## How do you find the number of arrangements?

The formula for combinations is generally n! / (r! (n — r)!), where n is the total number of possibilities to start and r is the number of selections made. In our example, we have 52 cards; therefore, n = 52.

## How do you find the number of repetition combinations?

The number of k-element combinations of n objects, with repetition is Cn,k = Cn+k-1,k = (n + k − 1 k ) = ((n k )) . It is also the number of all ways to put k identical balls into n distinct boxes, or the number of all functions from a set of k identical elements to a set of n distinct elements.

**How many combinations of 4 items are there?**

I.e. there are 4 objects, so the total number of possible combinations that they can be arranged in is 4! = 4 x 3 x 2 x 1 = 24.

**How many combinations of 3 items are there?**

* 3!) = 10 different ways. This generalises to other combinations too and gives us the formula #combinations = n! / ((n – r)! * r!)

### How do you find distinguishable permutations?

To find the number of distinguishable permutations, take the total number of letters factorial divide by the frequency of each letter factorial. Basically, the little n’s are the frequencies of each different (distinguishable) letter. Big N is the total number of letters.

### How do you do Permute letters?

To calculate the amount of permutations of a word, this is as simple as evaluating n! , where n is the amount of letters. A 6-letter word has 6! =6⋅5⋅4⋅3⋅2⋅1=720 different permutations.

**What are the permutations of 5 numbers?**

(For k = n, nPk = n! Thus, for 5 objects there are 5! = 120 arrangements.)

**What’s the difference between permutations and combinations?**

A permutation is an act of arranging the objects or numbers in order. Combinations are the way of selecting the objects or numbers from a group of objects or collection, in such a way that the order of the objects does not matter.

## What is 4C2 combination?

We know that the formula used to solve the combination expressions is given by: nCr = n!/[r! (n – r)! Substituting n = 4 and r = 2 in the above formula, 4C2 = 4!/ [2!

## How many combinations of arrangements are possible?

The second space can be filled by any of the remaining 3 letters. The third space can be filled by any of the 2 remaining letters and the final space must be filled by the one remaining letter. The total number of possible arrangements is therefore 4 × 3 × 2 × 1 = 4!