((1 + 40) x 20)/40 ie 20.5
You can't get a permutation of a number, but you can get the permutation of a set of numbers. Take the set {1, 2, 3, 4, 5}. If you move the 1 over to the right 1 or 3 spaces and keep the other numbers in the same order, you have an odd permutation: {2, 1, 3, 4, 5} and {2, 3, 4, 1, 5} in this case. If you move the 1 over to the right 2 or 4 spaces, you have an even permutation: {2, 3, 1, 4, 5} and {2, 3, 4, 5, 1} in this case. If two of the numbers in the set are both oddly permutated, the set as a whole is evenly permutated, kind of like how an odd number multiplied by an odd number is an even number. For example, {2, 1, 4, 3, 5} is an even permutation. However, and I'll let you figure out why on this one, {2, 1, 4, 5, 3} is an odd permutation. There are much more formal ways to describe permutations and many different ways to write them out, as you can see by following the link below, but what I've stated above should get you started with at least a basic comprehension.
This is a Permutation problem; nPr. This is: n!/(n-r)! or 45!/(45-5)! = 146611080.
The square numbers between 1 and 20 are 1,4,9 and 16.
5/100 ie 1/20
(1 + 20) x (20/2) ie 210
You do not have to figure the permutation. You simply rearrange the order of the numbers that you are presented with. The permutations of the number set 1, 2, 3 include 1, 3, 2, and 2, 1, 3.
((1 + 40) x 20)/40 ie 20.5
That depends how many are taken for each permutation. If you take 1 number at a time, then there are 20 permutations. 2 at a time . . . . . (20 x 19) = 380 3 at a time . . . . . (20 x 19 x 18) = 6,840 4 at a time . . . . . (20 x 19 x 18 x 17) = 116,280 5 at a time . . . . . (20 x 19 x 18 x 17 x 16) = 1,860,480 etc.
One possible permutation of these numbers to make 24 is (7 + 1) / (2/6).
You can't get a permutation of a number, but you can get the permutation of a set of numbers. Take the set {1, 2, 3, 4, 5}. If you move the 1 over to the right 1 or 3 spaces and keep the other numbers in the same order, you have an odd permutation: {2, 1, 3, 4, 5} and {2, 3, 4, 1, 5} in this case. If you move the 1 over to the right 2 or 4 spaces, you have an even permutation: {2, 3, 1, 4, 5} and {2, 3, 4, 5, 1} in this case. If two of the numbers in the set are both oddly permutated, the set as a whole is evenly permutated, kind of like how an odd number multiplied by an odd number is an even number. For example, {2, 1, 4, 3, 5} is an even permutation. However, and I'll let you figure out why on this one, {2, 1, 4, 5, 3} is an odd permutation. There are much more formal ways to describe permutations and many different ways to write them out, as you can see by following the link below, but what I've stated above should get you started with at least a basic comprehension.
Only 1, since the order of the digits does not matter in a compbination. In a permutation, yes, but not combination.
These are the prime numbers 1-20 2,3,5,7,11,13,17,19
This is a Permutation problem; nPr. This is: n!/(n-r)! or 45!/(45-5)! = 146611080.
The numbers 1-20 are on a dart board so there are twenty numbers.
1=1
20 and 1. 20 + 1 = 21 20 - 1 = 19