Answered

Get the most out of your questions with the extensive resources available on IDNLearn.com. Get thorough and trustworthy answers to your queries from our extensive network of knowledgeable professionals.

In how many ways can the elements of [n] be permuted if 1 must precede 2 and 3 is to precede 4?

Sagot :

No. of ways in the elements of n be permuted is  n!/4

Number of elements = n

Number of face in which the element of n be permuted if 1 must proceed 2 and 3 is to precede 4.

n =  4

The possible permutation will

(1,2,3,4), (1,3,2,4),(1,3,4,2),(3,4,1,2),(3,1,2,4),(3,1,4,2)

There are 6 possible outcomes

For n= 5

With 1,2,3,4 we add 5 in the arrangement with above terms

The possible number of permutation

5×6 = 30

When n = 6

30×6 = 180.

So for an element

If 1 precede 2 and 3 is yo precede 4 is

[tex]\frac{n*(n-1)*(n-2) ........ 6*5*6*4!}{4!}[/tex]

= n! ×6/4!

= n!/4

No. of ways in the elements of n be permuted is  n!/4

To know more about permutation:

https://brainly.com/question/4416319

#SPJ4

Thank you for using this platform to share and learn. Keep asking and answering. We appreciate every contribution you make. Thank you for visiting IDNLearn.com. We’re here to provide accurate and reliable answers, so visit us again soon.