Today's topic forms the base of a chunk of problems in Permutations and Combinations. We will break it down into distinct cases and divulge into each one separately. So for an effective learning the topic will be presented in 2 parts. Let's start right away with the part 1.
Case 1 : Objects(Letters) are IDENTICAL while the Groups(Envelopes) are DISTINCT.
Consider 4 distinct envelopes having E1,E2,E3,E4 letters in them with distinct notation used to represent distinct envelopes(while number of letters in them may be same) and 6 identical letters.
We know that :
E1 + E2 + E3 + E4 = 6 = which can be seen as 1 + 1 + 1 + 1 + 1 + 1, six identical objects.
For such an equation,
[When envelopes may be empty]
Case 2 : Objects are DISTINCT as well as groups are DISTINCT.
Consider the case of 4 different envelopes to be filled with 6 distinct letters. Each letter distinctly possesses the option to choose from 4 distinct envelopes.
Hence, total number of permutations possible are : 4*4*4*4*4*4 = 4^6 ways.
"NOTE : We do not restrict the possibility of some of the envelopes being empty in this case. We look at that restrictive arrangement in part 2 as it involves a different concept when both objects and groups are different.
Also,
Here the order in which letters(objects) are put into envelopes(groups) wasn't of any significance.
But what happens when it is of significance? Let's find out with a suitable example.."
Consider the case of 6 rings to be put on your 4 fingers(excluding your thumb). Now the order in which the rings are put on your fingers would hold significance as some of fingers are bound to have more than 1 rings on them. Hence, relative positioning of rings does matter here. Now this can be viewed with 2 perspectives :
Outlook 1 : Look it as the first ring R1 will have 4 options(any of the fingers). The second ring will have 5 options(3 fingers plus two above or below the already occupied finger) and third ring will have 6 options..and so on.
So we get, for 6 rings, total permutations as : 4*5*6*7*8*9.
Outlook 2 : Consider it as the Case 1 only where order of arrangement matters and the 6 rings(objects,letters) are no more identical. So we need to permute the rings in 6! ways.
Hence we would get total permutations as : C(6 + 4 - 1,4 - 1) * 6! = C(9,3) * 6! = P(9,6)
Doesn't matter how you look at it, both would produce same results as should have been the case.
This brings us to the close of Part 1 of this topic. In part 2, we will look at the other possibilities when it comes down arranging objects into groups that have not yet been explored.
Cheers!
AS
Case 1 : Objects(Letters) are IDENTICAL while the Groups(Envelopes) are DISTINCT.
Consider 4 distinct envelopes having E1,E2,E3,E4 letters in them with distinct notation used to represent distinct envelopes(while number of letters in them may be same) and 6 identical letters.
We know that :
E1 + E2 + E3 + E4 = 6 = which can be seen as 1 + 1 + 1 + 1 + 1 + 1, six identical objects.
For such an equation,
[When envelopes may be empty]
Number of integral non negative solutions = C(6 + 4 - 1, 4 - 1) = C(9,3) and,
[When envelopes can't be empty]
[When envelopes can't be empty]
Number of integral positive solutions = C(6 - 1,4 - 1) = C(5,3)
Consider the case of 4 different envelopes to be filled with 6 distinct letters. Each letter distinctly possesses the option to choose from 4 distinct envelopes.
Hence, total number of permutations possible are : 4*4*4*4*4*4 = 4^6 ways.
"NOTE : We do not restrict the possibility of some of the envelopes being empty in this case. We look at that restrictive arrangement in part 2 as it involves a different concept when both objects and groups are different.
Also,
Here the order in which letters(objects) are put into envelopes(groups) wasn't of any significance.
But what happens when it is of significance? Let's find out with a suitable example.."
Consider the case of 6 rings to be put on your 4 fingers(excluding your thumb). Now the order in which the rings are put on your fingers would hold significance as some of fingers are bound to have more than 1 rings on them. Hence, relative positioning of rings does matter here. Now this can be viewed with 2 perspectives :
Outlook 1 : Look it as the first ring R1 will have 4 options(any of the fingers). The second ring will have 5 options(3 fingers plus two above or below the already occupied finger) and third ring will have 6 options..and so on.
So we get, for 6 rings, total permutations as : 4*5*6*7*8*9.
Outlook 2 : Consider it as the Case 1 only where order of arrangement matters and the 6 rings(objects,letters) are no more identical. So we need to permute the rings in 6! ways.
Hence we would get total permutations as : C(6 + 4 - 1,4 - 1) * 6! = C(9,3) * 6! = P(9,6)
Doesn't matter how you look at it, both would produce same results as should have been the case.
This brings us to the close of Part 1 of this topic. In part 2, we will look at the other possibilities when it comes down arranging objects into groups that have not yet been explored.
Cheers!
AS
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