Question 1023425: There are two families attending a concert together. Each family consists of 1 male and 2 females. In how many ways can they be seated in a row of six seats if
(a) There are no restrictions?
(b) Each family is seated together?
(c) The members of each gender are seated together?
Answer by mathmate(429)   (Show Source):
You can put this solution on YOUR website!
Question:
There are two families attending a concert together. Each family consists of 1 male and 2 females. In how many ways can they be seated in a row of six seats if
(a) There are no restrictions?
(b) Each family is seated together?
(c) The members of each gender are seated together?
Solution:
Altogether 6 persons.
Use notation: P(n,r)=n!/(n-r)! and note that 0!=1.
(a) no restrictions, so permutation of 6 "objects" out of 6 is P(6,6)=6!=720.
(b) each family is seated together, then there are P(3,3) for each family (ABC,ACB,BAC,BCA,CAB,CBA), and P(2,2) for arranging the families (FG,
GF) so together,
N=P(3,3)P(3,3)P(2,2)=3!3!2!=72
(c) There are 4! ways to arrange the ladies, and 2! ways to arrange the gents, for a total of 4!2!=48 ways.
However, the gents can be placed before, in the middle, or the end of the ladies (3 ways), so total = 48*3=144 ways.
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