**Permutation and Combination Methods shortcut tricks**

Permutation and combination methods related math calculation are given in bank exams so its important to learn for exams We can’t ignore it. So it is very very important to you improve your maths skills for banking exams in permutation and combination methods. Anything we learn in our school days was basics and that is well enough for passing our school exams.

Now the time has come to learn for our competitive exams. For this we need our basics but also we have to learn something new. Permutation and Combination shortcut trick or Permutation and Combination tricks or Permutation and Combination formula or Permutation and Combination problems or Permutation and Combination examples or Permutation and Combination methods or shortcut tricks of Permutation and Combination That’s where shortcut tricks are comes into action.

Here is some general rule and formula are given,In maths exam papers there are two or three question are given from this chapter.This type of problem are given in Quantitative Aptitude which is a very essential paper in banking exam.Under below given some more example for your better practice.

Shortcut Tricks are very important things in competitive exam. Time takes a huge part in competitive exams. If you know time management then everything will be easier for you. Most of us miss this thing. Here in this page we give few examples on **Permutation and Combination shortcut tricks**. These shortcut tricks cover all sorts of tricks on Permutation and Combination. Visitors please read carefully all shortcut examples. You can understand shortcut tricks on Permutation and Combination by these examples.

Before starting anything just do a math practice set. Write down twenty math problems related to this topic on a page. Using basic math formula do first ten maths of that page. You also need to keep track of Timing. After finish write down total time taken by you to solve those ten maths. Now go through our page for permutation and combination shortcut trick. After this do remaining ten questions and apply shortcut formula on those math problems. Again keep track of the time. You will surely see the improvement in your timing this time. But this is not all you need. You need more practice to improve your timing more.

Math section in a competitive exam is the most important part of the exam. That doesn’t mean that other topics are less important. But if you need a good score in exam then you have to score good in maths. Only practice and practice can give you a good score. All you need to do is to do math problems correctly within time, and you can do this only by using shortcut tricks. Again it does not mean that you can’t do maths without using shortcut tricks. You may do math problems within time without using any shortcut tricks. You may have that potential. But so many other people may not do the same. So Permutation and Combination shortcut tricks here for those people. Here in this page we try to put all types of shortcut tricks on Permutation and Combination. But we may miss few of them. If you know anything else rather than this please do share with us. Your little help will help others.

Now we will discuss some basic ideas of **Permutation and Combination Methods**. On the basis of these ideas we will learn trick and tips of shortcut permutation and combination. If you think that * how to solve permutation and combination questions using permutation and combination shortcut tricks*, then further studies will help you to do so.

**General Rule and Formula**

Permutation and combination methods related math calculation are given in bank exams so its important to learn for exams.

**Factorial Notation :** If we consider that n be a positive integer. Then, Factorial n, we denoted as ⌊n or n!.

n! is defined as :

- n! = n ( n – 1 ) ( n – 2 ) …………….3.2.1.

**Ex.**5! = ( 1 x 2 x 3 x 4 x 5 ) = 120;**Ex.**4! = (1 x 2 x 3 x 4 ) = 24;**Ex.**3 = ( 1 x 2 x 3 ) = 6;

- 1! = 1 [factorial 1 is always 1]
- 0! = 1 [factorial 0 is always 1]

**Permutation :** The different arrangements of a given number or things by taking some or all at a time , are called Permutations.

**Note :**In Short and basic think of permutation is arrangement, given number or letter, that how we arrange it. we can arrange it taking some number or letter at a time or we can arrange it taking all at a time.- All permutations (or arrangements) made with the letters of a, b, c by taking two at a time are ( ab, ba, ac, ca, bc, cb )
- All permutations made with the letters a, b, c, taking all at a time are : ( abc, acb, bac, bca, cab, cba ).

**Number of Permutations : **Number of all permutations of n things, taken r at a time, is given by :

^{n }p _{r} = n ( n – 1 ) ( n – 2 )……(n – r + 1 ) = n! / ( n – r ) !

**Ex. 5**p_{2}

= ( 5 x 4 ) = 20.**Ex.**8^{ }p_{ 3}

= ( 8 x 7 x 6 ) = 336;

**Note:**Number of all permutation that is n things, we can take all at a time = n!

**Examples:** 6! / 2! = 1 x 2 x 3 x 4 x 5 x 6 / 1 x 2 = 360.

**Here is some Permutation and Combination important link which provided you better understanding.**

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best trick

in which cases we get answers like 16power 6 When 6 Digit number to be formed

when u hv 16 digits out of which u hv to choose 6 digit no. when repetition is allowed.

I have four questions, I tried a lot but couldn’t get the answer.

1. “PROMISE” In how many ways it can be arranged such that no two vowels should come together

2. How many numbers are there between 100 and 1000 in which all the digits are distinct.

3.A person has 10 different colour suits and 15 different colour ties. He has to take 4 suits and 10 ties on a tour. If a green tie is not to be selected in case he selects blue suit, find the number of ways in which he can select his dresses.

4. There are 8 pairs (Husband and Wife) in a mixed double tennis tournament which is to be organised in such a way that a particular pair can’t play in a particular game. Find the number of ways it can be arranged.

can you help?

240

Q1 Ans. 7p7 – 3p2*4p4

279*144 =40066

Ritik patidar class 11

Convert the following into factorials:

(1) 2.4.6.8.10

take 2 common from it.

so, we have

2(1.2.3.4.5) =2*5!

10!/945

In how many ways can the letters of ENGINEERING be arranged so that 3N’s come together but the 3E’s do not come together

How to formulate the equation in permutation and combination if k and answer is already given, how to find ‘n’?

C(8,k)=28

P(7,k)=840

Need your help!