There are 12 guests at a dinner party. Supposing that the master and mistress of the house have fixed seats opposite one another and that there are two specified guests who must always, be placed next to one another; the number of ways, be placed next company can be placed is
There are 12 guests at a dinner party. Supposing that the master and mistress of the house have fixed seats opposite one another and that there are two specified guests who must always, be placed next to one another; the number of ways, be placed next company can be placed is
The correct option is A 20⋅10!
Let us first seat the 1212 guests. 22 guests are always beside each other hence considering 1111 guests, the no. of combinations possible are 11!11!. But the 22 people can be arranged in 22 ways, so total no. of combinations is given by 2∗11!2∗11!.
Now the master can be seated between any 22 people except the pair to be seated together. Hence, the no. of possibilities to seat him becomes 1111. The mistress is always opposite to him hence she would not contribute to the no. of total ways.
This gives,
The total no. of ways =2∗11∗11!=2∗11∗11!
The Answer given:
First we seat the 22 specified people in 2∗102∗10 ways and the remaining 1010 people can be arranged in 10!10!ways. So total no. of ways =2∗10∗10!
Let us first seat the 1212 guests. 22 guests are always beside each other hence considering 1111 guests, the no. of combinations possible are 11!11!. But the 22 people can be arranged in 22 ways, so total no. of combinations is given by 2∗11!2∗11!.
Now the master can be seated between any 22 people except the pair to be seated together. Hence, the no. of possibilities to seat him becomes 1111. The mistress is always opposite to him hence she would not contribute to the no. of total ways.
This gives,
The total no. of ways =2∗11∗11!=2∗11∗11!
The Answer given:
First we seat the 22 specified people in 2∗102∗10 ways and the remaining 1010 people can be arranged in 10!10!ways. So total no. of ways =2∗10∗10!
