**Question:**

** In an examination, a student has to answer 4 questions out of 5 questions; questions 1 and 2 are however compulsory. Determine the number of ways in which the student can make the choice.**

**Solution:**

We know that,

nCr

$=\frac{\mathrm{n} !}{\mathrm{r} !(\mathrm{n}-\mathrm{r}) !}$

According to the question,

Total number of questions =5

Number of questions to be answered =4

Compulsory questions are question number 1 and 2

Hence, the number of ways in which the student can make the choice = 3C2

3C2 = 3!/(2!1!) =3 ways