231k views
3 votes
There are 7 brothers and 6 sisters. If all siblings sat in a row and the girls requested to be seated next to one another, how many different arrangements are possible?

User Pindatjuh
by
8.2k points

1 Answer

3 votes

Answer: 29,030,400

==============================================================

Step-by-step explanation:

Let's say the mother will temporarily take the place of all the sisters. Wherever the mother sits will represent the block of girls.

Taking out the 6 sisters and replacing them with the mother leads to 7+1 = 8 people in a line.

There are 8! = 8*7*6*5*4*3*2*1 = 40,320 different ways to arrange 8 people.

----------

Again, the mother's position is where the sisters block will go. So let's say the mother was in seat #2. This would mean one brother would take seat #1, and all of the sisters would take the next six seats, until we reach seat #7 is when another brother would take the next seat.

Within any given permutation (the 40320 mentioned), there are 6! = 6*5*4*3*2*1 = 720 different ways to arrange just the girls in that girls block/group.

All together, there are 40320*720 = 29,030,400 different ways to arrange the 13 siblings where all the girls are seated together.

User Hoy Cheung
by
9.2k points
Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.