How many 4 digit nos that do not contain the digits 3 or 6 are there ?
Actual solution states that the first digit has 7 possibilities which excludes 0,3,& 6. And the other 3 digits have 8 possibilities. So the total possibilities are 7 * 8 * 8 * 8==3584. This solution is perfect but why can't the other 3 digits have deceasing possibilities. The first digit has 7 possibilities. The second digit has 8 possibilities. So after selecting the second digit the third digit can have only 7 possibilities. similarly 4th digit could have 6 possibilities.
Why can't this be possible ? Could someone please clarify ?
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