Math, asked by svvk5797, 1 year ago

How many permutations can be formed using the latters of statistics?

Answers

Answered by Anonymous
2
There's a technique in probabilistic reasoning that is very useful in questions like this where you have "...keep xxfixed..." or "...at least 1 in each container...".

The trick is to first "set up" the scenario by distributing the fixed restrictions and then counting the possibilities from what you have remaining to work with. For example applied to your question you need it to start and end with an S. So from our metaphorical bag of letters, let us take out two S and place them one on each end. Now all that we need to count are the permutations of the remaining 8 letters in our bag (I'll leave this up to you).

This generalises nicely to the problem of putting rr objects from your collection of nn into rrparticular locations leaving you with the smaller problem of just permuting n−rn−r objects.

As an other unrelated example to get my message across about the technique (because ultimately, it's the technique that you need to take away from this) consider:

Ways to distribute nn objects into kk boxes such that no box is empty. Objects indistinguishable/distinguishable and each box can hold any number of objects.

Using the technique, put exactly one object in each box and then count the number of ways to distribute n−kn−k objects into kkboxes. For the case where the objects are distinguishable you need to also count the number of ways to choose and permute the kk objects you use to fill.

Similar questions