jasonchung55 wrote:

Hi Sandy, can you explain why we couldn't divide the product of the first two letters by two? Identical letters would count two identical outcomes right? (e.g., AA)

Look at this:

Group 1

AA, AB,............. AZ ----- 26 items

BA, BB, .............BZ ----- 26 items

ZA, ZB, ............. ZZ ---- 26 items

Here total number of combinations possible is \(26*26\)

Similarly, we have 3 digits

Group 2

000, 001, ...... 009 ---- 10 items

010, 011, .......019 ---- 10 items

.

.

.

.

990, 991,992 .....999 ---- 10 items

Here total number of combinations is \(10*10*10\)

Each item in group 1 has the complete group 2 associated with it.

So

AA and

000, 001, ...... 009 ---- 10 items

010, 011, .......019 ---- 10 items

.

.

.

.

990, 991,992 .....999 ---- 10 items

is one part of the set of solutions. Also,

AB and

000, 001, ...... 009 ---- 10 items

010, 011, .......019 ---- 10 items

.

.

.

.

990, 991,992 .....999 ---- 10 items

So on. Thus the total number of combinations is \(26*26*10*10*10\)

If you look at this

AA, AB,............. AZ

BA, BB, .............BZ

CA, CB, CC ..........CZ

ZA, ZB, .............ZZ

Here AA is not counted twice. Hence the answer is correct!

_________________

Sandy

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