I will form a random 5 digit number using only the integers 1 through 9 and allowing no integer to be used mor

I will form a random 5 digit number using only the integers 1 through 9 and allowing no integer to be used more than once in any 5 digit number. Find the probability that one of the 5 digit number has exactly two of the integers 1, 2, or 3 (and the rest bigger than 3)I will form a random 5 digit number using only the integers 1 through 9 and allowing no integer to be used mor
The total number of ways to form a 5-digit number using digits {1, ... , 9} but with no repeats would be:


9 x 8 x 7 x 6 x 5 = 15,120 total ways





But you need numbers that meet your criteria (exactly two digits from {1,2,3} and the remainder from {4,5,6,7,8,9}):





Let's imagine picking out 2 of the lower digits in the set {1, 2, 3}.


That is 3 choose 2 = 3 ways.





Also imagine picking out 3 of the remaining higher digits in the set {4, 5, 6, 7, 8, 9}.


That is 6 choose 3 = 20 ways





So all together there are 60 ways to pick the digits for the number that meets the requirements.


Then there are 5 factorial (5! = 120 ways) to rearrange those digits.





60 * 120 = 7,200 numbers meeting the restricted criteria.





Probability = 7200 / 15120


= 0.476190476


= 47.62%