Question

For a group photograph, 3 boys and 2 girls stand in a line in all possible ways. How many photos could be taken if each photo corresponds to each such arrangement?

Answer 1

Hint: In the above problem, we have to find the number of ways to arrange 3 boys and 2 girls in a line. We can say that we have 5 persons all are different from each other so arranging 5 persons in 5 gaps which is equal to the factorial of 5. We have taken the 5 persons as different because no two boys are identical and no two girls are identical.

Complete step-by-step answer:
We have given 3 boys and 2 girls and we have to arrange them in one line in all possible ways.
We have 3 boys and 2 girls, which means we have 5 people who are different from each other. You may argue that 3 boys can be considered the same and 2 girls can be considered the same which is the wrong proposition because no two girls are alike and no two boys are alike.
Now, we have to arrange 5 different persons in a line means we have 5 gaps and we have to make 5 persons stand in those 5 gaps in such a way that each gap is occupied by 1 person.
In the below, we have shown 5 gaps,
_ _ _ _ _
In the first gap any 5 different persons can come.
5 _ _ _ _
Now, one person is standing in the first gap so we are left with 4 people so 4 possibilities are there to fill the second gap.
5 4 _ _ _
Similarly, in the third gap 3 possibilities are there, in the second gap 2 possibilities are there and the last gap has only 1 possibility because then only 1 person is left.
5 4 3 2 1
Now, multiply these possibilities i.e. multiplying 5, 4, 3, 2 and 1 with each other which will give us the total number of possibilities.
$\begin{align}
  & 5.4.3.2.1 \\
 & =120 \\
\end{align}$
Hence, there are 120 possible photos that can be clicked.

Note: In the above solution, you might have thought why we have multiplied all the possibilities we can add them too. The reason is that the possibilities in each gap is independent with respect to other gaps and if we consider possibilities of occupying each gap as an event then we know that when events are independent to each other then we multiply the events.
We can also extract an important information from solving this problem is that if we have to arrange “n” different objects in the same number of gaps as the objects then the possible ways of arranging them is n!. We can say n! because in the above solution we have 5 different persons and we have to arrange them in 5 gaps so the possible arrangement was 5!