Permutations and Combinations
7.0 Restricted Arrangement
7.0 Restricted Arrangement
The number of arrangements of $n$ objects taken $r$ at a time, such that $m$ objects,
- always included, $$ = {\;^{n - m}}{C_{r - m}}r!$$
- never included, $$ = {\;^{n - m}}{C_r}r!$$
Question 20. There are $12$ students in a committee. Find the possible number of ways in which $6$ students can be selected if,
(A) Two of them must be included. Also find the number of ways in which they can be arranged.
(B) Two are not to be included.
Solution:
(A) Two are always included. Then the possible ways of selection are, $$\begin{equation} \begin{aligned} = {\;^{12 - 2}}{C_{6 - 2}} \\ = {\;^{10}}{C_4} \\\end{aligned} \end{equation} $$
The number of ways of arranging is, $$ = {\;^{10}}{C_4} \times 6!$$
(B) Two are excluded. Then the possible ways of selection are, $$\begin{equation} \begin{aligned} = {\;^{12 - 2}}{C_6} \\ = {\;^{10}}{C_6} \\\end{aligned} \end{equation} $$
Note:
1. $^n{C_r}{ = ^n}{C_{n - r}}$
2. If $^n{C_r} = {\;^n}{C_k}$, then $k = n - r = r$
3. $^n{C_r}{ + ^n}{C_{r - 1}} = {\;^{n + 1}}{C_r}$
4. $^n{C_r} = \;\frac{{{n^{n - 1}}{C_{r - 1}}}}{r}$
5. $\frac{{{}^n{C_r}}}{{{}^n{C_{r - 1}}}} = \frac{{n - r + 1}}{r}$
