a club has 10 members, a new board needs to be choosen, the board consists of 1 chairman and 2 secretaries. how many ways can they be choosen?
ten choices for chairman, then \( 9 \choose 2\) choices for secretary
so by counting principle it is \[10\times \dbinom{9}{2}\] 9 choose 2 for the secretary because presumably you cannot tell them apart, i.e. order does not matter
is it clear how to compute \(\dbinom{9}{2}\)?
btw if you are wondering "why choose the chair first?" you can choose the two secretaries first and then the chair. number of ways to do that would be \[\dbinom{10}{2}\times 8\] and you should convince yourself that these are the same
i don't understand, isn't this just simply probabilty with the P thing? its been some time thats why i dont remember...
it is not a probability but it is a topic used in probability this is a question that asks "how many"
ok.. then how to i solve the long bracket things?
the answer to a probability question "what is the probability of thus and such" is a number between zero and one a question like this "how many ways" has an answer that is a positive integer in this case is it 360 generally they use the counting principle
you compute \(\dbinom{9}{2}\) read "nine choose two" by \[\dbinom{9}{2}=\frac{9\times 8}{2}=9\times 4=36\]
so i would do the same witht he follow question? how many anagrams does the word negen have?
not quite
negen has two e's that you cannot tell apart if these letters were all different, you would have \(5!=5\times 4\times 3\times 2\) different arrangements, but since you cannot tell the e's apart you need to divide this by 2 and get \[\frac{5!}{2!}=\frac{5\times 4\times 3\times 2}{2}=5\times 4\times 3\]
speaking of random, this is really a random collection of questions is this some on line class?
thank you! i wasn't sure when to use the exclamations, i used them in the first question, however; where did you get the 8 from from nine choose 2?
i am geting my questions from a dutch medicince exam past paper.. my exam is tmw :/
@satellite73
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