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How many ways can an adviser choose $3$ students from a class of $10$?

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Last updated date: 22nd Feb 2024
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IVSAT 2024
Answer
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Hint: In this problem we need to calculate the number of ways to select the students from a class. First, we will consider that number of students as $n$ and the number of students to choose as $r$. In permutation and combinations, we have the formula for number of ways to choose $r$ things from $n$ is $C\left( n,r \right)={}^{n}{{C}_{r}}$. Now we will use this formula and substitute all the formulas we have in the problem. After that we will simplify the calculated value by using the formula ${}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$. Here we will use the formulas $n!=n\left( n-1 \right)\left( n-2 \right)\left( n-3 \right).....3\times 2\times 1$ and $n!=n\left( n-1 \right)!=n\left( n-1 \right)\left( n-2 \right)!=....$ to simplify the above fraction and to get the required result.

Complete step by step solution:
Given that,
There are $10$ students in the class.
We are assuming that the number of students is $n=10$.
From the class the advisor has to pick $3$ students.
We are assuming that the number of students to choose is $r=3$.
We know that the number of ways to choose $r$ things from $n$ is $C\left( n,r \right)={}^{n}{{C}_{r}}$. Hence, we can write
$\Rightarrow C\left( 10,3 \right)={}^{10}{{C}_{3}}$
We have the formula ${}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$. Applying this formula in the above equation, then we can write
$\begin{align}
  & \Rightarrow C\left( 10,3 \right)=\dfrac{10!}{3!\times \left( 10-3 \right)!} \\
 & \Rightarrow C\left( 10,3 \right)=\dfrac{10!}{3!\times 7!} \\
\end{align}$
From the formula $n!=n\left( n-1 \right)!=n\left( n-1 \right)\left( n-2 \right)!=....$, we are going to write the value $10!$ as $10!=10\left( 10-1 \right)\left( 10-2 \right)\left( 10-3 \right)!=10\times 9\times 8\times 7!$ in the above equation, then we will get
$\Rightarrow C\left( 10,3 \right)=\dfrac{10\times 9\times 8\times 7!}{3!\times 7!}$
Cancelling the $7!$ which is in both numerator and denominator, then we will have
$\Rightarrow C\left( 10,3 \right)=\dfrac{720}{3!}$
Using the formula $n!=n\left( n-1 \right)\left( n-2 \right)\left( n-3 \right).....3\times 2\times 1$ in the above equation, then we will get
$\begin{align}
  & \Rightarrow C\left( 10,3 \right)=\dfrac{720}{3\times 2\times 1} \\
 & \Rightarrow C\left( 10,3 \right)=120 \\
\end{align}$
Hence the advisor can choose $3$ students from a class of $10$ in $120$ ways.

Note:
In this problem they don’t have mentioned whether the students are for the same task or different task. So, we assumed that the students are for the same task and used combinations. If they have mentioned that the students are for different tasks then we need to use the permutations and we have to calculate the value of $P\left( n,r \right)=P\left( 10,3 \right)$.
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