Question

A family is taking a picture, with everyone in the family standing in a row. In the family, there are two sets of twins. In how many ways can the 9 people in the family be arranged such that nobody is sitting next to their twin?

Answer 1

Answer:

376320.

Step-by-step explanation:

Given:

There are two sets of twins in a family.

Total family member = 9

Solution:

Let family members = AA BB C D E F G

AA and BB are two set of twins .

First of we will find the arrangements of 5 family member   ( 9 - 4 = 5 )

No of ways 5 family member can seat =$^{5}P_{5}$

Now, we will find the arrangements of BB.

As given that nobody is sitting next to their twin, means A twins cannot seat together.

Therefore, they can seat in these blank places given below:

......B.......B......C......... D...... E........ F........ G.......... in $^{8}P_{2}$

Similarly,  B twins cannot seat together.

Therefore, they can seat in these blank places given below:

......A.......A......C......... D...... E........ F........ G.......... in $^{8}P_{2}$

Total  number of arrangements of all family members = $^{5}P_{5}$$\times$ $^{8}P_{2}$ $\times$$^{8}P_{2}$

                                                                                         $=\frac{5!}{5-5!} \times\frac{8!}{8-2!} \times\frac{8!}{8-2!}\\=\frac{5!}{0!}\times\frac{8!}{6!} \times\frac{8!}{6!}\\\\=5!\times\frac{8\times7\times6!}{6!} \frac{8\times7\times6!}{6!}$

                                                                                           $=5\times4\times3\times2\times1\times56\times56\\=376320$

Therefore, Total  number of arrangements of all family members is 376320.