Combinatorics

Question

A Sequence of nos. $<1, 2, 3, 4, ..... 10 $> is permuted randomly. What is the probability that all odd nos. appear before all even nos. I am getting $\frac{5! * 5!}{10!}$, is it correct

Comments

yup u are correct

---

Ya since all odd numbers are before all even numbers , so odd numbers compulsorily occupy the first 5 places then the even numbers occupy the last 5 places.And each of these group can permute amongst themselves.

So the probability calculated by u is correct.

Answer 1

since  all odd nos.  should appear before all even nos. we can have this way
 

1st   permutation of (1,3,5,7,9)= 5!  and then permutation of (2,4,6,8,10) =5!
and total way we can arrange <1,2. . . . 10> = 10 !
so probability should be = $\frac{5!*5!}{10!}$
u r right