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!
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