## Permutation and combination shortcut formula for bank exams

Here we are sharing

**important formulas and shortcuts of permutation and combination questions**.In**IBPS PO,SBI PO,RBI assistant/officer exams****permutation and combination**questions are asked repeatedly. In**IBPS clerk**examinations also some simple permutation and combination questions are asked.Here you can find**permutation and combination easy tricks**to solve these questions.We are sure that this article will help you**solve permutation and combination for bank po much faster**.Only**important formulas from permutation and combination**sections are given.We have also provided few permutation and combination problems with solutions.This will help you prepare for various bank's probationary officer exams.### Permutation and combination important formulas and facts

__Permuation formulas and shortcuts__**Factorial**:Let n be a positive integer.Then factorial of n denoted by n! is defined as

*n!=1*2*3*……………..(n-2)*(n-1)*n**Note:0!=1*

**Permutation**: Different arrangement of a given number of things by taking some or all at a time.

**:All arrangements made with letters a,b,c by taking two at a time are ab,ba,bc,cb,ca,ac.**

*Example*
All arrangements made with letters a,b,c by taking all at a time are abc,bca,cab,cba,acb,bac.

**Number of permutation**:Total number of possible arrangements(permutation) of n things, taken r at a time, is given by:

*
*

*=*

^{n}P_{r}

**n!/(n-r)!=n(n-1)(n-2)………….(n-r+1)**
Example 1.Arrangement of 3 items taken 2 at atime

^{3}

*P*

_{2}=

*3!/(3-1)!=3*2=6***. Arrangement of 4 items taken all at a time**

*Example2*

^{4}P_{4}**=4!/(4-4)!=4!/0!=4*3*2*1=24**###
**Permutation :Important point to note**

- Number of all permutation of n things all at a time is n!
- If there are n objects ,m numbers are alike Then number of permutations of these objects is :
*n!/(m!)* - If there are n objects ,p1 numbers are alike of one kind,p2 objects are alike of another kind,p3 are alike of third kind and so on and pr are alike of rth kind,such that
Then number of permutations of these objects is :*p1+p2+p3+……pr=n**n!/(p1!.p2!…..pr!)*

**
**

__Combination shortcuts and formulas__
Each of different groups or selection which can be formed by taking some or all of a number of object, is called a combination.

Suppose we want to select two students from a group of three students namely A,B and C.Then, possible selections are AB,BC and CA.

Note AB and BA represents same selection. But in permutation/arrangement AB and BA represents two different arrangements.

If we want to select ‘all at a time ‘, then there is only one possibility ABC.

Number of combinations:The number of all combination of n things, taken r at a time is:

*=*

^{n}C_{r}

*n!/((r!)(n-r)!)=[n(n-1)(n-2)….upto r factors]/r!*

^{n}*C*

_{r=}^{ n}C_{(n-r)}

^{n}*C*

_{n=1}

^{n}*C*

_{0=1}**:**

*Example**=*

^{ 10}C_{3}

*10!/(3!)(10-3)!=(10*9*8)/(1*2*3)=120*### Permutation and Combination Practice Questions

**.How many arrangements are possible using all the letters of the word BIHAR?**

*Question1**Solution*:The word BIHAR contains 5 different letters.

Required number of arrangements =

^{5}P_{5}=5!=120**.How many word can be formed using all the letters of the word DAUGHTER so that vowels always come together?**

*Question2**Solution:*There are 8 different letters in the given word,but it is given that vowels should come together.Treat vowels as a single entity ‘AUE’

So total number of letters is 5 different letters+ vowel entity=6

Total arrangement possible using 6 letters=

^{6}P_{6}=6!=720
Vowels can be arranged in

^{3}P_{3}ways.^{3}P_{3}=6
Total number of arrangement=720*6=4320

**.In how many ways a football team of 11 players can be selected from a group of 15 candidates?**

*Question3**Solution*:Number of ways of selection=

^{15}C

_{11}=

^{15}C

_{15-11}=

^{15}C

_{4}=(15*14*13*12)/(1*2*3*4)=1365

**.In how many ways, a committee of 5 members can be selected from a group of 6 men and 5 ladies ,consisting of 3 men and 2 ladies?**

*Question4**Solution*:3 men out of 6 and 2 ladies out of 5 are to be chosen.

So number of ways selections can be made is =

^{6}C_{3}*^{5}C_{2}=200
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