Question

In how many ways, can 5 ladies draw water from 5 taps, assuming that no tap remains unused?

Answer 1

Hint: We can solve the problem by finding the number of options for each lady differently and then multiply all possible options to get the desired result. When one tap is assigned to one lady then this tap can’t be assigned to another lady.

Complete step by step solution:
It is given in the problem that there are $5$ ladies that can draw water from the five taps. We have to find the number of ways that they can draw water from the tap.
Now, we have 5 different taps and 5 ladies.
When the first lady comes to draw the water from the tap, she will have 5 options to choose and she can choose any one of them. Hence, she has 5 options.
Now, when the second lady comes to draw water from the tap, she has the remaining 4 options as one is already chosen by the first lady. Hence, the second lady has 4 options. So, the second lady can choose any one from the remaining 4 taps.
Similarly, the third lady has three options, the fourth lady has two options and the last lady has only one option.
So, the required number of ways to draw the water from the tap is given as:
Required number of ways$ = 5 \times 4 \times 3 \times 2 \times 1$
Required number of ways$ = 120$
There are $120$ways to draw the water from $5$ taps.

Note: One can also solve this question by directly using the formula of permutation. When there are $n$ ways to choose $r$ quantities then the permutation is given as:
$^n{P_r} = \dfrac{{n!}}{{\left( {n - r} \right)!}}$
In the given problem, there are $5$ taps and $5$ ladies, so using the permutation, it can be express as:
\[^5{P_5} = \dfrac{{5!}}{{\left( {5 - 5} \right)!}}\]
\[{ \Rightarrow ^5}{P_5} = \dfrac{{5 \times 4 \times 3 \times 2 \times 1}}{{\left( 0 \right)!}}\]
\[{ \Rightarrow ^5}{P_5} = 120\]
So, there are $120$ ways to draw the water from the tap.