Permutations and combinations are both related mathematical concepts. Because they are related concepts, most of the time they are used with each other or switched or swapped with each other without realizing it. As mathematical concepts, they serve as precise terms and language to the situation they are describing or covering.

“Combination” is defined as the selection of objects, symbols, or values from a wide variety like a large group or a certain set with underlying similarities. In a combination, the importance is made on the choice of the objects or values themselves. One combination comprises one value plus another value (as a pair) with or without additional values (or as a multiple).

Values or objects in a combination do not require order or arrangement. The combination can also be random in nature. Also, the values or objects can be considered as alike or the same in comparison with each other. A combination, with relation to permutation, can be several in numbers while permutation can be less or single in comparison.

On the other hand, permutation is also the selection of objects, values, and symbols with careful attention to the order, sequence, or arrangement. Aside from giving an emphasis on these three things, permutation gives the values or objects’ destinations by virtue of assigning them into a specific placement with each other. For example, a certain value or a combination of values can be assigned as the first, second, and so on.

With respect to a combination, a permutation is basically an ordered or arranged combination. A permutation also deals with a number of ways to arrange, rearrange, and order the objects and symbols. One permutation is equal to a single arrangement or order. One arrangement or permutation is distinctly different from another arrangement or permutation.

Permutations and combinations are often used as word problems in mathematical textbook exercises. Another application is in data preparation and probability in research. Using “permutation” and “combination” can easily help to predict something with the given data.

Permutation has the formula: P(n,r). Meanwhile, finding the combination requires this particular mathematical method -

The (n,r) in the second permutation formula (which also applies when finding the combination) represents two things–the value of “n” is the initial number mentioned while the second value (which is r) is the times that the decreasing and succeeding value will be multiplied to the value of “n.”

Summary:

1.“Permutation” and “combination” are related mathematical concepts. “Combination” is any selection or pairing of values within a single criteria or category while “permutation” is an ordered combination.

2. Combinations do not place an emphasis on order, placement, or arrangement but on choice. Values can be single or paired. On the other hand, permutations place a high emphasis on the three aforementioned characteristics. Aside from these three, a permutation also gives the destination of each value (or paired value).

3. A number of permutations can be derived from a single combination. Meanwhile, one permutation calls for a single arrangement.

4. Permutations are often regarded as ordered elements while combinations are looked upon as sets.

5. A single permutation is distinct and different on its own and from each arrangement while a combination is often alike in comparison with other combinations.

6. Both “permutation” and “combination” are often used in math word problems and probabilities in statistics and research.

**Formulas**

For example,

..It is to be kept in mind

1. 0! = 1,

2. Number of permutations is greater than or equal to number of Combinations.

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My argument is that because divisors of the Mersenne number

can’t be < p if p is a prime number. Therefore if 2p +1 is a divisor of it has no divisors as p is > the square root of 2p + 1. This will therefore make 2p + 1 a prime number.Is this proof correct?

]]>**Every integer >5 can be expressed as the sum of 3 primes**

Because I find that really **beautiful**, but then I realised that this was still just a conjecture.

Whoops!

]]>The plane is infinitely extending and not on a sphere.

Antipodal points don't exist.

Lines have at least one point at infinity.

The natural units of length and area are fixed, but currently unknown; it is unknown if a natural unit of angle measurement exists.

Polar lines don't exist.

The area of a triangle is unbounded.

A representation I came up with is where lines seem to bend in parabolic shapes away from some chosen reference point, with the bending getting more extreme the farther the line is from the point; lines that go through the reference point appear straight, and how extremely a line bends away from the point is defined by the natural unit of length. There are problems with this representation though:

If you have two distinct lines, there will be two points at which they will intersect, but if you move the reference point the intersection points will also move.

A pair of perpendicular lines with at least one of the lines on the reference point intersect in one place while perpendicular lines off the reference point intersect in two places.

Line segments move around in unknown ways when the reference point moves.

