tertre

 We break this up into cases. To create 6, with three numbers, we can use: 6-0-0, 5-1-0, 4-2-0, 4-1-1, 3-3-0, 3-2-1, and. 2-2-2.

6-0-0: Only one way

5-1-0: Take the ball that is alone, so 6 ways.

4-2-0: 6C2=15 ways for the two balls in their own box.

4-1-1: Again 15 ways, pick two of the six balls, but they cannot go in the box of 4.

3-3-0: 5C2=10 ways, by selecting two of the remaining five balls.

3-2-1: 6*10=60 ways by picking any of the six balls to be by itself.

2-2-2: 15 ways? First, you think they are distinguishable, yet you overcount by a factor 3!=6.

Thus, adding them all up, we get \(1+6+15+15+10+60+15=\boxed{122}.\)

See this link: https://web2.0calc.com/questions/some-questions-please-elaborate-thanks#r1

Yes, that is the link ! Read the solutions carefully, guest. 

Meant 4).

3) You can have (Even, Even), (Even, Odd), (Odd, Even). The last (Odd, Odd) will result in odd no matter what...

So, we have 50*50*3=7500 ways?
 

No problem ! 

a): Label the integers: \(x-5, x-4, x-3, x-2, x-1, x, x+1, x+2, x+3, x+4, x+5=11x\) . And, this number is a multiple of \(11\) , so it is definitely divisible by \(11.\)

Try the second one on your own!

A "slick way" to answer this question, is to label the three sides of the triangle, \(x, y, z\) . Then, we have \(x^2+y^2+z^2=1800\), but we know that \(x^2+y^2=z^2\) in a right triangle, so we have \(z^2+z^2=1800\) or \(2z^2=1800\), and \(z^2=900, z=30\) .

Thus, the length of the hypotenuse is \(\boxed{30}\) .

For, problem 2, I suppose you are including the numbers in between.

I suggest you using a strategy called "complementary counting," subtracting the unwanted from the total.

We can first check the multiples of nine are between 1 and 1000, which is from 9 all the way up till 999, and that is 111 integers.

There are a total of 1000 numbers, and the answer should be \(1000-111=\boxed{889}\) integers.

Also, remember that tan A= sin A/ cos A.

Thus, we have \(2*\frac{\sqrt{3}}{2}*\frac{\sqrt{3}}{3}=\frac{3}{3}=\boxed{1}.\)

Thank you...it's my fault, could have been logged in ...!