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Question

Four Nigerians (A, B, C, D), three Chinese (#,* &), and three Greeks (α,β,γ)(\alpha, \beta, \gamma) are lined up at the box office, waiting to buy tickets for the World's Fair. (a) How many ways can they position themselves if the Nigerians are to hold the first four places in line; the Chinese, the next three; and the Greeks, the last three? (b) How many arrangements are possible if members of the same nationality must stay together? (c) How many different queues can be formed? (d) Suppose a vacationing Martian strolls by and wants to photograph the ten for her scrapbook. A bit myopic, the Martian is quite capable of discerning the more obvious differences in human anatomy but is unable to distinguish one Nigerian (N) from another, one Chinese (C) from another, or one Greek (G) from another. Instead of perceiving a line to be BβAD#&Cαγ,B * \beta A D \# \& C \alpha \gamma, for example, she would see NCGNNCCNGG. From the Martian's perspective, in how many different ways can the ten funny-looking Earthlings line themselves up?

Solution

Verified

(a) Since their places are fixed, hence they can only arrange among themselves.

number of arrangements = 4!3!3!4!\cdot 3!\cdot 3! = 864

(b) Apart from their arrangement in part (a). Now the three nationalities can shuffle in line in 3!3! ways

number of arrangements = 3!4!3!3!3!\cdot 4!\cdot 3!\cdot 3! = 5184

(c) Since all individuals are different.

Hence number of different queues = 10!=362880010! = 3628800

(d) Since people of one nationality look identical to Martian

Hence number of different queues = 10!4!3!3!=4200\dfrac{10!}{4!\cdot 3!\cdot 3!} = 4200

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