Question

Let A = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}. Then the number of subsets of A containing exactly two elements is
(a) 20
(b) 40
(c) 45
(d) 90

Answer 1

Hint: Here, we will find the number of subsets of using the formula for the combination of K objects chosen from a total of n objects at a time. It is given as: $^{n}{{C}_{k}}=\dfrac{n!}{k!\times \left( n-k \right)!}$

Complete step-by-step answer:

A set is a well defined collection of objects. Each object in a set is called an element of the set. A set is usually denoted by a capital letter, such as A, B or C. An element of a set is usually denoted by a small letter such as x, y or z. A set that contains no elements is called a null or empty set.

The various ways in which objects from a set may be selected, generally without replacement to form a subset is known as combination when the order of selection is not a factor. The total number of ways of selecting or choosing K objects from a set of n objects at a time is given as:

$^{n}{{C}_{k}}=\dfrac{n!}{k!\times \left( n-k \right)!}............\left( 1 \right)$

Here, the set given to us is A = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}.

We have to make subsets of A containing exactly two elements.

Number of ways of choosing 2 elements from a total of 10 elements from the given set A will be given according to equation (1) as:

$\begin{align}

  & ^{10}{{C}_{2}}=\dfrac{10!}{2!\times \left( 10-2 \right)!} \\

 & {{\Rightarrow }^{10}}{{C}_{2}}=\dfrac{10!}{2!\times 8!} \\

 & {{\Rightarrow }^{10}}{{C}_{2}}=\dfrac{10\times 9\times 8!}{2!\times 8!} \\

 & {{\Rightarrow }^{10}}{{C}_{2}}=\dfrac{10\times 9}{2}=45 \\

\end{align}$

Hence, option (c) is the correct answer.


Note: Students should note here that we have to choose 2 elements from the set of 10 elements. We are not concerned with the order or arrangement of the elements. The formula for $^{n}{{C}_{k}}$ must be remembered.