## Notes

On various topics from mental math to quick measurement estimates.

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• #### Combinations (Groups - Order does not matter)

• Number of ways "n" things can be ordered
• $$x = n!$$
• Number of way "n" things can be ordered if taking "r" at a time and order in the pair is not imporant (a,b)=(b,a)
• $$x = {n!\over {(n-r)!r!}}$$
• Example:
• The amount of unique number pairs in the set {1,2,3,4}, where order doesn't matter (1,2) = (2,1)
• For pairs of two:
• $$r = 2$$
• With four items:
• $$n = 4$$
• $$x = {4!\over {(4-2)!2!}} = 6$$
• #### Permutations (Lists - Order does matter)

• Number of way "n" things can be ordered if taking "r" at a time and order in each pair is imporant (a,b) != (b,a)
• $$x = {n!\over {(n-r)!}}$$

There is a simple process that allows someone to calculate the cube root of any integer number from 0^3 to 99^3 relatively quickly in their head. To find the integer cube root of a number up to 6 digits (0^3 to 99^3) in your head, memorize the cubes of the integers 1 through 9 shown in the table and follow the steps below.

#### Process:

1) The tens digit of the cube root answer can be determined by truncating the last 3 numbers off the given cube and seeing where it falls within the memorized table. The tens digit of the answer is the number corresonding to cube that is just smaller than the given truncated cube.

2) The ones digit can be found by taking the last digit of the given cube and finding the cube of that digit, by using the same memorized table. The last digit of this new cube is the ones digit of the cube root answer.

#### Example: What is the cube root of 389,017?

1) 390,017 -> 390 (With the last three digits truncated) -> 390 falls between 7^3(=343) and 8^3(=512). Therefore tens digit of the cube root answer is 7.

2) The last digit of the given cube is 7 -> 7^3 = 343 -> The last digit of 343 is 3. Therefore the ones digit of the cube root answer is 3.

The final answer is 73 and the cube root of 389,017 is in fact, 73.

##### Table to Memorize
Integer Cube
1 1
2 8
3 27
4 64
5 125
6 216
7 343
8 512
9 729
10 1000

An observer can estimate roughly how far away an object is if the approximate size of the object, or another near by object, is known. The ratio of a typical persons arm length to the distance between their eyes is about 10. Using this information and trigonemetry the distance to the referenced object can be estimated. First, the observer holds up their thumb and aligns it with the center of the referenced object with one eye closed. The observer then alternates between closing one eye and then the other such that only one eye is open at a time. The observer then estimates the distance the referenced object appears to jump back and forth. The estimated distance to the object is 10x this number.

#### Example:

If an observer looks off in the distance and sees a person and appears to be about 6 ft tall. The obsever then lines up their thumb over their view of the person with on eye closed. Then by alternating one closed then the other, they see that the referenced person moves back and forth about 12 ft (Using the person's 6ft height as reference for what 12 ft looks like at that distance. In this example, the distance from the observer to the person would be about 10x the percieved back and forth movement, or about 120 ft away.