I have been given a truth table. I want find a boolean expression for it. However , I am not able to come up with one. Is there a specific way to go about , in order to get it done ?
Also, Can this solution be possible using a single ssi 7400 chip ? (max of 4 nand gates)
$$C \wedge (A \vee \neg B)$$
By looking at the TT you can easily see that $\phi(A,B,C) \Rightarrow C$ wich means that $\phi \equiv C \wedge ?$. The rest is simple.
If you only want to come up with some formula, regardless of the number of conjuctives, the TT can be directly translated to a formula in CNF. (It has $2^n$ clauses each with $n$ literals for a total of $(n-1)2^n + 2^n - 1 = n2^n - 1$ gates. This case would result in $23$ gates compared to the two we actually needed)
To arrive at a circuit using 4 NAND gates we remember (using $\sqcap$ for NAND):
$$A \sqcap A = \neg A \\ A \wedge B = \neg (A \sqcap B) = (A\sqcap B)\sqcap(A\sqcap B)\\ A \vee B = \neg A \sqcap \neg B$$
Where the middle one only needs two NAND gates by reusing the result of $A\sqcap B$.
$$\begin{align*} C \wedge (A \vee \neg B) & \stackrel{(3)}\equiv C \wedge (\neg A \sqcap B) \\ & \stackrel{(1)}\equiv C \wedge ((A\sqcap A) \sqcap B) \\ & \stackrel{(2)}\equiv (C\sqcap((A\sqcap A) \sqcap B)) \sqcap (C\sqcap((A\sqcap A)\sqcap B)) \end{align*}$$
So this is possible with 4 NAND gates on your chip.