Any 6 from 15 = 15!/(9! x6!) = (15 x 14 x 13 x 12 x 11 x 10)/ (6 x 5 x 4 x 3 x 2) = 5005.
(If the order doesn't matter then multiply this by 720)
No. The number of permutations or combinations of 0 objects out of n is always 1. The number of permutations or combinations of 1 object out of n is always n. Otherwise, yes.
The number of R-combinations in a set of N objects is C= N!/R!(N-R)! or the factorial of N divided by the factorial of R and the Factorial of N minus R. For example, the number of 3 combinations from a set of 4 objects is 4!/3!(4-3)! = 24/6x1= 4.
Probably 222= 4,194,304Of course, you could add a string of factorials to make the number incredibly huge, but then there is no limit222!! for example, or 222!!! and so on.Probably 222= 4,194,304Of course, you could add a string of factorials to make the number incredibly huge, but then there is no limit222!! for example, or 222!!! and so on.Probably 222= 4,194,304Of course, you could add a string of factorials to make the number incredibly huge, but then there is no limit222!! for example, or 222!!! and so on.Probably 222= 4,194,304Of course, you could add a string of factorials to make the number incredibly huge, but then there is no limit222!! for example, or 222!!! and so on.
Yes
Often, when evaluating theoretical probability of something, say a lottery game or a card game, you consider permutations and combinations, in order to assess the number of outcomes that are possible versus the number that are desired. When you work with permutations and combinations, you often work with factorials. In short, the factorial of a number is the product of that number and all of the numbers successively smaller by one until you get to one, i.e. 5 factorial is 5 x 4 x 3 x 2 x 1 or 120.
To find the number of combinations possible for a set of objects, we need to use factorials (a shorthand way of writing n x n-1 x n-2 x ... x 1 e.g. 4! = 4 x 3 x 2 x 1). If you have a set of objects and you want to know how many different ways they can be lined up, simply find n!, the factorial of n where n is the number of objects. If there is a limit to the number of objects used, then find n!/(n-a)!, where n is the number of objects and n-a is n minus the number of objects you can use. For example, we have 10 objects but can only use 4 of them; in formula this looks like 10!/(10-4)! = 10!/6!. 10! is 10 x 9 x 8 x ... x 1 and 6! is 6 x 5 x ... x 1. This means that if we were to write out the factorials in full we would see that the 6! is cancelled out by part of the 10!, leaving just 10 x 9 x 8 x 7, which equals 5040 i.e. the number of combinations possible using only 4 objects from a set of 10.
Multiply: 9! (9 factorials) (9) (8) (7) (6) (5) (4) (3) (2) (1)
No. The number of permutations or combinations of 0 objects out of n is always 1. The number of permutations or combinations of 1 object out of n is always n. Otherwise, yes.
The number of R-combinations in a set of N objects is C= N!/R!(N-R)! or the factorial of N divided by the factorial of R and the Factorial of N minus R. For example, the number of 3 combinations from a set of 4 objects is 4!/3!(4-3)! = 24/6x1= 4.
Logic Of Strong number: Take anumber.First findout the factorials of all the digits of the number.Then sum the factorials of all the digits.If that sum is equal to the entered number then that number is said to be a strong number.
Peterson Number:145 = 1! + 4! + 5!number=sum of (factorials of digits)
Probably 222= 4,194,304Of course, you could add a string of factorials to make the number incredibly huge, but then there is no limit222!! for example, or 222!!! and so on.Probably 222= 4,194,304Of course, you could add a string of factorials to make the number incredibly huge, but then there is no limit222!! for example, or 222!!! and so on.Probably 222= 4,194,304Of course, you could add a string of factorials to make the number incredibly huge, but then there is no limit222!! for example, or 222!!! and so on.Probably 222= 4,194,304Of course, you could add a string of factorials to make the number incredibly huge, but then there is no limit222!! for example, or 222!!! and so on.
5! = 120 ! means factorial. A factorial is the product of of the positive integers and equals the number of different combinations of a number. A factorial can be work out quite simply. Take the number 5. 5! = 5x4x3x2x1 = 120 So simply place the number you are trying to find out the combinations for first and then times it by all the numbers below. Some more examples would be: 8! = 8x7x6x5x4x3x2x1 = 4320 3! = 3x2x1 = 6 10! = 10x9x8x7x6x5x4x3x2x1 = 3,628,800 6! = 6x5x4x3x2x1 = 720 * * * * * An interesting introduction on factorials but totally misses the point of the question. A factorial generates permutations, not combinations! For combinations, abc is the same as acb, cab, bac, etc. The number of combinations of that you can make out of 5 things *including the null combination - ie nothing) is 25 = 32.
I guess the expected answer is 97210. Using factorials and exponents very much greater number can be obtained. For example, 97210 is a number with 6880 digits. And that is without using factorials.
Yes
Without using exponents or factorials, that's it.
Often, when evaluating theoretical probability of something, say a lottery game or a card game, you consider permutations and combinations, in order to assess the number of outcomes that are possible versus the number that are desired. When you work with permutations and combinations, you often work with factorials. In short, the factorial of a number is the product of that number and all of the numbers successively smaller by one until you get to one, i.e. 5 factorial is 5 x 4 x 3 x 2 x 1 or 120.