Question

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

Answer 1

Hint:We will solve this question by assigning taps to different ladies one by one and then we will multiply them to get the final answer. Like the 1st lady can use any of the 5 taps, the 2nd lady then can use any of the 4 remaining taps, like that we will get to the final answer.

Complete step-by-step answer:
Let’s start solving this question,
Now we have 5 different taps and 5 ladies.
Now the 1st lady will have 5 options to choose and she can choose any one of them.
Hence, it has 5 options.
Now for the 2nd lady she has the remaining 4 options as one is already chosen by the 1st lady.
Hence, it has 4 options.
Until this point we can say that for each 5 options the 1st lady will pick the 2nd lady will have 4 different options in each case.
Hence, the total number of possible ways till now is $5\times 4$ .
Similarly, the 3rd , 4th and 5th lady will have 3, 2, 1 options.
Hence, the total possible case will be $5\times 4\times 3\times 2\times 1=120$

Note:One can also solve this question by directly using the formula of permutation, ${}^{n}{{P}_{r}}=\dfrac{n!}{\left( n-r \right)!}$ and then n and r both will be 5. And from this we will get the same answer which is 120 possible ways. Or one can first choose 5 taps from 5 taps then arrange them to get the answer.