Permutation and combination Basics
Permutation and combination are fundamental concepts in mathematics, particularly in the field of combinatorics, which deals with counting and arranging objects. These concepts are used to determine the number of ways in which objects can be selected or arranged from a set. Let's explore each concept separately:
Permutation:
A permutation is an arrangement of objects in a specific order. It is concerned with the order of selection or arrangement of items from a set. There are two main types of permutations:
Permutations with Repetition: In this case, you have a set of objects, and you want to arrange them in a specific order, allowing for repetition of objects. For example, arranging the letters A, B, and C with repetition would include arrangements like AABB, ABAC, etc.
Permutations without Repetition: In this case, you have a set of distinct objects, and you want to arrange them in a specific order without repeating any object. For example, arranging the letters A, B, and C without repetition would include arrangements like ABC, BCA, CAB, etc.
The formula for calculating the number of permutations of 'n' distinct objects taken 'r' at a time without repetition is given by:
Combination:
A combination is a selection of objects from a set without regard to the order in which they were selected. It is concerned with choosing a subset of objects from a larger set. There are two main types of combinations:
Combinations with Repetition: In this case, you have a set of objects, and you want to choose 'r' of them without regard to order, allowing for repetition of objects. For example, choosing two ice cream flavors from a menu with repetition would include combinations like Chocolate-Vanilla, Vanilla-Vanilla, etc.
Combinations without Repetition: In this case, you have a set of distinct objects, and you want to choose 'r' of them without regard to order and without repeating any object. For example, choosing two students from a class without repetition would include combinations like Alice and Bob, but not Alice and Alice.
The formula for calculating the number of combinations of 'n' distinct objects taken 'r' at a time without repetition is given by:
Permutation and Combination Problem
"MATHEMATICS"
As you'll be able to see there are some letters(like M, T, and A) during this word that is obtaining continual. So, when choosing the letters for arrangement we must always take into account all the cases. At first, I'm Manibhushan making a case for a way to choose letters for various cases and so later a way to prepare them.
We have eleven letters in "MATHEMATICS"
in which there are
2 M's, 2 T's, two A's and alternative letters H, E, I, C, S ar single
Selection of the four letters(Combination)
first case: 2 alike and alternative 2 alike
In this case we have a tendency to ar gonna choose the 2 letters that ar alike. we've got 3 selections M,T,A. Out of those, we've got to pick 2(Because we've got to pick four letters and choosing two alike letters means that choosing four letters). So, it will be wiped out 3C2 ways that
second case: 2 alike, 2 totally different
1 alike letter(which can mean 2 letters) will be chosen in 3C1 ways and alternative two totally different letters will be chosen in 7C2 ways.(as there'll be seven totally different letters).
So, 3C1*7C2 ways that
Third case: All are totally different
This can be wiped out in eightC4 ways as there are 8 totally different letters(M, T, A, H, E, I, C, S)
Arrangement:(permutation)
For the primary case, there'll be 2 alike letters. So, the arrangement of those letters will be wiped out 4!/2!*2! ways that
For the second case, there'll be one alike letter and 2 totally different letters which might be organized in 4!/2! ways that
For the third case, all letters are totally different thus it will be organized in 4! ways that
So, the ultimate answer is going to be
(3C2*4!/2!*2!) + (3C1*7C2*4!/2!) + (8C4*4!)
= 18+ 756 + 1680 ways that
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