# basic combinatorics(permutations) question

How many ways are there to seat six different boys and six different girls along one side of a long table with $12$ seats? How many are ways if boys and girls alternate sits?

MY try:

For first question, we start with boys, We know there are $P(12,6) = \frac{12!}{6!}$ ways for the boys to sit, and for the girls, there are $P(6,6) = \frac{6!}{6-6!} = 6!$ ways since we have already used $6$ sits. Therefore to answer the first question, there are $\frac{12!}{6!} 6! =12 !$ ways to do this.

For the second question, Im stuck, can someone help me? thanks

## 1 Answer

For the first question, why make it so complicated? 12 people, 12 seats, P(12,12) ways.

For the second problem, break it into two cases:
(a) How many ways to seat them BGBGBGBGBGBG?
(b) How many ways to seat them GBGBGBGBGBGB?

For part (a), there are P(6,6) ways to seat the boys in seats 1, 3, 5, 7, 9, 11, and P(6,6) ways to seat the girls in seats 2, 4, 6, 8, 10, 12, so the total number of arrangements is P(6,6)P(6,6).

• how about for part $b$? just the same $P(6,6)P(6,6)$ ? Oct 23, 2013 at 20:35
• Yes, it's just the same.
– bof
Oct 23, 2013 at 20:39
• so in total we have $P(6,6)P(6,6)P(6,6)P(6,6)$ ways to do this assigment right? Oct 23, 2013 at 20:40
• NO!! We ADD the answers to (a) and (b)!! Here's ANOTHER way to figure it: You can put any of the 12 kids in seat 1, boy or girl. After you've filled seat 1, there are 6 choices for seat 2 (because boys and girls must alternate), then 5 choices for seat 3, 5 choices for seat 4, etc. The total number of arrangements is 12*6*5*5*4*4*3*3*2*2*1*1 which is the same as 2*6*5*4*3*2*1*6*5*4*3*2*1 = 2(6!)(6!).
– bof
Oct 23, 2013 at 21:04