Answer:
Check the explanation
Step-by-step explanation:
in this particular type of problems,we have to go with the cases.we know there would be a 50 odd-even pairs between 1 and 100.
1st case: 1 is placed at the head position of given set S.so there is a chance of only one even number beside 1.so there are 50×98!
2nd case: 1 is placed at tail position of S.so there is again a chance of one even number beside 1. so there are
50×98!
3rd case: 1 is placed neither at head nor at tail.1 can be at rest 98 places.lets fix the place of 1.then the choices are difference between permuting all the numbers and 1 surrounded by two odd numbers.
i.e., 99! - 49p2 ×97!
so total choces are = 2×50×98! + 99! - 49p2 × 97!
= 7448 × 97! + 99!
The number of permutations are there of S in which the number 1 is next to at least one even number = [tex]7448 \times 97! + 99![/tex]
In this particular type of problem, we have to go with the cases. we know there would be 50 odd-even pairs between 1 and 100.
1st case:
1 is placed at the head position of given set S. so there is a chance of only one even number besides 1.
[tex]50\times98![/tex]
2nd case:
1 is placed at the tail position of S. so there is again a chance of one even number besides 1. so there are
[tex]50\times98![/tex]
3rd case:
1 is placed neither at the head nor at tail. 1 can be at rest 98 places. let us fix the place of 1. then the choices are the difference between permuting all the numbers and 1 surrounded by two odd numbers.
[tex]99! - 49p2 \times 97![/tex]
so total choices are
= [tex]2\times50\times98! + 99! - 49p2 \times 97![/tex]
= [tex]7448 \times 97! + 99![/tex]
Learn morne about permutation:
https://brainly.com/question/12468032
