Question

In a packet there are m different books, n different pens and p different pencils. The number of selections of at least one article of each type from the packet is
(a) \[{{2}^{m+p+n}}-1\]
(b) $ \left( m+1 \right)\left( n+1 \right)\left( p+1 \right)-1 $
(c) \[{{2}^{m+p+n}}\]
(d) $ \left( {{2}^{m}}-1 \right)\left( {{2}^{n}}-1 \right)\left( {{2}^{p}}-1 \right) $

Answer 1

Hint: Here, we will use the fundamental principle of counting to find the required answer. We will first find the total number of ways in which each of these items can be selected and then we will find their product. There are two possible outcomes for each of the objects, that is selection or rejection. There is also a possibility where all the objects are rejected so it will be subtracted from the total number of ways in each case. We have my books and we have to select at least 1 article of each type. So, for books, we can write the number of ways of selecting all books as m times, which is . Considering rejection, and subtracting 1 from it, we will get the total number of ways for selecting books. We will do the same thing for pens and pencils also. Then multiplying all the results, we will get our answer.

Complete step-by-step answer:
So, we are given with m different books, n different pens and p different pencils. Now each of the object, i.e. book, pen or pencil can either be selected or not selected.
So, each of these objects will have two outcomes, i.e., selection or rejection.
Therefore, the number of ways of selecting all the books will be = $ 2\times 2\times 2\times 2\times .......m $ times = \[{{2}^{m}}\]
Now, in all these ways, there is one way in which all the books are rejected, which is not valid. Hence, it needs to be subtracted.
So, total number of ways of selecting books = $ \left( {{2}^{m}}-1 \right) $
Number of ways of selecting all the pens = $ 2\times 2\times 2\times 2.......n $ times = $ {{2}^{n}} $
Here also the one way in which all the pens are rejected is to be subtracted. Hence, number of ways of selecting pens = $ \left( {{2}^{n}}-1 \right) $
Number of ways of selecting pencils = $ 2\times 2\times ......p $ times = $ {{2}^{p}} $
Here also the one way where all the pencils are rejected is to be subtracted. Hence, number of ways of selecting pencils = $ \left( {{2}^{p}}-1 \right) $
Therefore, by the fundamental principle of counting, total number of ways of selecting at least one article of each type from the packet is:
 $ =\left( {{2}^{m}}-1 \right)\left( {{2}^{n}}-1 \right)\left( {{2}^{p}}-1 \right) $
Hence, option (d) is the correct answer.

Note: Students should keep in mind that there is always one way where all the objects are rejected. So, while finding the total number of ways, we always have to subtract 1 from the total ways obtained. Here, students could do a mistake that they subtract 1 at last, that is after finding the final answer but it would be wrong because we are given that we have to find number of ways of selecting one article of each type and hence 1 has to be subtracted from the total number of ways of selecting each type of object.