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Hi Ralph, You have a six digit code where the digits can be 1, 2, 3 or 4. The first digit can be 1, 2, 3 or 4 and regardless of your choice for the first digit the second digit can also be either 1, 2, 3 or 4. Thus there are $4 \times 4$ possible choices for the first two digits. The choices are 11, 12, 13, 14, 21, 22, 23, 24, 31, 32, 33, 34, 41, 42, 43 and 44. Now the third digit. Regardless of your choice for the first two digits the third digit can also be either 1, 2, 3 or 4. Hence there are $4 \times 4 \times 4$ choices for the first three digits. They are 111, 112, 113 and so on. I hope this helps, | |||||||||||||||
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