Question

shen is putting 11 colored light bulbs into a string of lights. there are 3 yellow light bulbs, 4 pink 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)?

Explanation:

Step1: Use permutation formula for multi - sets

The formula for permutations of a multi - set is \(\frac{n!}{n_1!n_2!n_3!}\), where \(n\) is the total number of objects, and \(n_1,n_2,n_3\) are the number of objects of each type. Here, \(n=3 + 4+4=11\) (total number of light - bulbs), \(n_1 = 3\) (number of yellow bulbs), \(n_2=4\) (number of pink bulbs), \(n_3 = 4\) (number of white bulbs).

Step2: Calculate the factorial values

We know that \(n!=n\times(n - 1)\times\cdots\times1\). So, \(11! = 11\times10\times9\times8\times7\times6\times5\times4\times3\times2\times1=39916800\), \(3! = 3\times2\times1=6\), \(4! = 4\times3\times2\times1 = 24\).

Step3: Substitute into the formula

\(\frac{11!}{3!4!4!}=\frac{39916800}{6\times24\times24}\)
First, calculate \(6\times24\times24=6\times576 = 3456\)
Then, \(\frac{39916800}{3456}=11550\)

Answer:

\(11550\)