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Jack
10-10-2013, 11:33 AM
In how many different ways can 2 black boxes, 4 white boxes, 3 blue boxes and 2 brown boxes be placed in a row?

(A) 48
(B) 576
(C) 1200
(D) 69300

Thanks

miranda
10-10-2013, 11:34 AM
In this case we have permutations since the order matters. Additionally we must divide the number of permutations by 2! , 4!, 3! and 2! Since there are 2 black boxes, 4 white boxes, 3 blue boxes and 2 brown boxes.
Therefore the answer is \frac{^11P_{11}}{(2!) \cdot(4!) \cdot(3!) \cdot(2!)}= \frac{11!}{576}=69300 (D).

Jack
10-10-2013, 11:39 AM
Thank you!!