I will form a random 3 digit number using only the integers 1 through 9 and allowing no integer to be used mor

I will form a random 3 digit number using only the integers 1 through 9 and allowing no integer to be used more than once in any 3 digit number. Find the probability that one of the 3 digit number has exactly two of the integers 1, 2, or 3 (and the rest bigger than 3).I will form a random 3 digit number using only the integers 1 through 9 and allowing no integer to be used mor
OK. The probability of that is (3/9)(2/8)(6/7)(3C2), or 3/14.I will form a random 3 digit number using only the integers 1 through 9 and allowing no integer to be used mor
The possible combinations that the 3 digit number has exactly 2 of the integers 1, 2, or 3 are 3*2*6=36, the total possible combinations are 9*8*7=504. So 36/504=0.0714285714 or 7%
36 is the numerator. The denominator is the numer of ways of choosing 3 out of 1 through 9, which is 9C3.


9 x 8 x 7/3 x 2 x1=84


36/84 ?