In this article, we will be discussing Permutations and Combinations formulas, important concepts, shortcut tricks and tips to solve permutations and combinations problems.

**Permutations and Combinations Basics**

Before we discuss Permutations and Combinations formulas, we are going to have a look at what the words Combination and Permutation mean.

In simple terms, if the order doesn't matter, then we have a combination and if the order does matter then we have a permutation. A lot of times you might get confused between permutation and combination, so refer the **d****ifferences Between Permutation and Combination discussed below.**

Permutations |
Combinations |

Permutation refers to several ways of arranging a set of objects in sequential order. | Combination implies several ways of choosing items from a large pool of objects, such that their order is irrelevant. |

Permutations take into account the order, placement and position of the objects/things. | Combinations do not take into account characteristics like order, placement, position, etc. |

A permutation is nothing but an ordered combination. | Combination implies unordered sets or pairing of values within specific criteria. |

A single combination can be obtained from a single permutation. | Many permutations can be derived from a single combination. |

**For example: **Suppose, there is a situation where you have to find out the total number of possible samples of two out of three objects A, B, C.

**Solution:** In this question, first of all, you need to understand whether the question is related to permutation or combination and the only way to find this out is to check whether the order is important or not.

- If the order is significant, then the question is related to permutation, and possible samples will be, AB, BA, BC, CB, AC, CA. Where AB is different from BA, BC is different from CB and AC is different CA.
- If the order is irrelevant, then the question is related to the combination, and the possible samples will be AB, BC, and CA.

**Permutations and Combinations Formulas**

**1)** **Permutations Formula:** Permutations can also be termed as ordered choices or arrangements. Each of the arrangements that can be made by selecting ‘r’ things out of ‘n’ things can be termed as a permutation. In permutations, the order in which the items are arranged is significant.

Consider arranging r things selected from n things. There are n possibilities for the first choice, (n – 1) possibilities for the second choice,(n – 2) possibilities for the third choice and so on. In other words, the available choices reduce by 1 after every selection.

- Therefore, ways to arrange r things selected out of n things are
**n x (n-1) x (n-2) ........ r terms = n x (n-1) x (n-2)... (n-r+1) = n!/(n-r)!** - Hence the generalize formulas for permutations is

** **

**2) Similar items in permutations: **While arranging ‘n’ things from which ‘p’ things are of one kind and ‘q’ things are of the second kind, with the rest of the things being distinct, the number of different arrangements will be

**3)** The total number of ways in which ‘n’ things can be arranged in ‘r’ ways with repetition allowed is equal to **n**^{r }ways.

**4)** **Circular Arrangements: **

- If 'n' objects are arranged in a circular way and if the clockwise and anti-clockwise arrangement is different, then the formula is
**(n-1)!**ways. - When there is no difference between clockwise and anticlockwise arrangements. In those cases, the total possible arrangements are half of the original ways of arrangements, i.e
**(n-1)!/2**

**5) Combinations Formula: **Combinations can also be termed as selections. Hence, each of the selection of ‘r’ items made out of a set of ‘n’ items is called a combination. Generally, **r things are selected from n things in ^{n}C_{r} ways. **That is

**It can be inferred that **

** = **

**6)** If an event or a trial has ‘m’ outcomes and for every outcome of the previous event, a second event has ‘n’ outcomes, then the number of different ways by which both events can be conducted will be the product of m and n.

**Permutations and Combinations Questions & Solutions**

**Question 1:**** **In how many ways you can arrange the letters in the given set of words:

- Tiger
- Lion
- FACE
- Sun

**Solution:** A word with N letters can, in general, be arranged in N! ways. This means the above-given words can be arranged in the below-mentioned ways.

**Question 2: **Consider a collection of 5 distinct items, out of which 3 have to be selected and arranged. They can be considered as a sequence of three choices, each of which results in the selection of one of the remaining things.

**Solution:**

**First choice**- There are 5 possible outcomes.**Second choice**- There will be 4 possible outcomes because 1 item has already been chosen. We don’t know what the possible outcomes are because they depend on the outcome of the first selection, but regardless of which item was picked first, 4 items remain as possibilities for the second choice.**Third choice**- 3 possible outcomes, Therefore, to arrange 5 items in 3 places, the number of possible arrangements is

= 5 ways x 4 ways x 3 ways = 5!/(5-3)! = ^{5}P_{3}

**Question 3:**** **In how many ways can you arrange the letters of the word **"BBC".**

**Solution: **The possible ways to do it are: BBC, BCB, CBB, BBC, BCB, CBB. If you notice, there are repitions in the above mentioned possible arrangements. Now, we need to remove these repetitions and then calculate the possible arrangements.

**Question 4:** Two students are to be selected from 10 students in a class for a scholarship. In how many ways can this be done?

**Solution: **In this case, the order of the students in the pair is not important as they will anyway be given the same scholarship. Hence, this will be a combination or selection problem. To find the number of combinations, the number of arrangements is calculated and is divided by 2! or 2.

**Question 5: **In how many ways can a committee of 1 man and 3 women can be formed from a group of 3 men and 4 women?

**Solution:**

- No. of ways 1 man can be selected from a group of 3 men = 3C1 = 3! / 1!*(3-1)! = 3 ways.
- No. of ways 3 women can be selected from a group of 4 women = 4C3 = 4! / (3!*1!) = 4 ways.

**Question 6: **Suppose you now have the following lot – three red balls and three balls of different colors. Then what are the possible ways in which you can select 3 balls from this lot.

**Solution: **Firstly separate the balls into two lots – the identical balls (say, lot 1) and the distinct balls (lot 2) and then make cases as given below.

- All the three balls from lot 1 = 1 way
- Two from lot 1 and one from lot 2 = 1 x
^{3}C_{1}ways - One from lot 1 and two from lot 2 = 1 x
^{3}C_{2}ways - All the balls from lot 2 =
^{3}C_{3}ways - Add all the cases: 1 + 3 + 3 + 1 = 8 ways

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