# no. of mapping from from $A\rightarrow B$ such that $f(i)<f(j)\;\forall \; i<j,$ is

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

In part (a), you overlooked the fact that $f(i) < f(j)~\forall i < j$. This means that $f(1) < f(2) < f(3) < f(4)$.

Therefore, if $f(1) = 2$, you are forced to conclude that $f(2) = 3$, $f(3) = 4$, and $f(4) = 5$.

If $f(1) = 1$, there are four possibilities. They are

$f(2) = 2$, $f(3) = 3$, $f(4) = 4$

$f(2) = 2$, $f(3) = 3$, $f(4) = 5$

$f(2) = 2$, $f(3) = 4$, $f(4) = 5$

$f(2) = 3$, $f(4) = 4$, $f(4) = 5$

Thus, there are only five functions from $\{1, 2, 3, 4\} \rightarrow \{1, 2, 3, 4, 5\}$ such that $f(i) < f(j)~\forall i < j$.

In part (b), you will have more functions since you must account for functions such as $\{(1, 2), (2, 2), (3, 3), (4, 3)\}$. Start by considering the case $f(1) = 5$.

• Thanks N. F. Taussig. got it. – juantheron Sep 29 '14 at 2:40

One number is left out of the image. once you fix that number there is only one possibility for the function.

• Ohoo Sorry Rene Schipperus, Would you like to explain me in detail, Thanks – juantheron Sep 28 '14 at 16:28

If $\displaystyle A = \left\{1,2,3,4\right\}$ and $\displaystyle \left\{1,2,3,4,5\right\}$ Then Total no. of mapping from from $A\rightarrow B$ such that

$f(i)\leq f(j)\;\forall \; i<j,$ is

My Solution:(Using Hint) We have to find a relation on function as $f(1)\leq f(2)\leq f(3)\leq f(4)\;,$

Where $f(1),f(2),f(3),f(4)\in \left\{1,2,3,4,5\right\}$

$\bullet \;$ If $f(1)<f(2)<f(3)<f(4)\;,$ Then no. of functions$\displaystyle = \binom{5}{4} = 5.$

$\bullet \;$ If $f(1)=f(2)<f(3)<f(4)\;,$ Then no. of functions$\displaystyle = \binom{5}{3} = 10.$

$\bullet \;$ If $f(1)<f(2)=f(3)<f(4)\;,$ Then no. of functions$\displaystyle = \binom{5}{3} = 10.$

$\bullet \;$ If $f(1)<f(2)<f(3)=f(4)\;,$ Then no. of functions$\displaystyle = \binom{5}{3} = 10.$

$\bullet \;$ If $f(1)=f(2)=f(3)<f(4)\;,$ Then no. of functions$\displaystyle = \binom{5}{2} = 10.$

$\bullet \;$ If $f(1)=f(2)<f(3)=f(4)\;,$ Then no. of functions$\displaystyle = \binom{5}{2} = 10.$

$\bullet \;$ If $f(1)<f(2)=f(3)=f(4)\;,$ Then no. of functions$\displaystyle = \binom{5}{2} = 10.$

$\bullet \;$ If $f(1)=f(2)=f(3)=f(4)\;,$ Then no. of functions$\displaystyle = \binom{5}{1} = 5.$

So Total no. of functions is $\displaystyle = 5+10+10+10=10+10+10+5=\boxed{\boxed{70}}$