Question

In a book stall, there are 4 copies of one book, 5 copies of another and single copy of 5 other different books. Then the number of ways that a person can purchase the books is
(a) 340
(b) 535
(c) 959
(d) 1002

Answer 1

First, we need to find out the total number of possibilities of purchasing both the book one and book two, then find the possibility of purchasing 5 different books with a single copy. Use the formula for ‘m’ and ‘n’ alike books and ‘k’ different books, \left\\{ \left[ \left( \text{m + 1} \right)\left( \text{n + 1} \right) \right]{{2}^{\text{k}}} \right\\}-1 to find the required result.

Complete step by step answer:
Here, in a book stall we have 4 copies of one book, 5 copies of another book and 5 single copies of different books. We will consider if duplicate books have also been purchased.
For Book one:
There is a possibility of purchasing the book one is 0, 1, 2, 3, 4 copies.
Therefore, if m = 4 copies, the possibility of purchasing the book one = (m + 1).
Now, for Book two:
There are 5 copies of book two, hence the possibility of purchasing the book two is 0, 1, 2, 3, 4, 5 copies.
Therefore, if n = 5 copies, then the possibility of purchasing the book two = (n + 1).
Now for the remaining books, since they are all single copies of 5 different books, we can say the possibility of purchasing these books = ${{2}^{\text{k}}}$, where k = total number of distinct books.
The person will purchase from the alike copies of (m + 1) Book one, (n + 1) Book two and distinct 5 books of single copy =\left\\{ \left[ \left( \text{m + 1} \right)\left( \text{n + 1} \right) \right]{{2}^{\text{k}}} \right\\}-1.
Number of ways a person can purchase $=\left[ \left( 4+1 \right)\left( 5+1 \right){{2}^{5}}-1 \right]$
$\begin{aligned} & =\left[ \left( 5 \right)\left( 6 \right)\left( 32 \right)-1 \right] \\\ & =\left[ \left( 30 \right)\left( 32 \right)-1 \right] \\\ & =\left[ 960-1 \right] \\\ & =959 \\\ \end{aligned}$

Therefore, the total number of ways that the person can buy the books from the book stall is 959.

Note: In this question, the formula \left\\{ \left[ \left( \text{m + 1} \right)\left( \text{n + 1} \right) \right]{{2}^{\text{k}}} \right\\}-1, has ‘– 1’ to ensure that the possibility is not zero, and the person does not leave the book stall empty handed. Permutations is the calculation of the number of ways a particular set can be arranged but Combinations is the calculation of the number of ways a particular set can be selected.