# num_permutations: number of permutations for creating subsequences of k elements

A function to calculate the number of permutations for creating subsequences of k elements out of a sequence with n elements.

from mlxtend.math import num_permutations

## Overview

Permutations are selections of items from a collection with regard to the order in which they appear (in contrast to combinations). For example, let's consider a permutation of 3 elements (k=3) from a collection of 5 elements (n=5):

• collection: {1, 2, 3, 4, 5}
• combination 1a: {1, 3, 5}
• combination 1b: {1, 5, 3}
• combination 1c: {3, 5, 1}
• ...
• combination 2: {1, 3, 4}

In the example above the permutations 1a, 1b, and 1c, are the "same combination" but distinct permutations -- in combinations, the order does not matter, but in permutation it does matter.

The number of ways to combine elements (without replacement) from a collection with size n into subsets of size k is computed via the binomial coefficient ("n choose k"):

To compute the number of permutations with replacement, we simply need to compute $n^k$.

## Example 1 - Compute the number of permutations

from mlxtend.math import num_permutations

c = num_permutations(n=20, k=8, with_replacement=False)
print('Number of ways to permute 20 elements'
' into 8 subelements: %d' % c)

Number of ways to permute 20 elements into 8 subelements: 5079110400

from mlxtend.math import num_permutations

c = num_permutations(n=20, k=8, with_replacement=True)
print('Number of ways to combine 20 elements'
' into 8 subelements (with replacement): %d' % c)

Number of ways to combine 20 elements into 8 subelements (with replacement): 25600000000


## Example 2 - A progress tracking use-case

It is often quite useful to track the progress of a computational expensive tasks to estimate its runtime. Here, the num_combination function can be used to compute the maximum number of loops of a permutations iterable from itertools:

import itertools
import sys
import time
from mlxtend.math import num_permutations

items = {1, 2, 3, 4, 5, 6, 7, 8}
max_iter = num_permutations(n=len(items), k=3,
with_replacement=False)

for idx, i in enumerate(itertools.permutations(items, r=3)):
# do some computation with itemset i
time.sleep(0.01)
sys.stdout.write('\rProgress: %d/%d' % (idx + 1, max_iter))
sys.stdout.flush()

Progress: 336/336


## API

num_permutations(n, k, with_replacement=False)

Function to calculate the number of possible permutations.

Parameters

• n : int

Total number of items.

• k : int

Number of elements of the target itemset.

• with_replacement : bool

Allows repeated elements if True.

Returns

• permut : int

Number of possible permutations.

Examples

For usage examples, please see http://rasbt.github.io/mlxtend/user_guide/math/num_permutations/