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The number of $4$ digit numbers strictly greater then $4321$ that can be formed using the digits $0,1,2,3,4,5$ (Repetition of digits is allowed)

What I tried:

If $4$-digit number starts with $5,$ then total possibilities $6\cdot 6 \cdot 6=6^3$.

If $4$-digit number starts with $4,$ then total possibilities $3\cdot 4\cdot 5=60$.

Total ways $=216+60=276$.

But my answer is wrong.

Help me to solve it please.

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  • $\begingroup$ Hello. How did you do the calculation for numbers beggined with $4$? Note that all beggined with $45,44$ are good, but not all beggined with $43$. $\endgroup$
    – Quiet_waters
    Dec 21, 2019 at 14:24

1 Answer 1

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The "fast" answer:

The number of 4-digit numbers greater than 4321 using digits in $\{0,1,...,5\}$ is equal to

$$5555_{(6}-4321_{(6}=1234_{(6}=310$$

where the subscript $5555_{(6}$ means that the number is written in base 6. See here for details on numbering in bases (radices) different from 10, and here for details on how to do arithmetic in other bases.

If you are not familiar with non-base-10 numbering:

You can get the same result by counting:

  • All numbers that start with $5$: in total $6\cdot 6\cdot 6=216$ numbers.
  • All numbers that start with $44$ or $45$: in total $2\cdot 6\cdot 6=72$ numbers.
  • All numbers that start with $433$ or $434$ or $435$: in total $3\cdot 6=18$ numbers.
  • All numbers tat start with $4322$ or $4323$ or $4324$ or $4325$: in total $4$ numbers.

Grand total: $$216+72+18+4=310$$

Note that the actual math done in both methods is the same:

$$1234_{(6}=1\cdot6\cdot 6\cdot 6 + 2\cdot 6\cdot 6 + 3\cdot 6 +4 =310$$

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