wishmasterdj wrote:
Can someone tell me my mistake here?
I went for ways to choose 3 numbers from 1-9 -> 9C3
Then I tried to fill the 5 digit slots->
1. For the non-repeated number, we have 5 spots
2. For the repeated number, we have 4 and 3 spots (and then divide it by 2! since the order of the same digit wont matter)
3. For the other repeated number, we have 2 and 1 spot (again, divide it by 2! since the order of the same digit wont matter)
so 9C3 * 5 *4*3*2*1 / (2! * 2!) = 2520
Hi
wishmasterdj,
This problem bugged me an awful lot as well, and I was initially stuck on the same approach as yours, so let me give it a shot!
Firstly, let's use the letters "AABBC" to represent the two sets of repeated digits (A & B) as well as the single digit (C).
I believe your approach is best summarized as the following:
1. Find the (unordered) number of possible combinations of 3 digits from a set of 9 digits. \((9C3 = \frac{9!}{3!*6!} = 84)\)
2. Find the number of configurations of the letters AABBC. \((\frac{5!}{2!*2! = 30})\)
This is what you did when filling the 5 slots. So far you've chosen 3 numbers (ex. "123") and you've ordered the letters (ex. "AABBC"), but you also need to consider that the final number will change depending on how you assign the numbers to the letters.
Ex. Assigning ABC = 123 (1 assigned to A, 2 assigned to B, 3 assigned to C) is going to give you the number 11223, while assigning ABC = 321 is going to give you the number 33221.
3. The missing (but also flawed) step: Find the number of ways you can slot the 3 different numbers into the 3 different letters, aka. "how many ways can you order 3 items". \(3! = 6\)
Unfortunately, this approach is flawed, because when you multiply these numbers together you end up with \(84*30*6 = 15120\), twice the actual amount.
This issue arises because A and B share the
same properties (both digits are represented twice).
Example of this issue:
Case 1Chose three numbers: 123
Order the 5 digits : AABBC
Assign the 3 digits : ABC = 123
Result = 11223
Case 2Chose three numbers: 123
Order the 5 digits :
CCBBAAssign the 3 digits :
ABC = 321Result = 11223
Different configurations, same result.To tackle this issue, use the method mentioned by
TestPrepUnlimited:
1. Select the
ordered digits, this step both selects the 3 digits and assigns them to the three letters.
Use 9C2 to treat A and B as identical, you're doing this because A and B have the
same properties. \((9C2 = \frac{9!}{2!*7!} = 36)\)
That leaves you with 7 possible digits for C, leading to the equation: \(9C2 * 7 = \frac{9!}{2!*7!} * 7 = 36 * 7\)
2. Find the number of configurations of the letters AABBC. \((\frac{5!}{2!*2!} = 30)\)
Final calculations = \(36*7*30 = 7560\)
I hope this made sense to you, I consider myself quite decent at combinatorics, but this one bugged me for a while.