# Algorithm for finding a permutation combination

Source: Internet
Author: User
That is true
Randomly extract 3 numbers from the 36 characters of 0~9,a~z to calculate how many permutation combinations are available

36*36*36=46656

36*36*36

Forget it, I wrote it.

`/** * Permutation combination algorithm * C (\$n, \$m) * Number of \$n elements * \$m number of elements taken from \$n * \$num total number of combinations **/function get_combination (\$n, \$m) {if (Is_int (\$n) &am p;& Is_int (\$m)) && (\$m <= \$n)) {\$a = 1;//initialize for (\$i =1; \$i <= \$m; \$i + +) {\$a = \$n * \$a; \$b = 1* \$i; \$n--;} \$num = \$a/\$b; return \$num;}}`

Combination formula: C (n,m) =p (n,m)/m!=n!/((n-m)!*m!)
Do you think you are right?

Combination formula: C (n,m) =p (n,m)/m!=n!/((n-m)!*m!)
Do you think you are right?

You can try.

36*35*34

Calculated by the combined formula is 7140
Half less than you.

Combination formula: C (n,m) =p (n,m)/m!=n!/((n-m)!*m!)
Do you think you are right?

You can try.

echo get_combination (3,3);
Output 2
It's not right, is it?

echo get_combination (3,3);
Output 2
It's not right, is it?

Yes, just looked.
The formula you wrote is also wrong.
Two numbers from 2 numbers equals 1.

Right
The combination of two numbers from 2 numbers is 1.
The arrangement is 2.
P (n,m) =n (n-1) (n-2) ... (n-m+1) =n!/(N-M)! (stipulated 0!=1).

Add the following 3 numbers can be the same

For example, extracting a random two number from an array can be
11
22
12
21st

Right
The combination of two numbers from 2 numbers is 1.
The arrangement is 2.
P (n,m) =n (n-1) (n-2) ... (n-m+1) =n!/(N-M)! (stipulated 0!=1).

All right, I'm going to go around.
N^m is OK.
Thank you.
It's been a tangle of permutations and combinations.
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