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?
Answer
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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.
Complete step-by-step answer:
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.
Hence, the answer is 35.
Note:
There is a possibility that one may commit a mistake while finding the number of ways in which the student can choose 3 courses from the non - compulsory. It is possible that he/she chose these 3 courses from 9 courses instead of the 7 courses. But since 2 of these 9 courses are compulsory, we have to choose the 3 non-compulsory courses from the 7 remaining courses.
Complete step-by-step answer:
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.
Hence, the answer is 35.
Note:
There is a possibility that one may commit a mistake while finding the number of ways in which the student can choose 3 courses from the non - compulsory. It is possible that he/she chose these 3 courses from 9 courses instead of the 7 courses. But since 2 of these 9 courses are compulsory, we have to choose the 3 non-compulsory courses from the 7 remaining courses.
Last updated date: 21st Sep 2023
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