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How many four-digit natural numbers not exceeding the number $4321$ can be formed using the digits $1, 2, 3, 4$ if repetition is allowed?

This is the question and I am solving it like this:

Total no's can be formed - no's not allowed $4 \cdot 4 \cdot 4 \cdot 4-11=256-11 =245$, but the answer is $229$. What is wrong in this method?

And my second method is this: Total no. = no's digit do not repeat + no's in which $2$ digits are same + no's in which $3$ digit are same + no's in which all digit are same - no's not allowed

No repeat$=4!$

2 no's same$= 144$

3 same $=48$

4 same $=4$

Sum $=220-11=209$

What is wrong in both methods?

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  • $\begingroup$ Here is a similar question asked here math.stackexchange.com/questions/3483656/… $\endgroup$
    – Asher2211
    Jan 24, 2022 at 11:14
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    $\begingroup$ Welcome to MSE. A question should be written in such a way that it can be understood even by someone who did not read the title. $\endgroup$ Jan 24, 2022 at 11:15
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    $\begingroup$ It is hard to discern what you did wrong since you have not explained how you obtained the $11$ numbers which are not allowed, the number with two identical digits, or the numbers with three identical digits. $\endgroup$ Jan 24, 2022 at 11:18

1 Answer 1

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There are actually $27$ numbers larger than $4321$. They are:

Numbers of the form $44\square\square$: There are $1 \cdot 1 \cdot 4 \cdot 4 = 16$ such numbers.

Numbers of the form $433\square$ or $434\square$: There are $1 \cdot 1 \cdot 2 \cdot 4 = 8$ such numbers.

Numbers of the form $432\square$: There are $3$ such numbers, namely $4322, 4323, 4324$.

Hence, you should have $4^4 - (16 + 8 + 3) = 256 - 27 = 229$ in your first method.

In your second method, your forgot to count numbers in which two digits each appear twice, which is why your four cases do not add up to $4^4 = 256$. There are $\binom{4}{2}$ ways to select the two digits which each appear twice and $\binom{4}{2}$ ways to select two positions for the smaller of those digits. Hence, there are $$\binom{4}{2}\binom{4}{2} = 36$$ numbers which each appear twice.

Thus, you should have a total of $24 + 144 + 36 + 48 + 4 = 256$ numbers, of which $27$ are too large, again giving $256 - 27 = 229$ admissible numbers.

A direct count: We wish to count four-digit numbers formed using the digits $1, 2, 3, 4$ which are at most $4321$ when repetition of digits is permitted.

Numbers of the form $1\square\square\square$, $2\square\square\square$, or $3\square\square\square$: $3 \cdot 4 \cdot 4 \cdot 4 = 192$

Numbers of the form $41\square\square$ or $42\square\square$: $1 \cdot 2 \cdot 4 \cdot 4 = 32$

Numbers of the form $431\square$: $4$

The number $4321$: $1$

Hence, the numbers of admissible numbers is $192 + 32 + 4 + 1 = 229$.

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