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? 
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?

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 $192 + 32 + 4 + 1 = 229$.
