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
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