The angle of intersection points might change with the position of the reference point but I don't know for sure.

“ If a Sophie Germain prime p is congruent to 3 (mod 4), then it’s matching safe prime 2p + 1 will be a divisor of the Mersenne number 2^p - 1.”

And:

“ Fermat's little theorem states that if p is a prime number, then for any integer a, the number a^p − a is an integer multiple of p.”

However if an integer, 2p + 1, where p is a prime number, is a divisor of the Mersenne number 2^p - 1, then 2p + 1 is a safe prime and p it’s matching Sophie Germain prime.

Divisors of the Mersenne number 2^p - 1 can’t be < p if p is a prime number. Therefore if 2p +1 is a divisor of 2^p - 1 it has no divisors as p is > the square root of 2p + 1. This will therefore make 2p + 1 a safe prime and p it’s matching Sophie Germain prime.

For example 11 which is prime, (11*2) + 1 = 23. 2^11 - 1 is divisible by 23 making 11 a Sophie Germain prime and 23 it’s matching safe prime.

]]>Hi Βεν,

Nice contribution! Yes, the repeated iteration of the cosine function converges: actually, it converges to thefixed pointof the cosine function,

That makes so much sense! I also just tried

and it converged to its fixed point, 0. Tan has a fixed point but it is not attractive; sinh becomes a vertical line on 0 which means there is no attractive point, 0; cosh has no fixed point but it converges to ∞; 0 is tanh's attractive fixed point; cot has no attractive point due to it being tan with the x axis shifted; sec is related to tan because it shares half of its discontinuous points with tan suggesting it has no attractive point, and it does have no attractive point; csc is sec with the x axis shifted so it has no fixed point.]]>Welcome to the forum.

Yes, you should be doing your chem engg project. How long did this take?

forum rules wrote:

No Swearing or Offensive Topics. Young people use these forums, and should not be exposed to crudeness.

My first reaction was to delete it completely but then we lose all your posts. I think you should do the right thing and edit it yourself to something more wholesome.

Thanks,

Bob

]]>Mersenne Number= 2^11 -1

Factors = 23 and 89

2^11 -1 takes eleven goes to get to zero if I continually minus one and divide by 2. Watch what happens to 23 and 89 acting as remainders as I do the same to them.

2^11 -1

2^10 -1 First go

2^9 -1 Second go

2^8 -1 Third go

2^7 -1 Fourth go

2^6 -1 Fifth go

2^5 -1 Sixth go

2^4 -1 Seventh go

2^3 -1 Eighth go

2^2 -1 Ninth go

2-1 Tenth go

1-1=0 Eleventh go

23

(23-1)/2 = 11 First go

(11-1)/2 = 5 Second go

(5-1)/2 = 2 Third go

23+2 = 25, (25-1)/2 = 12 Fourth go

23+12 = 35, (35-1)/2 = 17 Fifth go

(17-1)/2 = 8 Sixth go

23+8 = 31, (31-1)/2 = 15 Seventh go

(15-1)/2 = 7 Eighth go

(7-1)/2 = 3 Ninth go

(3-1)/2 = 1 Tenth go

(1-1)/2 = 0 Eleventh go

89

8(89-1)/2 = 44 First go

89+44 = 133, (133-1/2) = 66 Second go

89+66 = 155, (155-1)/2 = 77 Third go

(77-1)/2 = 38 Fourth go

89+38 = 127, (127-1)/2 = 63 Fifth go

(63-1)/2 = 31 Sixth go

(31-1)/2 = 15 Seventh go

(15-1)/2 = 7 Eighth go

(7-1)/2 = 3 Ninth go

(3-1)/2 = 1 Tenth go

(1-1)/2 = 0 Eleventh go

They both take eleven goes to get to zero too.

Working backwards it can be seen that this must be the case. Starting with zero I multiply by 2 and add one continually until I get to 2^11 -1. The same must be done to the** remainders **which will then become 23 and 89 or zero, the factors of 2^11 -1.

