Question

There are nine students (5 boys and 4 girls) in the class. In how many ways:
(i) One student (either girl or boy) can be selected to represent the class.
(ii) A team of two students (one girl and one boy) can be selected.
(iii) Two medals can be distributed (no one gets both).
(iv) One prize for Maths, two prizes for Physics and three prizes for Chemistry can be distributed (No student can get more than one prize in the same subject and prizes are distinct).

Answer 1

Hint: We will be using the concepts of permutation and combination to solve the problem. We will be using the fact that among n items r items can be elected in $^{n}{{C}_{r}}$ ways.

Complete step-by-step solution -
Now, we have been given that there are a total of nine students in a class.
The number of girls G=4.
The number of boys B=5.
Now in (i) part we have to find the ways in which one student either boy or girl, can be selected to represent the class.
Now, we know that the ways in which r items can be selected among n is $^{n}{{C}_{r}}$.Therefore, we have ways in which one student is selected in $^{9}{{C}_{1}}=9$ .
Now in (ii) part we have to find the ways in which a team of two students (one girl and one boy) can be selected. So,
No. of ways = No. of ways to select a girl × No. of ways to select a boy
Now, we have to select 1 girl among 4 and 1 boy among 5 which we know can done as
$^{4}{{C}_{1}}$ = No. of ways of selecting a girl.
$^{5}{{C}_{1}}$ = No. of ways of selecting a boy.
Total ways ${{=}^{4}}{{C}_{1}}{{\times }^{5}}{{C}_{1}}$
$\begin{aligned} & =4\times 5 \\\ & =20 \\\ \end{aligned}$
Now, in (iii) part, we have to distribute medals such that no one gets the both.
So, we can distribute a medal to any one of 9 students and the other medal to only 8 students as we have to exclude the one who has already received a medal. Therefore,
The way to distribute two medals ${{=}^{9}}{{C}_{1}}{{\times }^{8}}{{C}_{1}}$
$\begin{aligned} & =9\times 8 \\\ & =72 \\\ \end{aligned}$
Now in (iv) part we have to distribute one prize for Math, two prizes for Physics and three prizes for Chemistry. We can do this by distributing the prizes for mathematics to any of 9 students for Physics to any two of 9 students and Chemistry to any of three of 9 students.
Therefore, we have total ways
${{=}^{9}}{{C}_{1}}\times \left( ^{9}{{C}_{2}}\times 2! \right)\times \left( ^{9}{{C}_{3}}\times 3! \right)$
Now we know that $^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$
$\begin{aligned} & =9\times \left( \dfrac{9!}{7!\times 2!}\times 2! \right)\times \left( \dfrac{9!}{6!\times 3!}\times 3! \right) \\\ & =9\times \left( 8\times 9 \right)\times \left( 7\times 8\times 9 \right) \\\ & =326592 \\\ \end{aligned}$

Note: To solve these types of question one must have a basic understanding of permutation and combination one must also know that $^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$ ,
$^{n}{{P}_{r}}=\dfrac{n!}{\left( n-r \right)!}$ ,
$^{n}{{C}_{1}}=n$.
Here we should have knowledge about the differences between permutation and combination. There is a need to be aware about the multiplication theorem and addition theorem of combination.