Question

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

Answer 1

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

$\frac{n*(n-1)*(n-2) ........ 6*5*6*4!}{4!}$

= n! ×6/4!

= n!/4

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

To know more about permutation:

brainly.com/question/4416319

#SPJ4