# Easy Permutation-Combination Solved QuestionAptitude Discussion

 Q. A class photograph has to be taken. The front row consists of 6 girls who are sitting. 20 boys are standing behind. The two corner positions are reserved for the 2 tallest boys. In how many ways can the students be arranged?
 ✖ A. 6! × 1440 ✔ B. 18! × 1440 ✖ C. 18! ×2! × 1440 ✖ D. None of these

Solution:
Option(B) is correct

Two tallest boys can be arranged in $2!$ ways, rest $18$ can be arranged in $18!$ ways.
Girls can be arranged in $6!$ ways.

Total number of ways in which all the students can be arranged,

$= 2! × 18! × 6!$

$= 18! ×1440$

## (6) Comment(s)

May
()

Why aren't the two tallest boys chosen first before permuting them.

$^{20}C_2 \times 2! \times 18! \times 6!$

Soni
()

Because, 2 tallest boys is a fact. You just have to pick them and need not FIND them. So no selection out of 20 boys, i.e. no term containing $^{20}C_2$.

Priya
()

No Suyog it does not include GIRLS.

In the question it says that 'the front row consists of 6 girls who are sitting. 20 boys are standing behind'.

This makes total number of students 26 and factor of 18! includes only BOYS.

Hope this helps.

Suyog
()

Here, in answer, we have taken 18! for their arrangement(including girls), then why to take arrangement of 6 girls again?

Ram
()

why cant this be $2!*24!$

Sulokit
()

See the back row has 20 positions. The extreme left and right positions r fixed for 2 boys.

Now only 18 positions and 18 boys r left so 18! ways arrangement. Now those two boys can swap their positions.

So 2! ways.