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-In this particular type of question we need to use the method of combination step by step. The selection of n people from a group of r people is given by $^n{C_r}$ . If the selection is from different groups , just multiply their number of ways of individual selection .

Complete step-by-step answer:
(i) There are 5+4=9 students, of whom one needs to be selected.
This is given by :
$^9{C_1} = \dfrac{{9!}}{{\left( {9 - 1} \right)!1!}} = 9ways$ ( since $^n{C_r} = \dfrac{{n!}}{{\left( {n - r} \right)!r!}}$ )

(ii) One girl has to be selected from a group of 4, while one boy should be selected from among 5, in order to form the team. This is given by:
$^4{C_1}{ \times ^5}{C_1} = 4 \times 5 = 20ways$
(iii) In order to distribute 2 medals among 9 students, we can first select 2 students ( in $^9{C_2}ways$) and determine the ways in which the 2 medals can be distributed among them ( in 2! ways) .
This is given by
$^9{C_2} \times 2! = 72ways$
(iv) One prize for Math can be distributed in
$^9{C_1}ways$ .
Two prizes for Physics can be distributed in $^9{C_2} \times 2!ways$
and
Three prizes for Chemistry can be distributed in
$^9{C_3} \times 3!ways$
such that no student gets more than one prize in the same subject.
Therefore , Total number of ways
$\begin{gathered} { = ^9}{C_1} \times 1!{ \times ^9}{C_2} \times 2!{ \times ^9}{C_3} \times 3! \\\ = {9^3} \times {8^2} \times 7 \times 2 = 653184ways \\\ \end{gathered}$

Note-It is important to understand that for selection purposes we use combinations and for arranging purposes we use permutations . Remember that in these types of questions the basic formula for combination needs to be recalled . Also note that if the order doesn't matter then we have a combination, if the order does matter then we have a permutation. One could say that a permutation is an ordered combination .