Arrange numbers with multiple constraints you have the following string of numbers: $10,3,7,9,11$ and they are in their respective positions from left to right, so 10 is 1st, 3 is 2nd, 7 is third, 9 is fourth and 11 is 5th.
Now, how many ways are there to arrange these 5 numbers, considering that 10 must always be at the left of 7 (not necessarily next to each other) and 3 must always be at the left of 9 (not necessarily next to each other)?

my approach:
11 can stay in any position so 5 ways to arrange it.
10 can stay in 3 of the remaining 4 position, because at its right there must be the 7.
3 can stay in only 2 of the remaining 3 because at his right there must be the 9.
9 and 7 are obliged in their positions.
Number of ways: $5*3*2 = 30 $
What do you think?
 A: Answer is correct , for another approach:
The probability for  $10$ must always be at the left of $7$ (not necessarily next to each other) is $1/2$ and  the probability for   $3$ must always be at the left of $9$ (not necessarily next to each other) is $1/2$
So , $1/2 \times 1/2 = 1/4$ gives us the probability in which $10$ must always be at the left of $7$ (not necessarily next to each other) and $3$ must always be at the left of $9$ (not necessarily next to each other) in all arrangements of these $5$ numbers.
Hence ,it can be concluded that $1/4$ of the arragenments satisfy it , so $1/4 \times 5! =30$
A: After placing $11$, you mention that $10$ can be put in three of the remaining four positions, and $3$ can be put in only two of the remaining three positions. You finally state that $9$ and $7$ are obliged in their positions. But take these two cases -
11 10 3 9 7 and 11 10 3 7 9
After placing $11, 10$ and $3$, we still have two choices for $7$ and $9$.
So while the overall count is correct, the reasoning is not.
One of my approaches would be to place $11$ first and then as you mentioned there are three choices for $10$ but we observe that for every position of $10$ starting from left, the number of choices for $7$ reduces by one. So if $10$ takes the leftmost position after $11$ is placed, there are three choices for $7$; for the next position of $10$, there are two choices for $7$ and so on. Once we place $11, 10$ and $7$, positions of $3$ and $9$ are fixed.
So that gives $5 \cdot (3 + 2 + 1) = 30$ valid arrangements.
