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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
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