Aidan's method

**can** be used here but you need to add one generalization.

To begin with, there are 9 As out of 18 cards: probability the first card drawn is an A is 9/18. Now there are only 8 As left out of 17: probability the second card is drawn is an A is 8/17. Now there are 9 Bs out of 16 cards so the probability the third card is drawn is 9/16. Now there are 8 Bs out of 15 cards so the probability the fourth card is drawn is 8/15. Now there are 7 Bs out of 14 cards so the probability the fifth card is drawn is 7/14. Now there are 6 Bs out of 13 cards so the probability the sixth card is 6/13.

The probability of 2As and 4Bs

**in that order** is (9/18)(8/17)(9/16)(8/15)(7/14)(6/13).

Now the important point is that if we looked at 2As and 4Bs in a different order, we would get different fractions but the denominators would be exactly the same as in that first calculation whikle the numerators would be the same numbers in a different order! In other words, the product of the fractions, and so the probability, would be exactly the same. All we need to do to allow for all the different orders is to multiply that by the number of ways 2 As and 4Bs can be order: that is [tex]\frac{6!}{2! 4!}[/itex], the binomial coefficient. The probability of 2 As and 4 Bs

**in any order** is

Notice that the denominators of all the fractions, 17 down to 13, all multiplied together, is the same as 17!/12!. The numerator involves 9(8)= 9!/7! and 9(8)(7)(6)= 9!/5!. Putting those together should give the same answer Soroban got, corrected.