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let say there are three elements in a set {1,2,3}.find total #of 4 digit no. which are neither non decreasing nor non increasing.
in Combinatory
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set with 3 numbers and form 4-digit no.... it implies repetition of digits allowed...

then i have a doubt that 1 2 3 3 or 1 2 2 3 is increasing?
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@Shaik. Both are increasing.
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1 Answer

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Number of $4$ digit numbers having their digits formed by using the digits from $\{1,2,3\} = 3^4 = 81 = T$

Let,
Number of $4$ digit numbers having their digits in non-increasing order formed by using the digits from $\{1,2,3\} =A $
Number of $4$ digit numbers having their digits in non-decreasing order formed by using the digits from $\{1,2,3\} =B $

Required solution is $T - (A \cup B).$
$AUB = A+B - A \cap B$

We can form a 4 digit number by selecting $x_1 \ 1's, x_2 \ 2's \ and \ x_3 \ 3's$ such that
$x_1+x_2+x_3=4 \Rightarrow equation 1$    

Then each such solution can be arranged in non-decreasing order.
The number of solutions to equation 1 is given by $\Large\binom{4+3-1}{2}=15 =A$

Similarly,  number of $4$ digit numbers having their digits in non increasing order formed by using the digits from $\{1,2,3\} = B =15$

$A \cap B = 3$  $(1111,2222,3333)$
$(A \cup B)=15+15 - 3= 27$
$T - (A \cup B)=81-27 =54$
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