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# help

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A manager wants to assign 20 workers to four distinct construction jobs. These jobs require 6, 4, 3, and 7 workers respectively. In how many different ways can the manager assign the workers?

Mar 27, 2020

#1
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Take a look at this: https://www.algebra.com/algebra/homework/Permutations/Permutations.faq.question.1114944.html

Hope this helped!

Mar 27, 2020
#2
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Choose  any  6  of the 20  for  any  one  of  the  4 jobs =

C(20, 6) * C(4,1)  ways

Choose  any  4  of  the remaining 14   for  any  of the 3 remaining jobs

C(14, 4)  * C(3,1)   ways

Choose   any  3  of  the remaining  10  for either  of  the  2 remaining jobs

C(10, 3) * C(2,1)   ways

[ Note  the remaining 7 workers will automatically be assigned to the remaining job ]

So

C(20, 6) * C(4,1)  *  C(14, 4)  * C(3,1) *  C(10, 3) * C(2,1)  =

111,740,428,800  ways  ...... (yikes !!!!)   Mar 27, 2020
#3
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Isn't that TOO big, Chris?!!!

CalTheGreat  Mar 27, 2020
#4
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A manager wants to assign 20 workers to four distinct construction jobs.
These jobs require 6, 4, 3, and 7 workers respectively.
In how many different ways can the manager assign the workers?

$$\text{Let Job_1 = a} \\ \text{Let Job_2 = b} \\ \text{Let Job_3 = c} \\ \text{Let Job_4 = d} \\ \text{The word is \mathbf{aaaaaabbbbcccddddddd} (The word-length is 20 ),\\how many differnet words we can have ?}$$

multinomial coefficient:

$$\begin{array}{|rcll|} \hline \dfrac{20!}{6!\ 4!\ 3!\ 7!} &=& 4655851200 \\ \hline \end{array}$$

In 4655851200 different ways can the manager assign the workers. Mar 27, 2020