Aptitude 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,

$= 2! × 18! × 6!$

$= 18! ×1440$

**May**

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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**

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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**

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

**Ram**

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why cant this be $2!*24!$

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.

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

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