Eleven boys and girls wait to take their seats in the same row in a movie theater. There are exactly 11 seats in the row. They decided that after the first person sits down, the next person has to sit next to the first. The third sits next to one of the first two and so on until all eleven are seated. In other words, no person can take a seat that separates him/her from at least one other person. How many different ways can this be accomplished? Note that the first person can choose any of the 11 seats.

if first person desires to sit at end positions,it takes 2 ways and other 10 people will have only 1 way so 2 ways
if 1st prsn wishes t sit anywhere he/she has 9 ways[excepting 2 ends] and all others will have 2 ways[for each] so 9*2^10
total 2+9*2^10
11 years agoHelpfull: Yes(5) No(1)
Answer : 1024 different ways.

If there is just one person and one seat, that person has
only one option.

If there are two persons and two seats, it can be
accomplished in 2 different ways.

Hence , Formula for n persons is : 2^(n-1)

11 persons are there, therefore number of ways are :

2^(11-1) = 2^10 = 1024 ways
11 years agoHelpfull: Yes(4) No(5)
11c2 11 boys and 11girls may sit alternately
11 years agoHelpfull: Yes(2) No(7)
11*10*9*8...............
11!
11 years agoHelpfull: Yes(2) No(0)
can u please xplain in detail as it didnt mention dat they should b placed alternately
11 years agoHelpfull: Yes(1) No(1)
what the ans exact answer is?????
11 years agoHelpfull: Yes(1) No(1)
2+(9*2^9 *1)=4610
if 1st person sit at end positions then we have only 2 ways to arrange them.
if the 1st person want to sit at position other than end positions then
we have 9*2^9 ways to arrange 10 persons & only 1 way for last(11th) person.
11 years agoHelpfull: Yes(1) No(0)

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