Permutations and Combinations
6.0 Restricted Selection
6.0 Restricted Selection
The number of ways in which $r$ objects can be selected from $n$ different objects if $m$ particular objects
- always included, $$ = {\;^{n - m}}{C_{r - m}}$$
- never included, $$ = {\;^{n - m}}{C_r}$$
Question 19. A NSS camp is being arranged in a school for class $11$ students. The strength of the class being $62$.
(A) In how many ways can you select a team of $12$ if inclusion of both the male and female representative is must?
(B) In how many ways can you select a team of $15$ if four students want to be together ?
(C) In how many ways can you select a team of $23$ if exactly five boys are not to be included and six boys have to be included?
Solution: There are in total $62$ students.
(A) Since two have to be included, one has to choose $(10-2)$ students from the rest $(62-2)$ students.
Thus total selections, $$\begin{equation} \begin{aligned} = {\;^{62 - 2}}{C_{10 - 2}} \\ = {\;^{60}}{C_8} \\\end{aligned} \end{equation} $$
(B) Four students want to be together. Either they are not chosen, or together chosen.
When all four are not chosen, total ways of selection are, $$\begin{equation} \begin{aligned} = {\;^{62 - 4}}{C_{15}} \\ = {\;^{58}}{C_{15}} \\\end{aligned} \end{equation} $$
When all four are chosen, total ways of selection, $$\begin{equation} \begin{aligned} = {\;^{62 - 4}}{C_{15 - 4}} \\ = {\;^{58}}{C_{11}} \\\end{aligned} \end{equation} $$
Thus total ways of selecting is, $$ = {\;^{58}}{C_{11}}{ + ^{58}}{C_{15}}$$
(C) When $5$ boys are not included, there are $(62-5)$ students.
Out of this six are to be selected.
Thus total ways of selecting is, $$\begin{equation} \begin{aligned} = {\;^{62 - 5 - 6}}{C_{23 - 6}} \\ = {\;^{51}}{C_{17}} \\\end{aligned} \end{equation} $$
