Question 1151702
<br>
The number of 9-digit numbers using the digits 1-9 once each is 9!.<br>
The probability that the 1 precedes the 3 is 1/2.
The probability that the 1 precedes the 4 is 1/2.
The probability that the 2 precedes the 3 is 1/2.
The probability that the 2 precedes the 4 is 1/2.<br>
The number of 9-digit numbers using the digits 1-9 once each, in which the 1 and 2 precede the 3 and 4, is<br>
{{{(9!)*(1/2)^4 = 9!/16 = 22680}}}<br>