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.
Example:All arrangements made with letters a,b,c by taking two at a time are ab,ba,bc,cb,ca,ac.
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:
nPr =n!/(n-r)!=n(n-1)(n-2)………….(n-r+1)
Example 1.Arrangement of 3 items taken 2 at atime
3P2=3!/(3-1)!=3*2=6
Example2. Arrangement of 4 items taken all at a time
4P4=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 p1+p2+p3+……pr=n Then number of permutations of these objects is : 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:
nCr=n!/((r!)(n-r)!)=[n(n-1)(n-2)….upto r factors]/r!
nCr= nC(n-r)
nCn=1
nC0=1
Example: 10C3=10!/(3!)(10-3)!=(10*9*8)/(1*2*3)=120
Permutation and Combination Practice Questions
Question1.How many arrangements are possible using all the letters of the word BIHAR?
Solution:The word BIHAR contains 5 different letters.
Required number of arrangements = 5P5=5!=120
Question2.How many word can be formed using all the letters of the word DAUGHTER so that vowels always come together?
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=6P6=6!=720
Vowels can be arranged in 3P3 ways. 3P3=6
Total number of arrangement=720*6=4320
Question3.In how many ways a football team of 11 players can be selected from a group of 15 candidates?
Solution:Number of ways of selection=15C11=15C15-11=15C4=(15*14*13*12)/(1*2*3*4)=1365
Question4.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?
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 = 6C3*5C2=200
If you find this permutation and combination shortcuts for bank po exam useful ,share it with your friends.
simon
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