Question

12 persons are to be seated at a square table, three on each side. 2 persons wish to sit on the north side and two wish to sit on the east side. One other person insists on occupying the middle seat (which may be on any side). Find the number of ways they can be seated.

Answer 1

Let the persons sit on north side are $N_1$ and $N_2$

and the persons sit on east side are $E_1$ and $E_2$

and the person sits in middle is $M$

$Case:1$

Let $M$ sit on south or west side

No. of ways for $M$ can sit $=2$

No. of ways for $N_1$ and $N_2$ $=$ $C_2^3\times 2!$

No. of ways for $E_1$ and $E_2$ $=$ $C_2^3\times 2!$

And , no. of ways remaining 7 person can sit $=7!$

So, no. of ways all can sit $=$ ${2\times C}_2^3\times 2!\times C_2^3\times 2!\times 7!$

$Case:2$

Let $M$ sit on north side

No. of ways for $M$ can sit $=1$

No. of ways for $N_1$ and $N_2$ $=$ $2!$

No. of ways for $E_1$ and $E_2$ $=$ $C_2^3\times 2!$

And , no. of ways remaining 7 person can sit $=7!$

So, no. of ways all can sit $=$ $2!\times C_2^3\times 2!\times 7!$

$Case:3$

Let $M$ sit on east side

No. of ways for $M$ can sit $=1$

No. of ways for $N_1$ and $N_2$ $=$ $C_2^3\times 2!$

No. of ways for $E_1$ and $E_2$ $=$ $2!$

And , no. of ways remaining 7 person can sit $=7!$

So, no. of ways all can sit $=$ $2!\times C_2^3\times 2!\times 7!$

So, Total no. of ways $=$ (${2\times C}_2^3\times 2!\times C_2^3\times 2!\times 7!$)$+$($2!\times C_2^3\times 2!\times 7!$)$+$$(2!\times C_2^3\times 2!\times 7!)$

$=$ ($2\times 3\times 2!\times 3\times 2!\times 7!$)$+$($2!\times 3\times 2!\times 7!$)$+$$(2!\times 3\times 2!\times 7!)$

$=$ $(72\times 7!)+(12\times 7!)+(12\times 7!)$

$=$ $(72+12+12)7!$

$=$ $96(7!)$