Question

How many ways 9 students may be seated so that two of them are always together

Answer 1

+here the answer is!

hopefully it is true

Answer 2

Given -

1 There are total 9 students .

2 of them have to be seated together.

To Find -

No. of ways in which they are sitting arrangement can be done when two student always sit together .

Solution -

We have total 9 students . Let's consider those 2 students which want to sit together as single unit .

Now after considering those two students as single we have 8 students total .

$\boxed{\sf{Permutation = {}^{n} P_r = \frac{n !}{[n - r] ! } }}$

Now using this formula we'll arrange 8 students.

Here n = 8 as well as r = 8

${\sf{Permutation = {}^{8} P_8 = \frac{8 !}{[8 - 8] !} }}$

$\sf{\implies \frac{8 !}{0 !} = 8! }$

[because 0! = 1 ]

8! = $\sf{ 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 }$

$\underline{\sf{\implies 40320 = 8 ! }}$

Now those two student can also changed their sitting arrangement but together .

${\sf{Permutation = {}^{2} P_2 = \frac{2 !}{[2 - 2] !} }}$

$\underline{\sf{\implies 2! = 2\times 1 = 2}}$

No. of sitting arrangement = 8! × 2!

$\sf{\implies 8 \times 7 \times 6\times 5 \times 4 \times 3 \times 2 \times 2}$

${\underline{\sf{\implies 40320 \times 2 = 80640 \: ways }}}$