Question

How do you find the number of distinguishable permutations of the group of letters A, L, G, E, B, R, A?

Answer 1

Hint: Now we know that the number of arrangements of n objects where ${{a}_{1}}$ objects are of type 1. ${{a}_{2}}$ objects are of type 2 and so on ${{a}_{m}}$ objects are of type m then the number of arrangements is given by $\dfrac{n!}{{{a}_{1}}!{{a}_{2}}!...{{a}_{m}}!}$ . Hence using this we can find the number of distinguishable permutations of the group of letters A, L, G, E, B, R, A?

Complete step-by-step answer:
Now first let us understand the concept of permutations.
Permutations are nothing but the number of possible arrangements of certain elements.
Now the number of possible arrangements of n distinct elements is n! where $n!=n\left( n-1 \right)\left( n-2 \right)...3\times 2\times 1$ .
To understand this formula we can take an example in which we have to arrange n objects in n places.
Now we will count the number of possible ways to arrange n objects in n places and this will give a total number of arrangements.
Now let us first place the first object. Now we have n choices and hence can be placed in n ways.
Now after the object is placed consider the second object. Now it has n – 1 choices to be placed since we have already placed one object. Hence we have n – 1 choices.
Now similarly if we go on we get the total number of choices is n × (n - 1) × (n - 2) × … = n!
Now if we have m objects of type 1, r objects of type 2 and so on then the number of possible arrangements is given by $\dfrac{n!}{r!m!}$ .
Now consider the given example. We have 7 letters in which 2 are the same.
Hence the number of arrangements will be given by $\dfrac{7!}{2!}=\dfrac{7!}{2}$
Hence the number of arrangements of A, L, G, E, B, R, A is given by $\dfrac{7!}{2}$ .

Note: Now the number of ways of selecting r objects from n objects is given by $\dfrac{n!}{r!\left( n-r \right)!}$ and the number of ways of arranging r objects out of n objects is given by $\dfrac{n!}{\left( n-r \right)!}$ by substituting r = n we get the number of arrangements of n objects which is nothing but n! as 0! = 1.