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.
$${2^{m + n + p}} - 1$$
B.
$$\left( {m + 1} \right)\left( {n + 1} \right)\left( {p + 1} \right) - 1$$
C.
$${2^{m + n + p}}$$
D.
None of these
Answer :
$${2^{m + n + p}} - 1$$
Solution :
The required number of ways
= total number of ways of selecting any number of books from $$m$$ different books, any number of pens from $$n$$ different pens and any number of pencils from $$p$$ different pencils $$- 1$$
$$\eqalign{
& = \left( {^m{C_0} + {\,^m}{C_1} + ..... + {\,^m}{C_m}} \right)\left( {^n{C_0} + {\,^n}{C_1} + ..... + {\,^n}{C_n}} \right) \times \left( {^p{C_0} + {\,^p}{C_1} + ..... + {\,^p}{C_p}} \right) - 1 \cr
& = {2^m} \times {2^n} \times {2^p} - 1. \cr} $$