Question

The number of selections of four letter taken from the word “COLLEGE” must be
A. 12
B. 18
C. 27
D. 22

Answer 1

Hint: We will consider three different cases for finding the number of selections. Since there are 5 distinct letters in the given word. In the first case, consider that all the letters are different. In the second case, consider that two letters are the same, and two are different. In the third case, we will consider two letters are the same and the other two letters are also the same. We will write expressions for all the three cases and find its value from the expression of combination. Then we will add selections in all the three cases to get the final answer.

Complete step-by-step answer:
In the given word, that is COLLEGE.
For the selection of four letters, we have three different cases.
First of all, we will consider that all the letters are distinct. We have five distinct letters in the word COLLEGE that are C, O, L, E, G. Out of these five letters, we need to select 4. This can be expressed as:
${S_1} = {}^5{C_4}$
We know that combination can be expressed as
${}^n{C_r} = \dfrac{{\left| \!{\underline {\,
  n \,}} \right. }}{{\left| \!{\underline {\,
  {n - r} \,}} \right. \;\,\left| \!{\underline {\,
  r \,}} \right. }}$ where $n$ the total number of letters and $r$ is the letters that are required to be taken.
We will find ${S_1}$ it by using the above expression.
$\begin{array}{l}
{S_1} = {}^5{C_4}\\
{S_1} = \dfrac{{\left| \!{\underline {\,
  5 \,}} \right. }}{{\left| \!{\underline {\,
  {5 - 4} \,}} \right. \;\,\left| \!{\underline {\,
  4 \,}} \right. }}\\
{S_1} = \dfrac{{5 \times \left| \!{\underline {\,
  4 \,}} \right. }}{{\left| \!{\underline {\,
  4 \,}} \right. }}\\
{S_1} = 5
\end{array}$
Now we will consider the case in which two letters will be the same, but the other two will be distinct. Hence, we will choose two letters like LL and EE. Hence out of these two options we will choose 1. Now, from the remaining 4 letters, we will choose 2. This can be expressed as
${S_2} = {}^2{C_1} \times {}^4{C_2}$
We will again expand the above expression from the expression of combination. This can be expressed as:
$\begin{array}{l}
{S_2} = {}^2{C_1} \times {}^4{C_2}\\
{S_2} = \dfrac{{\left| \!{\underline {\,
  2 \,}} \right. }}{{\left| \!{\underline {\,
  {2 - 1} \,}} \right. \;\,\left| \!{\underline {\,
  1 \,}} \right. }} \times \dfrac{{\left| \!{\underline {\,
  4 \,}} \right. }}{{\left| \!{\underline {\,
  {4 - 2} \,}} \right. \;\,\left| \!{\underline {\,
  2 \,}} \right. }}\\
{S_2} = \dfrac{{2 \times \left| \!{\underline {\,
  1 \,}} \right. }}{{\left| \!{\underline {\,
  1 \,}} \right. }} \times \dfrac{{4 \times 3 \times 2}}{{\left| \!{\underline {\,
  2 \,}} \right. \,\,\left| \!{\underline {\,
  2 \,}} \right. }}\\
{S_2} = 2 \times 6\\
{S_2} = 12
\end{array}$
 Now we will consider the case in which two letters will be the same and the other two will also be the same. We have 2 choices EE and LL. This can be expressed as:
$\begin{array}{l}
{S_3} = {}^2{C_2}\\
{S_3} = \dfrac{{\left| \!{\underline {\,
  2 \,}} \right. }}{{\left| \!{\underline {\,
  {2 - 2} \,}} \right. \;\,\left| \!{\underline {\,
  2 \,}} \right. }}\\
{S_3} = 1
\end{array}$
Hence to find the total of selection we will add ${S_1}$ , ${S_2}$ and ${S_3}$ which can be expressed as
$\begin{array}{l}
{\rm{Total \; number \; of \; selection}} = {S_1} + {S_2} + {S_3}\\
{\rm{Total \; number \; of \; selection}} = 5 + 12 + 1\\
{\rm{Total \; number \; of \; selection}} = 18
\end{array}$
So, the correct answer is “Option B”.

Note: This question is of permutation and combination. We can define the selection of the objects from the set of objects. It does not consider the arrangement of the objects and deals with only selection. In contrast, permutation considers selection and arrangement both.