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I'm self learning combinatorics methods, I've found in my combinatorics book the following exercise:

"There are 25 workers in a corporation sharing 12 cutting machines. Every hour, some group of the workers needs a cutting machine. We never expect more than 12 workers to require a machine at any given time. We assign to each machine a list of the workers cleared to use it, and make sure that each worker is on at least one machine's list. If the number of names on each of the lists is added up, the total is 95. Show that it is possible that at some hour some worker might not be able to find a machine to use."

This seems some application of the pigeonhole principle, sadly I'm not able to solve it by myself. Can someone help me with some hint about how to approach this problem?

Thanks

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    The main challenge is to understand this text! – 2017-02-27
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    Each hour, we have a list of workers needing a machine, but we do not know the number of hours they work, right ? – 2017-02-27
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    What a relief! I was worried to be the only one which do not understand this problem.. But seems that it might be ill formed or something. The exercise is the number 17 from paragraph 2.19 coming from this [book](https://books.google.pl/books?id=szBLJUhmYOQC&pg=PA109&lpg=PA109&dq=There+are+25+workers+in+a+corporation+sharing+12+cutting+machines&source=bl&ots=8gTLF1jgSI&sig=qt10U0FETwrYULDFJAnHk4EgVuc&hl=it&sa=X&ved=0ahUKEwjctYzOjrLSAhXIlCwKHUBdAFcQ6AEIKDAC#v=onepage&q=There%20are%2025%20workers%20in%20a%20corporation%20sharing%2012%20cutting%20machines&f=false). – 2017-02-28
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    From the 25 and 12 reference, you get at least 3 people, share the same machine. From the 95 reference, you get at least 11 of the machines, have to have 8 people sharing them. from the 25 and 95 together, you get, that 20 of those 25 workers, are on at least 4 lists. – 2017-06-28

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