Shabnam has 2 school bags, 3 tiffin boxes, and 2 water bottles. Show how many ways she can carry these items.

Hint: First find the number of ways 1 school bag will be selected from 2 school bags. Then, find the number of ways to select 1 tiffin box out of 3 tiffin boxes. After that find the number of ways to select 1 water bottle out of 2 water bottles. Then to carry these items multiply the number of ways the school bag has been selected, the number of ways the tiffin box has been selected, and the number of ways water bottles must be selected.

Complete step by step answer:
Given:- Shabnam has 2 school bags, 3 tiffin boxes, and 2 water bottles.
There are 2 school bags out of which 1 bag has to be selected. So,
No of ways =$^n{C_r}$
Put, n= 2 and r= 1,
${ \Rightarrow ^2}{C_1} = \dfrac{{2!}}{{1!\left( {2 - 1} \right)!}}$
Simplify the terms,
${ \Rightarrow ^2}{C_1} = \dfrac{{2!}}{{1!1!}}$
Now, expand the factorial,
${ \Rightarrow ^2}{C_1} = \dfrac{{2 \times 1}}{{1 \times 1}}$
Cancel out the common factors to get the number of ways 1 school bags can be selected,
${ \Rightarrow ^2}{C_1} = 2$.........…..(1)
There are 3 tiffin boxes out of which 1 box has to be selected. So,
No of ways =$^n{C_r}$
Put, n= 3 and r= 1,
${ \Rightarrow ^3}{C_1} = \dfrac{{3!}}{{1!\left( {3 - 1} \right)!}}$
Simplify the terms,
${ \Rightarrow ^3}{C_1} = \dfrac{{3!}}{{1!2!}}$
Now, expand the factorial,
${ \Rightarrow ^3}{C_1} = \dfrac{{3 \times 2 \times 1}}{{2 \times 1 \times 1}}$
Cancel out the common factors to get the number of ways 1 school bags can be selected,
${ \Rightarrow ^3}{C_1} = 3$...........…..(2)
There are 2 water bottles out of which 1 bottle has to be selected. So,
No of ways =$^n{C_r}$
Put, n= 2 and r= 1,
${ \Rightarrow ^2}{C_1} = \dfrac{{2!}}{{1!\left( {2 - 1} \right)!}}$
Simplify the terms,
${ \Rightarrow ^2}{C_1} = \dfrac{{2!}}{{1!1!}}$
Now, expand the factorial,
${ \Rightarrow ^2}{C_1} = \dfrac{{2 \times 1}}{{1 \times 1}}$
Cancel out the common factors to get the number of ways 1 school bags can be selected,
${ \Rightarrow ^2}{C_1} = 2$............…..(3)
So, the total number of ways is obtained by multiplying the number of ways a school bag has been selected, the number of ways tiffin boxes have been selected, and the number of ways water bottles must be selected. Then,
$ \Rightarrow $Total Ways ${ = ^2}{C_1}{ \times ^3}{C_1}{ \times ^2}{C_1}$
Substitute the values from the equation (1), (2), and (3),
$ \Rightarrow $Total Ways $ = 2 \times 3 \times 2$
Multiply the terms on the right side to get the final result,
$\therefore $Total Ways $ = 12$

Hence, the total number of ways is 12.

Note: The students are likely to make mistakes when finding the number of ways of selection of the team.
Total Ways ${ = ^2}{C_1}{ \times ^3}{C_1}{ \times ^2}{C_1}$
A combination is a mathematical technique that determines the number of possible arrangements in a collection of items where the order of the selection does not matter. In combinations, you can select the items in any order.
The formula for determining the number of possible arrangements by selecting only a few objects from a set with no repetition is expressed in the following way:
$C\left( {n,r} \right) = \left( \begin{gathered}
  n \\
  k \\
\end{gathered} \right) = \dfrac{{n!}}{{k!\left( {n - k} \right)!}}$