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zzz [600]
3 years ago
9

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 m

any different arrangements are possible?
Mathematics
1 answer:
SashulF [63]3 years ago
7 0
<h3>Answer: 29,030,400</h3>

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

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.

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