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
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
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