Question

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

Hint: Let us assume that we have n different things in total. From the concept of permutations and combinations, the number of ways in which we can choose r things from these n things is given by the formula ${}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$ . Using this formula, we can solve this question.

Before proceeding with the question, we must know the formula that will be required to solve this question.
From permutations and combinations, if we are given n different things from which we are required to select r things, then the number of ways in which we can do so is given by the formula,
${}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$ . . . . . . . . . . . (1)

In this question, a student has to choose a programme of 5 courses if 9 courses are available and 2 specific courses are compulsory for every student. We are required to find the number of ways in which he can choose the courses.

Since 2 courses are compulsory, he/she has to choose both of them. Using formula (1), the number of ways in which this can be done is equal to,
\begin{align} & {}^{2}{{C}_{2}}=\dfrac{2!}{2!\left( 2-2 \right)!} \\ & \Rightarrow {}^{2}{{C}_{2}}=\dfrac{2!}{2!\left( 0 \right)!} \\ & \Rightarrow {}^{2}{{C}_{2}}=1 \\ \end{align}

From the remaining 7 courses, the student has to choose 3 courses. Using formula (1), the number of ways in which he can do so is equal to,
\begin{align} & {}^{7}{{C}_{3}}=\dfrac{7!}{3!\left( 7-3 \right)!} \\ & \Rightarrow {}^{7}{{C}_{3}}=\dfrac{7\times 6\times 5\times 4!}{\left( 3\times 2\times 1 \right)\left( 4 \right)!} \\ & \Rightarrow {}^{7}{{C}_{3}}=35 \\ \end{align}

The number of ways in which the student can choose a programme of 5 courses if 9 courses are available and 2 specific courses are compulsory for every student will be given by multiplying the above two obtained numbers and is equal to 35 $\times$ 1.