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That is impressive, but your 30 still does not make sense to me.  

Would you like to try and explain your logic there?

EDIT

Thanks for your input guest. I have worked it out using my own logic now.

30 makes perfect sense.

First triangle is 3 by 4, that makes a base of 5. 
Than you have right triangle with legs of 5 and 12, so diagonal is 13.

The formula for the diagonal is :

D= (length^2+width^+height^2)^1/2

D= (12^2*4^2*3^2)^1/2

  = (144+16+9)^1/2=  (169)^1/2= 13

The length of the diagonal is 13 cm

from https://www.enotes.com/homework-help/find-lenght-diagonal-rectangular-shaped-box-whose-175641

 fair 6-sided dice are rolled. What is the probability of all 6 numbers showing (1, 2, 3, 4, 5, 6)? The remaining 3 numbers can take any value. Thank you.

Here is an adaption of my first answer.

I have corrected all silly errors and I have corrected one logic error.

Now it is correct

Number of permutations in total = 6^9 = 10, 077,696

3 doubles the others different =        \( 6C3 *\frac{9!}{2!2!2!}=45360=20*45360 = 907200\)

-----------------------

The next one was the one that gave me the most trouble but it should not have

the 6C2 is because to of the numbers must be chosen as the duplicate ones.

Then i had to double it because one occured 3 times and the other only 2 and it could be either way around.

so

 1 double, 1 triple =       \(  6C2*2*\frac{9!}{3!2!}=30*30240=907200\)

-------------------------
1 quad = \(6*1*1*\frac{9!}{4!}=6*15120=90720 \)

907200+907200+90720 = 1905120 

1905120/10077696 = 0.1890432098765432

Approx   18.90%

Finally our answers agree and they also agree with Wolfram Alpha!

Micacle of miracles.

WOW!! I didn't know Wolfram did dice probabilities !! Believe or not their number agrees with mine !!

https://www.wolframalpha.com/input/?i=9+dice%2C+all+faces+show

Let  W  =  number of wins.

Let  L  =  number of losses.

Since he played  100  times:     W + L  =  100

Since he earned  30  dollars, he had  30  more wins than losses:  W  =  L + 30

Combining:  W  +  L  = 100  and  W  =  L + 30:

                (L + 30) + L  =  100

                       2L + 30  =  100

                                2L  =  70

                                  L  =  35

So, he had  65  wins (or hits).

Because ABCD is a parallelogram, angle(D)  =  60°.

Because AB || DC. the alternate interior angle below the letter A is also 60°.

Angle(A)  =  90° - x°.

Thus the three angles to the left of diagonal line through A are:  (90° - x°) + 60° + 2x°  =  180°.

Solving:     150° + x°  =  180°     --->      x  =  30°

If every box must have a ball then it would be 4! = 24 ways

Assuming that some boxes do not have to have any balls then I think the answer is 

7C3=35 ways

Let     x  =  mom's age (now)

Then  y  =  my age (now)

Seven years ago:         (x - 7)  =  7(y - 7)       --->     x - 7  =  7y - 49      --->     x - 7y  =  -42

Eight years from now:  (x + 8)  =  2(y + 8)     --->     x + 8  =  2y + 16     --->     x - 2y  =  8

Combining:     x - 7y  =  -42                                         --->       x - 7y  =  -42

                       x - 2y  =  8          --->   multiply by -1     --->      -x + 2y  =  -8

Adding down the columns:                                                            -5y  =  -50     

Dividing:                                                                                --->     y  =  10     (my age now) 

Substituting:   x - 2y  =  8     --->     x - 2(10)  =  8     --->     x  =  28  (mom's age now)

Mmm

You might be right, it does sound logical,

BUT

if it is meant to be 6C3 (as you did)  then 30 (which we both got) is also wrong, it should be 6C2=15

Triangle(BPS) is similar to Triangle(BAC)     --->     Area(BPS) / Area(BAC)  =  BP² / BA²  =  2² / 3²  =  4/9

     Consequently, Area(APSC)  =  5/9 · Area(ABC)

Similarly, Area(DWT) / Area(DAC)  =  4/9     --->     Area(WACT)  =  5/9 · Area(DAC)

Therefore, Area(APSCTW)  =  5/9 · Area(ABCD)     --->     Area(APSCTW)  =  5/9 · 180  =  20

We got the permutations right!. It is the combinations of the remaining 3 numbers that we disagree. You have 6.5.4 =120, while I count only 20! Should yours be 1.5.4 ? There are only 20 combinations possible with 6C3, but with repeats (such as 222, 555, 112, 334.....etc.), there are  56 such combinations, while your total is 126, hence the large discrepancy. 

Hi guest,

I made a number of careless errors in my first attempt.

I have now fixed the ones that I found.

Our techniques are basically the same.

But we have different answers for 3 doubles and 3 singles. Could you just check yours please?

Here is my attempt:

There are:[6+3-1]C3 =56 combinations of 3 remaining numbers with repeats as follows:
111, 222, 333, 444, 555, 666 = 6 - [3 of a kind.]

(112, 113, 114, 115, 116, 122, 133, 144, 155, 166, 223, 224, 225, 226, 233, 244, 255, 266, 334, 335, 336, 344, 355, 366, 445, 446, 455, 466, 556, 566)=30 - [2 of a kind ]

(123, 124, 125, 126, 134, 135, 136, 145, 146, 156, 234, 235, 236, 245, 246, 256, 345, 346, 356, 456) = 20 - [3 of different kind.]

123456111=9!/4! =15120 x 6 =90,720......The 6 is from  the above count.
123456112=9!/3!2!=30,240 x 30 =907,200......The 30 is from  the above count.
123456123=9!/2!2!2!=45,360 x 20 =907,200.......The 20 is also from the above{56 - 6 - 30 =20}.

Therefore, the probability is:
90,720 + 907,200 +907,200 =1,905,120 / 6^9 =18.90 % 

Thanks guest for the attempt.  

I have not come up with a likely answer but I am reasonably confident that yours is not correct.

Not counting yourself, in order to have a DISTINCT group of 4 people, you would have at least 12 people, because: 11 C 4 =330  Distinct combinations of 4 people. You need a minimum of 365. Therefore:

12 C 4 =495 Distinct combinations, which would be the minimum n.

∠ BAC = 120°             The law of cosines:      a² = b² + c² - 2ab * cosA  

AC = 150                         (BC)² = (AB)² + (AC)² - 2(AB)(AC) * cos(BAC) 

AB = 300                                                        BC = 396.86      

Please outline your working guest.

Since  a  and  d  are both positive, and these are sides of a right triangle,  a + 2d  must be the hypotenuse.

Pythagorean Theorem:  (a)²  +  (a + d)²  =  (a + 2d)²

--->     a² + a² + 2ad + d²  =  a² + 4ad + 4d²

--->     a² - 2ad - 3d²  =  0

--->     (a - 3d)(a + d)  =  0

--->     Either  a = 3d  or  a = -d

--->     Since  a  and  d  are both positive,  a  cannot equal  -d.

--->     When  a  =  3d     --->     a/d  =  3  

The probability that each of the first 6 rolls has a 1, 2, 3, 4, 5, or 6 is 6!/6^6.  The next three rolls can be anything, so the probability is 6!/6^6 = 5/324.

sin(x)^6 + cos(x)^6 = 1/4.

2. The possible slopes m are 2, 17, and -5.