# High School Statistics Problem 12 (Awarding B.A.)?

How many 4 digit pins can be made with the numbers "121234"?

Relevance

Taking what Leonard started, I am thinking that the digits can't be reused, but since there are duplicate numbers, his count of 360 possible outcomes includes duplicates:

1234 (using the first 1 and first 2)

1234 (using the first 1 and second 2)

1234 (using the second 1 and first 2)

1234 (using the second 1 and second 2)

It should still count as 1 occurrence instead of 4.

To get rid of the duplicates, since there are 2 1's and 2 2's, you divide the number of possible outcomes by (n!) for each. Since it's 2 each, divide by (2! * 2!) which is 4, so:

360 / 4 = 90 unique 4-digit PINs using the 6 digits shown.

• The actual answer is 102 as I explained in the follow-up version. The digits were all reduced by 1, but it doesn't change the number of PINs. See the link below:

• Anonymous
2 months ago

Combinations with repetition

1 2 1 2 3 4 - 6 elements where 1 and 2 are repeated

CR (n, r) = (n + r - 1)! / (r! (n-1)!)

n = 6, r = 4

CR (6, 4) = (6 + 4 - 1)! / (4!.* 5!)

CR (6, 4) = 9! / (5(4!)²)

CR (6, 4) = 126

• Your question is very unclear - can you reuse the digits? e.g, 1111?

if yes, then there are 4 unique digits, and there are 4 choices for each of the places for the 4-digit pin, so 4^4=256.

if you are supposed to think of your "121234" as 6 choices which cant be reused, then there are 6 choices for the 1st digit, 5 choices for the 2nd, ...., and then there would be 6x5x4x3=360 possibilities.

Or the question could be asking something else - who knows.

• Your Friendly Neighborhood Asian
Lv 5
2 months agoReport

My solution is (6P4) / (2!*2!) to account for the repetition