i missed one 2 2 y 2 x

schumi_gmat wrote:I think i got my answer -

Here are the 5P5/3! = 20 combinations

2 2 2 x y
2 2 x 2 y
2 2 x y 2
2 x 2 y 2
2 x y 2 2
x 2 y 2 2
x y 2 2 2
y x 2 2 2
y 2 x 2 2
2 y x 2 2
2 y 2 x 2
2 2 y x 2
2 2 2 y x
x 2 2 2 y
y 2 2 2 x
2 x 2 2 y
2 y 2 2 x
x 2 2 y 2
y 2 2 x 2

Let me know if this is correct way of doing it. Hence now the answer is 81 * 20/ 10^5

At first I though of offering this as an alternative explanation, but then I realized that there's a serious problem.

We can only use 5P5/3! if the only duplicate digits are the 2s. However, it's also possible that the other two digits could be the same, messing up our calculation.

For example, if our code contains 3 "2"s (2a, 2b and 2c) and 2 "4"s (4a and 4b), you'd have counted

(2a)(2b)(2c)(4a)(4b)

and

(2a)(2b)(2c)(4b)(4a)

as two different codes, even though they're identical.

Since some of the codes have 3 "2"s and 2 unique digits and others have 3 "2"s and 2 identical digits, there's no easy way to apply the permutations formula to this question.

for taking 3 2's we had 1*1*1 =1 ways
and for the rest of 2 digits we had 9P2 ways

and the total possible ways for choosing 5 digits are 10P5

so 9P2/10P5 = 1/420