Let A=1,2,3,4,5,6.

Question

Let $A={1,2,3,4,5,6}$. The number of one-one functions $f:A	o A$ such that $f(1)\ge 3,f(3)\le 4$ and $f(2)+f(3)=5$, is $____$ .

Answer: 72

Accepted Answer

We need the number of one-one functions $f:A	o A$, where

$A={1,2,3,4,5,6}$

and the conditions are:

$f(1)\ge 3$

$f(3)\le 4$

$f(2)+f(3)=5$

Since $f$ is one-one and domain and codomain are the same finite set, such a function is a permutation of $A$.

Now let us count carefully.

From

$f(2)+f(3)=5$

and since values are from $A={1,2,3,4,5,6}$, the possible pairs are:

$(f(2),f(3))=(1,4),(2,3),(3,2),(4,1)$

Also $f(3)\le 4$, which is already satisfied by all these cases.

Now we use the condition that $f$ is one-one, so all function values must be distinct.

We also have:

$f(1)\ge 3\, \implies \, f(1)\in {3,4,5,6}$

But $f(1)$ must be different from both $f(2)$ and $f(3)$.

Case 1: $(f(2),f(3))=(1,4)$
Then $f(1)\in {3,4,5,6}$, but it cannot be $4$ because function is one-one.

So,

$f(1)\in {3,5,6}$

Number of choices for $f(1)$ = $3$.

After fixing $f(1),f(2),f(3)$, the remaining $3$ elements of the domain can be mapped to the remaining $3$ elements of the codomain in

$3!=6$

ways.

So total in this case:

$3	imes 6=18$

Case 2: $(f(2),f(3))=(2,3)$
Then $f(1)\in {3,4,5,6}$, but it cannot be $3$.

So,

$f(1)\in {4,5,6}$

Number of choices for $f(1)$ = $3$.

Remaining mappings can be done in

$3!=6$

ways.

Total in this case:

$3	imes 6=18$

Case 3: $(f(2),f(3))=(3,2)$
Then $f(1)\in {3,4,5,6}$, but it cannot be $3$.

So again number of choices for $f(1)$ = $3$.

Remaining mappings:

$3!=6$

Total:

$3	imes 6=18$

Case 4: $(f(2),f(3))=(4,1)$
Then $f(1)\in {3,4,5,6}$, but it cannot be $4$.

So number of choices for $f(1)$ = $3$.

Remaining mappings:

$3!=6$

Total:

$3	imes 6=18$

Total number of functions
$18+18+18+18=72$

Hence, the required number of one-one functions is

$72$

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