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Mathematics 17 Online
OpenStudy (anonymous):

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?

OpenStudy (anonymous):

ten choices for chairman, then \( 9 \choose 2\) choices for secretary

OpenStudy (anonymous):

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

OpenStudy (anonymous):

is it clear how to compute \(\dbinom{9}{2}\)?

OpenStudy (anonymous):

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

OpenStudy (anonymous):

i don't understand, isn't this just simply probabilty with the P thing? its been some time thats why i dont remember...

OpenStudy (anonymous):

it is not a probability but it is a topic used in probability this is a question that asks "how many"

OpenStudy (anonymous):

ok.. then how to i solve the long bracket things?

OpenStudy (anonymous):

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

OpenStudy (anonymous):

you compute \(\dbinom{9}{2}\) read "nine choose two" by \[\dbinom{9}{2}=\frac{9\times 8}{2}=9\times 4=36\]

OpenStudy (anonymous):

so i would do the same witht he follow question? how many anagrams does the word negen have?

OpenStudy (anonymous):

not quite

OpenStudy (anonymous):

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 \[\frac{5!}{2!}=\frac{5\times 4\times 3\times 2}{2}=5\times 4\times 3\]

OpenStudy (anonymous):

speaking of random, this is really a random collection of questions is this some on line class?

OpenStudy (anonymous):

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?

OpenStudy (anonymous):

i am geting my questions from a dutch medicince exam past paper.. my exam is tmw :/

OpenStudy (anonymous):

@satellite73

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