ms arrangement

Consider Joe and Mary as one group. The arrangement is :
Choose 3 persons from the rest 8 person (8C3), and arrange the group and the 3 persons (4P4) in one of the row (2C1). Then arrange the rest 5 persons in the another row (5P5). Consider also the internal arrangement of Joe and Mary (2P2).

No. of arrangements
= 8C3 x 4P4 x 2C1 x 5P5 x 2P2
= (8!/5!3!) x 4! x (2!/1!1!) x 5! x 2!
= 56 x 24 x 2 x 120 x 2
= 645120

考慮其隻八人 8C4 *(4!)^2
Joe and Mary 有左右並第1,2,3,4,5 十個可能性
所以答案是
10*( 8C4) *(4!)^2

Assume the two rows (A and B) are:
A1, A2, ...A5

B1, B2, ...B5

If they seat next to each other, they can seat in :
A1, A2
A2, A3
A3, A4
A4, A5
B1, B2
B2, B3
B3, B4
B4, B5

(8 pairs)

and they can exhnage their seats, i.e. 16 ways of seatings.

For each of the 16 seatings, the remaining 8 people can seat randomly with ways = 10P8 =1814400 as order is concerned.


total no. of ways =16 x 1814400 =29030400


I am not sure about the answer.

2010-06-15 18:54:07 補充：
Corrections:
For each of the 16 seatings, the remaining 8 people can seat randomly with ways = 8P8 =40320 as order is concerned.

total no. of ways =16 x 40320 =645120

Sorry for the mistake.

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設a=joe, b=mary
先從ab自己分析
ab可以掉轉，所有已有2可能性
然後ab可以有8個可以位置

安放好ab之後，餘下8個人有3個人抽到第一位
可能性8C3
但3個人可以掉轉，有3!可能性

接住5個人抽5個，可能性有5C5
但5個人可以掉轉，有5!可能性

所以總可能性
2x8x(8C3)x(3!)x(5C5)x(5!)
=645120