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a) The country's prime minister has reached to $40$ candidates for possible mining. If the number of ministries is $20$, how many different ministerial schemes can the prime minister make, provided that no person will be called upon to take over one ministry?

b) A Megara toll worker is asked to choose his five days week in which to work. The only limitation is that it must be at least one day day off during the weekend. How many different options are there?

c) A group of $9$ diabetics visited a café. $2$ people ordered $\text{Greek coffee}$, $3$ people ordered $\text{Espresso}$ and $4$ people ordered $\text{Cappuccino}$ (of course without sugar). The waiter was confused with the order and simply served the drinks to randomly to the members of the group. What is the possibility that the waiter's confusion may be unnoticed, namely serve each member of the group with the drink he ordered?

I want to know if there is a technique to use on these problems or some way to understand what formula to use. Nevertheless my try:

a) I used the combination's formula cause we choose $20$ ministerial spots for $40$ people so ${40\choose 20}=137846528820$

b) If he gets the day off at saturday then he has $6^5$ different ways to arrange his work schedule. If he gets the day of at all weekend then he has only one way and that's it Monday-Friday.

c) I have no idea how to handle this problem.

Thanks for your time.

argiriskar
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  • Why do you think ${40 \choose 20}=2$? That's way off. ${40\choose 20}=\frac{40!}{20!20!}\neq\frac{40}{20}=2$ – SK19 May 06 '18 at 14:24
  • For general tips on how to solve combinatorical problems see the book referred to in this question: https://math.stackexchange.com/questions/1104996/big-list-of-guided-discovery-books – SK19 May 06 '18 at 14:26
  • Oh god sorry. You are right. I was trying to copy it right from my textbook. I will correct that thank you – argiriskar May 06 '18 at 14:35

1 Answers1

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In question (b), choosing to take Saturday off gives $\binom 65=6$ options (not $6^5$), and there are similarly $6$ option if choosing to take the other weekend day off, and $1$ of those options is common to both schemes (taking the whole weekend). This is a simple example of the inclusion-exclusion principle.

In question (c), you are effectively asking how many anagrams of $GGEEECCCC$ there are (say $q$). Only one of those will be correct, giving a probability of $1/q$ that a random distribution of drinks will be correct. This can be quickly established with a multinomial coefficient, or calulated by multiplying two binomial coefficient successively placing say $G$s and $E$s.

Joffan
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  • how about problem a) ? – argiriskar May 07 '18 at 14:04
  • it seems to me that thats a lot of ways to do that so i wasn't sure if it's right – argiriskar May 07 '18 at 14:16
  • My main problem with part (a) is that I think there may be a couple of typos, which I didn't want to assume. And the policy here is one question per question if you understand me, so responding to 2 seemed generous. – Joffan May 07 '18 at 14:27
  • All the questions were from the same category (combinatorics) and I didn’t want to spam. It won’t happen again – argiriskar May 07 '18 at 14:28
  • So the typos are "mining" and then I think probably you meant "more than one ministry".unless you meant that one ministry post will be left vacant. – Joffan May 07 '18 at 14:40