What is the formula to find the permutation of r objects given that there are n total object and out of them n1 and n2 are indistinguishable?

Jacoby Erickson

Jacoby Erickson

Answered question

2022-10-31

What is the formula to find the permutation of r objects given that there are n total object and out of them n1 and n2 are indistinguishable?
The question is I've n the total number of objects, out of them I've to find the permutation of r objects given that r < n. It's also given that there m and t are indistinguishable objects in n such that m + t = n. What will be the formula to find the permutation in such a case?
As an example, we've 8 marbles out of which 5 are red and 3 and blue, and the number of objects we have to choose is 2. The permutation we know would be equal to 4. But I don't don't the formula to evaluate this value of 4.

Answer & Explanation

Szulikto

Szulikto

Beginner2022-11-01Added 22 answers

Step 1
One way is dividing it into cases, eg for the numerical example, either you can have 2 red, 2 blue, or 1 red + 1 blue in two ordered ways
But this sort of computation quickly becomes very tedious.
So here is a way you can use where a lot of cases may arise. We know that when we multiply polynomials in x, the indices of x get added up.
And we also know that if one type, eg, M occurs, say, k times in the extracted letters, we need to divide the permutations by k!
Using these two ideas, we can represent each letter as a polynomial of x, the indices representing the number of times a letter can occur. You can regard x′s as only placeholders for the number of times a letter can occur.
Step 2
Putting it all together, for a somewhat more complex example, 4 letter permutations of PINEAPPLE (I,N,A,L occur once each, E occurs twice, P occurs thrice), Remember that P, for example, may occur 0,1,2,or3$ times
find 4! times the coefficient of x 4 in the expression ( x 0 + x 1 ) ( x 0 + x 1 ) ( x 0 + x 1 ) ( x 0 + x 1 ) ( x 0 + x 1 + x 2 2 ! ) ( x 0 + x 1 + x 2 2 ! + x 3 3 ! )
which of course would normally be written as 4 ! ( 1 + x ) 4 ( 1 + x + x 2 2 ! ) ( 1 + x + x 2 2 ! + x 3 3 ! )
and gives an answer of 626
You may like to countercheck by the case by case approach, eg
- 4 singles: ( 6 4 ) 4 ! = 360
- 2 singles and a double: ( 4 3 ) ( 2 1 ) 4 ! 2 ! = 240
- 1 single, 1 triple: ( 5 1 ) 4 ! 3 ! = 20
- 2 doubles: ( 2 2 ) 4 ! 2 ! 2 ! = 6

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