Permutation
Assume that I have 4 letters (A, B, C and D). Now if anybody asks me to write down all the permutations of 3 of these letters…..
ABC BAC CAB DAB
ACB BCA CBA DBA
ABD BAD CAD DAC
ADB BDA CDA DCA
ACD BCD CBD DBC
ADC BDC CDB DCB
ACB BCA CBA DBA
ABD BAD CAD DAC
ADB BDA CDA DCA
ACD BCD CBD DBC
ADC BDC CDB DCB
So, there are 24 permutations in total. Here the order is important. In other words ACB is different, BCA is different, CBA is different and ABC is different (even-though they all are formed with same group letters).
Combination
As we have already discussed, the collection of letters is important here, not the order. That means, if you have ABC in your set that’s enough. So you cant claim ABC, ACB, BAC, BCA, CAB, CBA… for combinations. These all are 1 combination of letters A, B and C.
So, from the given 4 letters (A, B, C and D), You can write the combination of 3 of those letters
ABC ABD ACD BCD
hope you have got the basic concept now.
Now lets have a look at the technical side, before going to calculate Permutations and Combinations, you should know the word Factorial.
Factorial : The factorial of a number, represented by n!, is the product of the natural numbers up to and including n
In simple words, the Factorial of thenumber n is the number of ways that the n elements of a group can be ordered.
So, if somebody ask you a question,how many different ways six people can sit at a table with six chairs,
you should say them its 6! = 6 × 5 × 4 × 3 × 2 × 1 = 720.
So, if somebody ask you a question,how many different ways six people can sit at a table with six chairs,
you should say them its 6! = 6 × 5 × 4 × 3 × 2 × 1 = 720.
Note : We treat 0! as 1
How to Calculate Permutations ?
There is a simple formula for calculating permutations.
Number of all permutations of nitems, taken r at a time, is given by:
nPr = n(n – 1)(n – 2) … (n – r + 1)
=
n above case n is 4 and r is 3
So, nPr = P(n,r) = 24
How to Calculate Combinations ?
Important Note : To calculate combinations,
- First you should calculate all the equivalent permutations.
- Later you should correct this list by cutting out duplicates / repetitions.
Now lets have a look at the mathematical formula for Combinations..
| nCr = | n! | = | n(n – 1)(n – 2) … … r | . |
| (r!)(n – r!) | r! |
n above case n is 4 and r is 3, so
nCr = C(n,r) = 4
That’s all for now friends. In our next post we shall discuss some practice problems on Permutations and Combinations.
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