Doremy

Thank you. That was a very clear explanation.

0.007

There are 25 ways to choose the first square. To select second square there are 16 ways. To select the third square there are 9 ways. So, number of ways to choose 3 square blocks is 25×16×9=3600.

I assume that the question states that the order doesn't matter, so then each collection of three blocks may be chosen in any of six different orders, which leaves the final answer as 3600/3! = 600

If order does matter the answer would be 3600 instead since no overcounting has been done.

Ah an inverse question.

set f(x) as y for simplicity.

y = 2/(x+1), the goal for finding the inverse of the function is to isolate x, so the first step: yx + y = 2. 

next, move y to the right hand side to reach yx = 2 - y. 

finally, divide y over to the right hand side to reach x = (2-y)/y

Now as the core concept of the inverse, we switch the positions of x and y by the time we reach this step.

y = (2-x)/x, and plug in 1/5 to get y = (2-1/5)/(1/5) and that is 9/5 * 5 = 9 

plug in 2x for every x in f(x) = x^2-3x+4

That leaves us with f(2x) = (2x)^2 -3(2x) + 4 = 4x^2 - 6x + 4

5421

5331

5322

53211

4431

4422

44211

4332

43311

43221

6321

432111

Total—12 partitions that work.

cards originally had by adam = x, cards originally had by claire = y.

7/4 y = x from sentence 1

1/2y = x - 35(I'm assuming your questions means that Adam did not give the postcards away to claire here)

7/4y - x = 0

1/2y - x = -35

equation 1 - equation 2 to get

5/4y = 35

Thus

y = 28, and

x = 49

Adam originally had 49 cards, and Claire originally had 28 cards

4 3 2 1, is the original partition of 10. However, we can have a hidden "0" at the end, making it 4,3,2,1,0 for our use in calculating the partitions of 12.

For 12, we can place two extra 1's into the five existing slots (4 3 2 1 0), in other words five choose two places. Note that five choose two = 10.

At this point the answer seems like 10 but there are two additional cases not counted, which are to place both extra 1's into the same place AND to create 2 new slots at the end instead of one. There are two ways to do so, leaving us with respectively 6,3,2,1 and 4,3,2,1,1,1

So the answer is 10+2 = 12. I was originally kinda stuck but figured it out later.

Idk if this is faster than just listing everything out but its probably more elegant

It should be C5,2 but there are 2 additional cases for when both are added on the same one. We can also use complementary counting and get 16 - 4(the repeated ones)

8 is the smallest number that satisfies (i) and (ii). 

next, 8+(3*5)x = 5 (mod7), 8+15x = 5 (mod 7), and 8+x = 5 (mod 7), it's clear than 8+4 = 12 so x = 4

Thus the smallest x can be is x = 4 (mod 7) and x = 4.

The answer is 8 + 15(4) = 68, the smallest number that satisfies the three conditions.

The fraction he shared is total - amount left for himself = (1-7/12) = 5/12.

Each of the four recieved one quarter of that, which is 5/12 /4 = 5/48 of the his original number of fishballs.

Thank you for your work! It helped me to arrive at a final answer of 12.