Permutations and Combinations
11.0 Possible Selection from $n$ Objects having Distinct and Identical Objects
11.0 Possible Selection from $n$ Objects having Distinct and Identical Objects
Let $p$ be alike, $q$ be of one kind, and so on, and $k$ be each distinct.
Ways of selecting atleast one from $k$ distinct objects, $$ = {2^k} - 1$$
Ways of selecting none from $k$ distinct object $=1$
Thus ways of selecting none or one distinct objects is, $$ = {2^k}$$
Now, ways of selecting none or one from set of ${n-k}$ objects, where $p$ are alike, $q$ are alike and so on is given as product of ways of selecting none or atleast one from each kind. $$ = [(p + 1)(q + 1)(r + 1)....]$$
Ways of selecting none or atleast one from given $n$ objects is, $$ = [(p + 1)(q + 1)(r + 1)....]{.2^k}$$
Ways of selecting none from $n=1$
Thus ways of selecting atleast one from $n$ objects which has both identical and distinct objects = Ways of selecting none or atleast one - Ways of selecting none
$$ = [(p + 1)(q + 1)(r + 1)....]{.2^k} - 1$$
Question 25. There are $37$ books in a shelf. If five are of one kind, eight of one kind, three of one kind, and eleven of one kind, four of the other and the rest are unique. Find the number of ways of selecting,
(A) atleast one,
(B) none?
Solution:
(A) Given that there are,
$5$ books of say kind $a$,
$8$ books of kind $b$,
$3$ of kind $c$,
$11$ of kind $d$ and
$4$ of kind $e$.
The rest are unique which is equal to 6.
The total ways of selecting atleast one, $$\begin{equation} \begin{aligned} = [(8 + 1)(5 + 1)(3 + 1)(11 + 1)(4 + 1)]{.2^6} - 1 \\ = [(9)(6)(4)(12)(5)]{.2^6} - 1 \\ = 829439 \\\end{aligned} \end{equation} $$
(B) Ways of selecting none is equal to $1$.
Question 26. In a library there are $n$ different books. One can borrow one or more books. If the number of ways in which book can be drawn is $16,383$. Find the total number of books available. What happens to the total number of selections, if $n \over 2$ books are identical?
Solution: Ways of selecting atleast one book,$$\begin{equation} \begin{aligned} = {2^n} - 1 \\ = 16,383 \\\end{aligned} \end{equation} $$
Thus, $${2^n} = 16,384$$
Taking log on both sides, $$n\log 2 = \log (16,384)$$
So, $$n(0.3010) = 4.2144$$
The total number of books is $14$.
If $\left( {\frac{n}{2}} \right)$ are identical. Then there are $7$ identical and seven unique books.
Thus the ways of selecting atleast one, $$\begin{equation} \begin{aligned} = [7 + 1]{2^7} - 1 \\ = 8 \times 128 - 1 \\ = 1023 \\\end{aligned} \end{equation} $$