Let the number of goes an odd number, y, takes to get to zero by continually minusing one and dividing by 2, (adding y when even) =z.

The z for 2^11 -1, 23 and 89 =11.

**Whatever z equals for y, 2^z -1 must be factorable by y. **When we use the method for 2^z -1, we never add 2^z -1 as it never equals an even number. This could potentially alter remainders for potential factors, but for Mersenne Numbers, this is not the case.

Does anyone know what “z” a composite can’t be?

]]>You can record such a parameterization of solutions.

]]>Did you want me to provide a solution or what? No square brackets (to hide) on my Kindle so you just have to close your eyes.

Also no diagram making, sorry.

This is 3D so I'll describe the diagram.

Let TB be the tower in the 'z' direction and BE be an easterly direction with the first observation at E (x direction).

Draw an oblique line NES to indicate the North/South direction (y) with ES= 42.4 ft.

As TEB = 45, BE = TB = h, the height of the tower.

And as TSB= 30, BS = root 3 h.

So in triangle BES we have BE = h, BS = root 3 h and ES = 42.4 with a right angle at E.

So it would be easy to use Pythag to get h (approx) 30 ft I think. But here's a construction for it. I have no paper nor instruments handy so the scale might not work.

Draw a verticle line BE' in the top right corner of your page. I'll make it 5cm to represent 10 ft. Construct a right angle at E' and extend the line right across the paper.

On a fresh sheet, construct an equilateral triangle sides 5 cm. Construct a perpendicular bisector so the height = 5 root 3 and set this as a radius. On the first diagram, centre B, make an arc to cut the horizontal line at S'

The triangle BE'S' is the right shape for BES but needs to be scaled up.

On E'S' produced Mark X so that E'X = 21.2 cm.

Draw XS parallel to BE' with BS' extended crossing it at S. SE parallel to S'E' will fix E and hence the correct size for triangle BES. h = BE.

Bob

]]>If I minus 1 and divide by 2 it will take p goes to get to zero.

Example:

2^11 -1

2^10 -1 First go

2^9 -1 Second go

2^8 -1 Third go

2^7 -1 Fourth go

2^6 -1 Fifth go

2^5 -1 Sixth go

2^4 -1 Seventh go

2^3 -1 Eighth go

2^2 -1 Ninth go

2-1 Tenth go

1-1=0 Eleventh go

Similarly if 2^p -1 has a factor of the form, n, it should take p goes to get to zero. Only, if minus 1 divided by 2 results in an even number, I should be allowed to add n to the number to continue. As this is what happens to n, as a remainder for 2^p -1, as each go takes place.

Examples:

23

(23-1)/2 = 11 First go

(11-1)/2 = 5 Second go

(5-1)/2 = 2 Third go

23+2 = 25, (25-1)/2 = 12 Fourth go

23+12 = 35, (35-1)/2 = 17 Fifth go

(17-1)/2 = 8 Sixth go

23+8 = 31, (31-1)/2 = 15 Seventh go

(15-1)/2 = 7 Eighth go

(7-1)/2 = 3 Ninth go

(3-1)/2 = 1 Tenth go

(1-1)/2 = 0 Eleventh go

89

(89-1)/2 = 44 First go

89+44 = 133, (133-1/2) = 66 Second go

89+66 = 155, (155-1)/2 = 77 Third go

(77-1)/2 = 38 Fourth go

89+38 = 127, (127-1)/2 = 63 Fifth go

(63-1)/2 = 31 Sixth go

(31-1)/2 = 15 Seventh go

(15-1)/2 = 7 Eighth go

(7-1)/2 = 3 Ninth go

(3-1)/2 = 1 Tenth go

(1-1)/2 = 0 Eleventh go

Both 23 and 89 take eleven goes to get to zero, as they must do, as they are factors of 2^11 -1, and 2^11 -1 takes eleven goes to get to zero also.

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