If $A = \left\{1,2,3,4\right\}$ and $B = \left\{1,2,3,4,5\right\},$ Then
$(a)\; :: $ Total no. of mapping from from $A\rightarrow B$ such that $f(i)<f(j)\;\forall \; i<j, $ is
$(b)\;\;::$ Total no. of mapping from from $A\rightarrow B$ such that $f(i)\leq f(j)\;\forall \; i<j, $ is
$\bf{My\; Trial\; solution::}$ for $(a)::$ Given $f(i)<f(j)\;\; \forall \; i<j$ and $i,j\in \{1,2,3,4\}$
and $f(i)\;,f(j)\in \left\{1,2,3,4,5\right\}$
Now if $i=1\;,$ Then value of $j=1,2,3$. So function as $f(1)<f(2)\;,f(1)<f(3)\;,f(1)<f(4)$
So no. of ways is $\displaystyle \binom{5}{2}\times 3=30$
Similarly if $i=2\;,$ Then value of $j=2,3$. So function as $f(2)<f(3)\;,f(2)<f(4)$
So no. of ways is $\displaystyle \binom{5}{2}\times 2=20$
Similarly if $i=3\;,$ Then $j=4.$ So function as $f(3)<f(4)$
So no. of ways is $\displaystyle \binom{5}{2}\times 1=10$
So Total no. of function is $ = 30+20+10 =60$
But Solution given as $ = 5$
I did not understand how can we get answer is $ = 5$
help me
Thanks