Hans is putting 11 colored light bulbs into a string of lights. There are 3 blue light bulbs, 4 red light bulbs, and 4 white light bulbs. How many distinct orders of light bulbs are there if two light bulbs of the same color are considered identical (not distinct)?
Question
Hans is putting 11 colored light bulbs into a string of lights. There are 3 blue light bulbs, 4 red light bulbs, and 4 white light bulbs. How many distinct orders of light bulbs are there if two light bulbs of the same color are considered identical (not distinct)?
Solution
To solve this problem, we can use the formula for permutations of a multiset. This formula is n! / (r1! * r2! * ... * rk!), where n is the total number of items, and r1, r2, ..., rk are the numbers of each type of item.
In this case, n = 11 (the total number of light bulbs), r1 = 3 (the number of blue light bulbs), r2 = 4 (the number of red light bulbs), and r3 = 4 (the number of white light bulbs).
So, the number of distinct orders of light bulbs is 11! / (3! * 4! * 4!).
First, calculate the factorial of 11, which is 1110987654321 = 39,916,800.
Next, calculate the factorial of 3, which is 321 = 6.
Then, calculate the factorial of 4, which is 432*1 = 24.
Finally, divide the factorial of 11 by the product of the factorials of 3, 4, and 4.
So, the number of distinct orders of light bulbs is 39,916,800 / (6 * 24 * 24) = 27,645.
Therefore, there are 27,645 distinct orders in which Hans can put the light bulbs into the string of lights.
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