factor

4x ( x - 5)  =  0

Only the second factor produces what we need

x  - 5   = 0

x  = 5

omg thx ur a hero :D

We have something like this  : https://www.desmos.com/calculator/jpriys8xex

The equation of the ellipse  is

(x - 4)^2          (y - 1)^2

______    +    _______    =   1

   16                     1

The focal  length ,c,  is given  by   √  [ 16 - 1 ]   =  √(15)  =  c

So.....the   distance between the  foci   =  2c   = 2 √15   units

I'm getting x = 10.

not correct

The distance between the foci is 6*sqrt(3).

5 ways to choose the first person in line

4 ways to choose the second

3 ways to choose the  third.....etc.

So

5 x 4 x 3 x 2 x 1   =    120   ways