In how many ways can a student choose a program of $5$ courses if $9$ courses are available and $2$ specific courses are compulsory for every student?

Number of compulsory courses $=2$

Now, remaining courses $=n=7$

And, Number of courses to be selected $=r=3$

$\therefore$ Number of ways of selecting $3$ courses out of $7$ courses ${=}^7{C}_3$

$=\frac{7!}{3!(7-3)!}$

$=\frac{7!}{3!4!}$

$=\frac{7\times 6\times 5\times 4!}{3\times 2\times 1\times 4!}$

$=35\text{ways}$

Hence, total number of ways $=35\text{ways}$.